OP Malhotra Class 12 Maths Solutions Chapter 20 Theoretical Probability Distribution Exercise 20 (D)

 

Question 1. The mean of a binomial distribution is 20 and standard deviation is 4. Find out n, p and q.
Answer:
Given mean of binomial distribution \( = 20 \)
So, we have \( np = 20 \) .......(1)
Standard deviation of binomial distribution \( = 4 \)
Squaring both sides, variance \( npq = 4^2 = 16 \) .......(2)
Now, divide equation (2) by equation (1):
\( \frac{npq}{np} = \frac{16}{20} \)
\( \implies q = \frac{4}{5} \)
We know that \( p + q = 1 \). So, we can find p:
\( p = 1 - q = 1 - \frac{4}{5} = \frac{1}{5} \)
Substitute the value of p into equation (1):
\( n \times \frac{1}{5} = 20 \)
\( \implies n = 20 \times 5 = 100 \)
Thus, the values are \( n = 100 \), \( p = \frac{1}{5} \), and \( q = \frac{4}{5} \). The binomial distribution depends on these key parameters.
In simple words: We used the given mean and standard deviation to find the probability of success (p), probability of failure (q), and the total number of trials (n) for the binomial distribution.

🎯 Exam Tip: Remember the formulas for mean (\(np\)) and variance (\(npq\)) of a binomial distribution. These are crucial for solving such problems quickly and accurately.

 

Question 2. Determine the binomial distribution whose
(i) mean is 9 and whose Standard deviation is \( \frac{3}{2} \)
(ii) mean is 10 and standard deviation is \( 2 \sqrt{2} \).
Answer:
(i) Let the required binomial distribution be \( (q + p)^n \).
Given mean of binomial distribution \( = 9 \)
So, \( np = 9 \) .......(1)
Standard deviation (S.D.) of binomial distribution \( = \frac{3}{2} \)
Squaring both sides, variance \( npq = (\frac{3}{2})^2 = \frac{9}{4} \) .......(2)
Now, divide equation (2) by equation (1):
\( \frac{npq}{np} = \frac{9/4}{9} \)
\( \implies q = \frac{1}{4} \)
Since \( p + q = 1 \), we find p:
\( p = 1 - q = 1 - \frac{1}{4} = \frac{3}{4} \)
Substitute the value of p into equation (1):
\( n \times \frac{3}{4} = 9 \)
\( \implies n = 9 \times \frac{4}{3} = 12 \)
Thus, the required binomial distribution is \( (\frac{1}{4} + \frac{3}{4})^{12} \). This represents the probabilities of different outcomes in 12 trials.

(ii) Let the required binomial distribution be \( (q + p)^n \).
Given mean of binomial distribution \( = 10 \)
So, \( np = 10 \) .......(1)
Standard deviation (S.D.) of binomial distribution \( = 2 \sqrt{2} \)
Squaring both sides, variance \( npq = (2 \sqrt{2})^2 = 4 \times 2 = 8 \) .......(2)
Now, divide equation (2) by equation (1):
\( \frac{npq}{np} = \frac{8}{10} \)
\( \implies q = \frac{4}{5} \)
Since \( p + q = 1 \), we find p:
\( p = 1 - q = 1 - \frac{4}{5} = \frac{1}{5} \)
Substitute the value of p into equation (1):
\( n \times \frac{1}{5} = 10 \)
\( \implies n = 10 \times 5 = 50 \)
Thus, the required binomial distribution is \( (\frac{4}{5} + \frac{1}{5})^{50} \). Understanding these parameters helps in predicting future events.
In simple words: For each part, we used the mean and standard deviation to find the individual probabilities (p and q) and the total number of trials (n), which together define the specific binomial distribution.

🎯 Exam Tip: Always remember that variance is the square of the standard deviation. This relationship is often used to simplify calculations and find the value of \(q\) quickly.

 

Question 3. If two dice are rolled 12 times, obtain the mean and the variance of the distribution of successes, if getting a total greater than 4 is considered a success.
Answer:
When two dice are rolled, the total number of outcomes is \( 6^2 = 36 \).
The sample space S consists of all pairs \( (x, y) \) where \( x, y \in \{1, 2, 3, 4, 5, 6\} \):
S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}
We consider "getting a total greater than 4" as a success.
The outcomes whose sum is NOT greater than 4 are: (1,1), (1,2), (1,3), (2,1), (2,2), (3,1). There are 6 such outcomes.
So, the number of favourable outcomes (sum > 4) = Total outcomes - Outcomes (sum \(\le\) 4)
Number of favourable outcomes = \( 36 - 6 = 30 \).
The probability of success \( p \) (getting a total greater than 4) is:
\( p = \frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{30}{36} = \frac{5}{6} \)
The probability of failure \( q \) is:
\( q = 1 - p = 1 - \frac{5}{6} = \frac{1}{6} \)
The number of trials \( n = 12 \).
The mean of the binomial distribution is given by \( np \):
Mean \( = np = 12 \times \frac{5}{6} = 10 \)
The variance of the binomial distribution is given by \( npq \):
Variance \( = npq = 12 \times \frac{5}{6} \times \frac{1}{6} = \frac{5}{3} \)
So, the mean is 10 and the variance is \( \frac{5}{3} \). These values help describe the center and spread of the distribution.
In simple words: First, we found the chance of getting a sum bigger than 4 when rolling two dice. Then, using that chance and the number of times we roll the dice, we calculated the average number of successes (mean) and how spread out the results would be (variance).

🎯 Exam Tip: When dealing with dice rolls, always list all possible outcomes to accurately identify favorable outcomes. This prevents miscalculations of probabilities.

 

Question 4. The sum of mean and variance of a binomial distribution is \( \frac{35}{16} \) for 5 trials. Find the distribution.
Answer:
Let the required binomial distribution be \( (q + p)^n \).
Given that the number of trials \( n = 5 \).
The sum of mean and variance of B.D. \( = \frac{35}{16} \)
So, \( np + npq = \frac{35}{16} \)
Factor out \( np \): \( np(1 + q) = \frac{35}{16} \)
Since \( n = 5 \), substitute this value:
\( 5p(1 + q) = \frac{35}{16} \)
\( p(1 + q) = \frac{35}{16 \times 5} = \frac{7}{16} \)
We know that \( p = 1 - q \). Substitute this into the equation:
\( (1 - q)(1 + q) = \frac{7}{16} \)
This simplifies to \( 1 - q^2 = \frac{7}{16} \)
Now, solve for \( q^2 \):
\( q^2 = 1 - \frac{7}{16} = \frac{16 - 7}{16} = \frac{9}{16} \)
Taking the square root, \( q = \sqrt{\frac{9}{16}} = \frac{3}{4} \) (since \( q \) must be positive).
Now find \( p \):
\( p = 1 - q = 1 - \frac{3}{4} = \frac{1}{4} \)
Thus, the required binomial distribution is \( (\frac{3}{4} + \frac{1}{4})^5 \). This distribution fully describes the probabilities of outcomes in 5 trials.
In simple words: We used the given sum of mean and variance, along with the number of trials, to find the chances of success (p) and failure (q). Then we wrote down the binomial distribution in its standard form.

🎯 Exam Tip: Always remember that \( p+q=1 \). This identity is very useful for simplifying equations and finding \( p \) or \( q \) when the other is known.

 

Question 5. The sum of mean and variance of a binomial distribution of 18 trials, is 10, find the distribution.
Answer:
Let the required binomial distribution be \( (q + p)^n \).
Given number of trials \( n = 18 \).
Mean of binomial distribution \( = np \)
Variance of binomial distribution \( = npq \)
It is given that the sum of mean and variance is 10.
So, \( np + npq = 10 \)
Factor out \( np \): \( np(1 + q) = 10 \)
Substitute \( n = 18 \):
\( 18p(1 + q) = 10 \)
\( p(1 + q) = \frac{10}{18} = \frac{5}{9} \)
We know that \( p = 1 - q \). Substitute this into the equation:
\( (1 - q)(1 + q) = \frac{5}{9} \)
This simplifies to \( 1 - q^2 = \frac{5}{9} \)
Now, solve for \( q^2 \):
\( q^2 = 1 - \frac{5}{9} = \frac{9 - 5}{9} = \frac{4}{9} \)
Taking the square root, \( q = \sqrt{\frac{4}{9}} = \frac{2}{3} \) (since \( q > 0 \)).
Now find \( p \):
\( p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3} \)
Thus, the required binomial distribution is \( (\frac{2}{3} + \frac{1}{3})^{18} \). This distribution provides all the probabilities for successes in 18 trials.
In simple words: We used the total number of trials and the combined sum of the average and spread of the distribution to figure out the individual chances of success and failure, and then wrote the binomial distribution.

🎯 Exam Tip: Always start by writing down the given information and the relevant formulas for mean and variance. This helps to set up the equations correctly and systematically solve for the unknown parameters.

 

Question 6.
(i) The sum of mean and variance of a binomial distribution is 15 and the sum of their squares is 117. Find the distribution.
(ii) The difference between the mean and variance of a binomial distribution is 1 and the difference of their squares is 11. Find the distribution.
Answer:
(i) Let the required binomial distribution be \( (q + p)^n \).
Mean of B.D. \( = np \)
Variance of B.D. \( = npq \)
Given that the sum of mean and variance is 15.
\( np + npq = 15 \)
\( np(1 + q) = 15 \) .......(1)
Also, the sum of squares of mean and variance is 117.
\( (np)^2 + (npq)^2 = 117 \)
\( n^2 p^2 + n^2 p^2 q^2 = 117 \)
\( n^2 p^2 (1 + q^2) = 117 \) .......(2)
Square equation (1):
\( (np(1 + q))^2 = 15^2 \)
\( n^2 p^2 (1 + q)^2 = 225 \) .......(3)
Divide equation (3) by equation (2):
\( \frac{n^2 p^2 (1 + q)^2}{n^2 p^2 (1 + q^2)} = \frac{225}{117} \)
\( \frac{(1 + q)^2}{1 + q^2} = \frac{225}{117} \)
\( \frac{1 + 2q + q^2}{1 + q^2} = \frac{25}{13} \)
\( 13(1 + 2q + q^2) = 25(1 + q^2) \)
\( 13 + 26q + 13q^2 = 25 + 25q^2 \)
\( 12q^2 - 26q + 12 = 0 \)
Divide by 2: \( 6q^2 - 13q + 6 = 0 \)
Factor this quadratic equation:
\( 6q^2 - 9q - 4q + 6 = 0 \)
\( 3q(2q - 3) - 2(2q - 3) = 0 \)
\( (3q - 2)(2q - 3) = 0 \)
So, \( q = \frac{2}{3} \) or \( q = \frac{3}{2} \).
Since \( 0 < q < 1 \) for a binomial distribution, we take \( q = \frac{2}{3} \).
Now find \( p \):
\( p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3} \)
Substitute \( p = \frac{1}{3} \) and \( q = \frac{2}{3} \) into equation (1):
\( n \times \frac{1}{3} (1 + \frac{2}{3}) = 15 \)
\( n \times \frac{1}{3} \times \frac{5}{3} = 15 \)
\( \frac{5n}{9} = 15 \)
\( n = \frac{15 \times 9}{5} = 3 \times 9 = 27 \)
Thus, the required binomial distribution is \( (\frac{2}{3} + \frac{1}{3})^{27} \). This distribution can model success and failure events over 27 trials.

(ii) We know that the mean of a binomial distribution is \( np \) and variance is \( npq \).
Given that the difference between mean and variance is 1.
\( np - npq = 1 \)
\( np(1 - q) = 1 \) .......(1)
Also, the difference of their squares is 11.
\( (np)^2 - (npq)^2 = 11 \)
\( n^2 p^2 - n^2 p^2 q^2 = 11 \)
\( n^2 p^2 (1 - q^2) = 11 \) .......(2)
Square equation (1):
\( (np(1 - q))^2 = 1^2 \)
\( n^2 p^2 (1 - q)^2 = 1 \) .......(3)
Divide equation (3) by equation (2):
\( \frac{n^2 p^2 (1 - q)^2}{n^2 p^2 (1 - q^2)} = \frac{1}{11} \)
\( \frac{(1 - q)^2}{(1 - q)(1 + q)} = \frac{1}{11} \)
\( \frac{1 - q}{1 + q} = \frac{1}{11} \)
\( 11(1 - q) = 1 + q \)
\( 11 - 11q = 1 + q \)
\( 10 = 12q \)
\( q = \frac{10}{12} = \frac{5}{6} \)
Now find \( p \):
\( p = 1 - q = 1 - \frac{5}{6} = \frac{1}{6} \)
Substitute \( p = \frac{1}{6} \) and \( q = \frac{5}{6} \) into equation (1):
\( n \times \frac{1}{6} (1 - \frac{5}{6}) = 1 \)
\( n \times \frac{1}{6} \times \frac{1}{6} = 1 \)
\( \frac{n}{36} = 1 \)
\( n = 36 \)
Thus, the required binomial distribution is \( (\frac{5}{6} + \frac{1}{6})^{36} \). This is a foundational concept in statistics for modeling discrete probability.
In simple words: For both parts, we used the given relationships between the mean, variance, and their squares to set up equations. Solving these equations helped us find the values of n, p, and q, which then allowed us to describe the binomial distribution.

🎯 Exam Tip: When solving for \(q\) from a quadratic equation, always check the condition \(0 < q < 1\). If a value falls outside this range, it's not a valid probability and should be discarded.

 

Question 7.
(i) Comment on the following : For a binomial distribution, mean = 7 and variance = 11.
(ii) Can the mean of a binomial distribution be less than variance? Bring out the fallacy, if any, in the statement. “The mean of a binomial distribution is 15 and its standard deviation is 5."
Answer:
(i) For a binomial distribution, mean \( = np \) and variance \( = npq \).
Given: Mean \( = 7 \implies np = 7 \) .......(1)
Given: Variance \( = 11 \implies npq = 11 \) .......(2)
Divide equation (2) by equation (1):
\( \frac{npq}{np} = \frac{11}{7} \)
\( \implies q = \frac{11}{7} \)
However, for a probability, \( q \) must be between 0 and 1 (i.e., \( 0 < q < 1 \)). Since \( \frac{11}{7} \approx 1.57 \), which is greater than 1, it is impossible for \( q \) to have this value. Therefore, such a binomial distribution cannot exist. The values for mean and variance must be consistent with probability rules.

(ii) For a binomial distribution, mean \( = np \) and variance \( = npq \).
The difference between the mean and variance is \( np - npq = np(1 - q) \).
Since \( p + q = 1 \), we have \( 1 - q = p \).
So, \( np - npq = np(p) = np^2 \).
As \( n \) is a natural number (\( n \ge 1 \)) and \( p \) is a probability (\( p > 0 \)), \( np^2 \) must be greater than 0.
Therefore, \( np - npq > 0 \implies np > npq \).
This means that the mean is always greater than the variance for a binomial distribution (unless \( p=0 \) or \( p=1 \), in which case variance is 0). Hence, the mean of a binomial distribution cannot be less than its variance.

Now, let's look at the statement: "The mean of a binomial distribution is 15 and its standard deviation is 5."
Given: Mean \( = 15 \implies np = 15 \) .......(1)
Given: Standard deviation \( = 5 \)
Variance \( = (\text{Standard Deviation})^2 = 5^2 = 25 \)
So, \( npq = 25 \) .......(2)
Divide equation (2) by equation (1):
\( \frac{npq}{np} = \frac{25}{15} \)
\( \implies q = \frac{5}{3} \)
Again, \( q \) must be between 0 and 1. Since \( \frac{5}{3} \approx 1.67 \), which is greater than 1, this value for \( q \) is impossible. Thus, the statement is fallacious. It's important that all parameters remain within their defined ranges for a valid distribution.
In simple words: (i) We checked if the given mean and variance values make sense for a binomial distribution. They didn't because the calculated chance of failure was more than 1, which is impossible. (ii) We explained that for a binomial distribution, the average (mean) is always bigger than the spread (variance). Then we showed that the example statement was wrong because it also led to an impossible chance of failure.

🎯 Exam Tip: For a valid binomial distribution, probabilities \(p\) and \(q\) must always be between 0 and 1. Also, remember that mean \( > \) variance for a binomial distribution, which serves as a quick check for consistency.

 

Question 8. If the random variable X follows binomial distribution with mean 3 and variance \( \frac{3}{2} \), find P(X < 5).
Answer:
Given mean of binomial distribution \( = 3 \)
So, \( np = 3 \) .......(1)
Given variance of binomial distribution \( = \frac{3}{2} \)
So, \( npq = \frac{3}{2} \) .......(2)
Divide equation (2) by equation (1):
\( \frac{npq}{np} = \frac{3/2}{3} \)
\( \implies q = \frac{1}{2} \)
Since \( p + q = 1 \), we find \( p \):
\( p = 1 - q = 1 - \frac{1}{2} = \frac{1}{2} \)
Substitute \( p = \frac{1}{2} \) into equation (1):
\( n \times \frac{1}{2} = 3 \)
\( \implies n = 3 \times 2 = 6 \)
Thus, the binomial distribution parameters are \( n = 6 \), \( p = \frac{1}{2} \), and \( q = \frac{1}{2} \).
The probability mass function for a binomial distribution is \( P(X=r) = ^n C_r p^r q^{n-r} \).
For this distribution, \( P(X=r) = ^6 C_r (\frac{1}{2})^r (\frac{1}{2})^{6-r} = ^6 C_r (\frac{1}{2})^6 = \frac{^6 C_r}{64} \).
We need to find \( P(X < 5) \). This means finding the probability that \( X \) is 0, 1, 2, 3, or 4.
It's easier to calculate \( P(X < 5) = 1 - P(X \ge 5) \).
Since \( n=6 \), \( P(X \ge 5) = P(X=5) + P(X=6) \).
\( P(X=5) = ^6 C_5 (\frac{1}{2})^6 = 6 \times \frac{1}{64} = \frac{6}{64} \)
\( P(X=6) = ^6 C_6 (\frac{1}{2})^6 = 1 \times \frac{1}{64} = \frac{1}{64} \)
So, \( P(X \ge 5) = \frac{6}{64} + \frac{1}{64} = \frac{7}{64} \)
Therefore, \( P(X < 5) = 1 - \frac{7}{64} = \frac{64 - 7}{64} = \frac{57}{64} \). This calculation gives the likelihood of having fewer than 5 successes in 6 trials.
In simple words: First, we used the average and spread to find the number of trials (n) and the chance of success (p). Then, we calculated the probability of getting less than 5 successes by subtracting the probability of getting 5 or 6 successes from 1.

🎯 Exam Tip: When asked for \( P(X < k) \) or \( P(X > k) \), sometimes it's simpler to calculate the complementary probability \( 1 - P(\text{complementary event}) \). For example, \( P(X < 5) = 1 - P(X \ge 5) \).

 

Question 9. The sum and product of the mean and variance of a binomial distribution are 24 and 128 respectively. Find the distribution. (NMOC)
Answer:
We know that the mean of a binomial distribution is \( np \) and the variance is \( npq \).
Given that the sum of the mean and variance is 24.
\( np + npq = 24 \)
\( np(1 + q) = 24 \) .......(1)
Given that the product of the mean and variance is 128.
\( (np)(npq) = 128 \)
\( n^2 p^2 q = 128 \) .......(2)
Square equation (1):
\( (np(1 + q))^2 = 24^2 \)
\( n^2 p^2 (1 + q)^2 = 576 \) .......(3)
Divide equation (3) by equation (2):
\( \frac{n^2 p^2 (1 + q)^2}{n^2 p^2 q} = \frac{576}{128} \)
\( \frac{(1 + q)^2}{q} = \frac{9}{2} \)
\( 2(1 + q)^2 = 9q \)
\( 2(1 + 2q + q^2) = 9q \)
\( 2 + 4q + 2q^2 = 9q \)
\( 2q^2 - 5q + 2 = 0 \)
Factor this quadratic equation:
\( 2q^2 - 4q - q + 2 = 0 \)
\( 2q(q - 2) - 1(q - 2) = 0 \)
\( (2q - 1)(q - 2) = 0 \)
So, \( q = \frac{1}{2} \) or \( q = 2 \).
Since \( q \) must be between 0 and 1, \( q = 2 \) is impossible.
Therefore, \( q = \frac{1}{2} \).
Now find \( p \):
\( p = 1 - q = 1 - \frac{1}{2} = \frac{1}{2} \)
Substitute \( p = \frac{1}{2} \) and \( q = \frac{1}{2} \) into equation (1):
\( n \times \frac{1}{2} (1 + \frac{1}{2}) = 24 \)
\( n \times \frac{1}{2} \times \frac{3}{2} = 24 \)
\( \frac{3n}{4} = 24 \)
\( n = \frac{24 \times 4}{3} = 8 \times 4 = 32 \)
Thus, the required binomial distribution is \( (\frac{1}{2} + \frac{1}{2})^{32} \). This distribution is useful for situations where there are two equally likely outcomes.
In simple words: We used the sum and product of the mean and variance to create two equations. Solving these equations for the chance of failure (q) and then for the chance of success (p) and the number of trials (n) helped us define the full binomial distribution.

🎯 Exam Tip: When solving simultaneous equations involving mean and variance, it's often effective to square the mean equation and then divide by the variance equation to isolate \(q\).

 

Question 10. If the mean of a binomial distribution is 24 and its standard deviation is 4, calculate the number of observations and the relative frequency of occurrence of the event.
Answer:
Let \( n \) be the number of observations.
Given mean of binomial distribution \( = 24 \)
So, \( np = 24 \) .......(1)
Given standard deviation \( = 4 \)
Variance \( = (\text{Standard Deviation})^2 = 4^2 = 16 \)
So, \( npq = 16 \) .......(2)
Divide equation (2) by equation (1):
\( \frac{npq}{np} = \frac{16}{24} \)
\( \implies q = \frac{2}{3} \)
Since \( p + q = 1 \), we find \( p \):
\( p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3} \)
Substitute \( p = \frac{1}{3} \) into equation (1):
\( n \times \frac{1}{3} = 24 \)
\( \implies n = 24 \times 3 = 72 \)
The number of observations is \( n = 72 \).
The relative frequency of occurrence of the event is the probability of success, \( p = \frac{1}{3} \). These calculations help in understanding the underlying process of the binomial distribution.
In simple words: We used the average and spread of the distribution to find the chance of failure (q), then the chance of success (p), and finally the total number of observations (n). The chance of success (p) also tells us the relative frequency of the event.

🎯 Exam Tip: The "relative frequency of occurrence of the event" in binomial distribution context refers to the probability of success, \(p\). Don't confuse it with the frequency from a frequency table.

 

Question 11. The mean number of success of a binomial distribution \( (p + q)^n \) is 240, where p is the probability of success. The standard deviation is 12 . Calculate the values of n, p and q.
Answer:
Given mean number of successes \( = 240 \)
So, \( np = 240 \) .......(1)
Given standard deviation \( = 12 \)
Variance \( = (\text{Standard Deviation})^2 = 12^2 = 144 \)
So, \( npq = 144 \) .......(2)
Divide equation (2) by equation (1):
\( \frac{npq}{np} = \frac{144}{240} \)
\( \implies q = \frac{3}{5} \)
Since \( p + q = 1 \), we find \( p \):
\( p = 1 - q = 1 - \frac{3}{5} = \frac{2}{5} \)
Substitute \( p = \frac{2}{5} \) into equation (1):
\( n \times \frac{2}{5} = 240 \)
\( \implies n = \frac{240 \times 5}{2} = 120 \times 5 = 600 \)
Thus, the values are \( n = 600 \), \( p = \frac{2}{5} \), and \( q = \frac{3}{5} \). These parameters are fundamental for any binomial probability calculation.
In simple words: We started with the average number of successes and the spread to find the chance of failure (q). Then, we found the chance of success (p) and the total number of trials (n) using these values.

🎯 Exam Tip: Always clearly state the formulas used for mean (\(np\)) and variance (\(npq\)) at the beginning of your solution. This shows a clear understanding of the concepts.

 

Question 12. If the mean of a binomial distribution is 960 and its standard deviation is 24, calculate the number of observations; and the probability of occurrence of the event.
Answer:
Let \( n \) be the number of observations.
Given mean of binomial distribution \( = 960 \)
So, \( np = 960 \) .......(1)
Given standard deviation \( = 24 \)
Variance \( = (\text{Standard Deviation})^2 = 24^2 = 576 \)
So, \( npq = 576 \) .......(2)
Divide equation (2) by equation (1):
\( \frac{npq}{np} = \frac{576}{960} \)
\( \implies q = \frac{3}{5} \)
Since \( p + q = 1 \), we find \( p \):
\( p = 1 - q = 1 - \frac{3}{5} = \frac{2}{5} \)
Substitute \( p = \frac{2}{5} \) into equation (1):
\( n \times \frac{2}{5} = 960 \)
\( \implies n = \frac{960 \times 5}{2} = 480 \times 5 = 2400 \)
The number of observations is \( n = 2400 \).
The probability of occurrence of the event is the probability of success, \( p = \frac{2}{5} \). This probability tells us how likely the event is in each trial.
In simple words: We used the given average and spread to first calculate the chance of failure (q), then the chance of success (p), and finally the total number of trials (n). The chance of success (p) is the event's probability.

🎯 Exam Tip: Pay attention to what the question asks for. Sometimes it's just parameters (n, p, q), other times it's specific probabilities or interpretations like "probability of occurrence of the event" (which is \(p\)).

 

Question 13. Razor blades is sold in packets of 5. The following is the frequency distribution of 100 packets according to the number of faulty blades in them. Find the number of faulty blades per packet. Assuming that the distribution is binomial, estimate the probability that a blade taken at random from any packet will be faulty.

Answer:
We first calculate the mean number of faulty blades per packet.
Total number of packets \( = 100 \).
Mean number of faulty blades per packet \( = \frac{\sum (\text{No. of faulty blades} \times \text{No. of packets})}{\sum (\text{No. of packets})} \)
Mean \( = \frac{(0 \times 80) + (1 \times 17) + (2 \times 2) + (3 \times 1)}{100} \)
Mean \( = \frac{0 + 17 + 4 + 3}{100} = \frac{24}{100} = 0.24 \)
Assuming the distribution is binomial, the mean number of faulty blades is also \( np \).
So, \( np = 0.24 \)
A packet contains 5 blades, so the number of trials \( n = 5 \).
Now, we can find the probability \( p \) that a blade taken at random from any packet will be faulty:
\( 5p = 0.24 \)
\( p = \frac{0.24}{5} = \frac{24}{500} = 0.048 \)
Thus, the estimated probability that a blade taken at random from any packet will be faulty is \( 0.048 \). This small probability suggests good quality control for the blades.
In simple words: We used the given table to find the average number of faulty blades per packet. Since each packet has 5 blades, we then used this average to calculate the chance that any single blade is faulty.

🎯 Exam Tip: When given frequency data for a binomial distribution, calculate the mean using \( \sum fx / \sum f \) first. Then equate this mean to \(np\) to find \(p\) (the probability of success/faulty item) or \(n\) (number of trials/items in a packet).

 

Question 14. A coin is tossed (i) 10 times, (ii) 100 times, (iii) 1000 times. Calculate in each case the expected number of heads and the standard deviation. Which would surprise you more 3 heads and 7 tails in 10 tosses or 300 heads and 700 tails in 1000 tosses? Justify your answer mathematically.
Answer:
For a coin toss, the probability of getting a head \( p = \frac{1}{2} \) and the probability of getting a tail \( q = \frac{1}{2} \).
Expected number of heads (Mean) \( = np \).
Standard deviation (S.D.) \( = \sqrt{npq} \).

(i) For \( n = 10 \) tosses:
Expected number of heads \( = np = 10 \times \frac{1}{2} = 5 \)
Standard deviation \( = \sqrt{10 \times \frac{1}{2} \times \frac{1}{2}} = \sqrt{\frac{10}{4}} = \sqrt{\frac{5}{2}} \approx 1.58 \)

(ii) For \( n = 100 \) tosses:
Expected number of heads \( = np = 100 \times \frac{1}{2} = 50 \)
Standard deviation \( = \sqrt{100 \times \frac{1}{2} \times \frac{1}{2}} = \sqrt{25} = 5 \)

(iii) For \( n = 1000 \) tosses:
Expected number of heads \( = np = 1000 \times \frac{1}{2} = 500 \)
Standard deviation \( = \sqrt{1000 \times \frac{1}{2} \times \frac{1}{2}} = \sqrt{250} = 5\sqrt{10} \approx 15.8 \)

To determine which outcome would be more surprising, we calculate how many standard deviations away from the mean each observed outcome is.
For 10 tosses: Observed heads = 3, Expected heads = 5.
Difference \( = 5 - 3 = 2 \).
Standard deviation \( \sigma = 1.58 \).
Number of standard deviations away \( = \frac{\text{Difference}}{\sigma} = \frac{2}{1.58} \approx 1.27 \sigma \).
Since \( 1.27\sigma \) is less than \( 2\sigma \), this outcome is not considered very surprising. It falls within the normal range of variation for an unbiased coin.

For 1000 tosses: Observed heads = 300, Expected heads = 500.
Difference \( = 500 - 300 = 200 \).
Standard deviation \( \sigma = 15.8 \).
Number of standard deviations away \( = \frac{\text{Difference}}{\sigma} = \frac{200}{15.8} \approx 12.66 \sigma \).
Since \( 12.66\sigma \) is much greater than \( 3\sigma \), this outcome would be highly surprising. It suggests that the coin might be biased, as such a large deviation from the expected mean is very unlikely with an unbiased coin. Therefore, 300 heads and 700 tails in 1000 tosses would be much more surprising.
In simple words: For each case, we found the average number of heads and the spread of the results. To see which outcome was more surprising, we checked how far each result was from its average, measured in units of standard deviation. A result many standard deviations away from the average is very surprising.

🎯 Exam Tip: To justify "surprise" in probability questions, calculate the number of standard deviations the observed value is from the mean. A deviation of more than \( 2\sigma \) or \( 3\sigma \) usually indicates a statistically significant (and thus surprising) event.

Examples:

 

Question 1. In a large collection of bulbs, 3 out of 5 are daffodils and the rest are tulips. If they are planted at random, calculate the probability that in a row of five plants (i) all are daffodils, (ii) at least four are daffodils.
Answer:
Let \( p \) be the probability that a bulb is a daffodil.
\( p = \frac{3}{5} \)
Let \( q \) be the probability that a bulb is a tulip (not a daffodil).
\( q = 1 - p = 1 - \frac{3}{5} = \frac{2}{5} \)
The number of plants in the row is \( n = 5 \).
Let \( X \) be the random variable denoting the number of daffodils planted out of 5 plants. All trials are independent.
This is a binomial variate with parameters \( n = 5 \), \( p = \frac{3}{5} \).
The probability mass function is \( P(X=r) = ^n C_r p^r q^{n-r} = ^5 C_r (\frac{3}{5})^r (\frac{2}{5})^{5-r} \).

(i) Required probability that all are daffodils is \( P(X = 5) \).
\( P(X = 5) = ^5 C_5 (\frac{3}{5})^5 (\frac{2}{5})^0 \)
\( P(X = 5) = 1 \times (\frac{3^5}{5^5}) \times 1 = \frac{243}{3125} \)

(ii) Required probability that at least four are daffodils is \( P(X \ge 4) \).
\( P(X \ge 4) = P(X = 4) + P(X = 5) \)
First, calculate \( P(X = 4) \):
\( P(X = 4) = ^5 C_4 (\frac{3}{5})^4 (\frac{2}{5})^1 \)
\( P(X = 4) = 5 \times \frac{3^4}{5^4} \times \frac{2}{5} = 5 \times \frac{81}{625} \times \frac{2}{5} = \frac{810}{3125} \)
We already calculated \( P(X = 5) = \frac{243}{3125} \).
So, \( P(X \ge 4) = \frac{810}{3125} + \frac{243}{3125} = \frac{1053}{3125} \). These probabilities allow us to predict the composition of the planted row.
In simple words: We first found the chance of planting a daffodil and a tulip. Then, for the first part, we calculated the chance that all five plants would be daffodils. For the second part, we added the chances of having exactly four daffodils or exactly five daffodils.

🎯 Exam Tip: When calculating "at least" probabilities, remember to sum the probabilities of all outcomes that meet the condition. Sometimes, calculating the complementary probability (1 minus "less than") is easier.

 

Question 2. Mohan and Sohan play 12 games of chess. Mohan wins 6 games and Sohan 4 games. 2 games end in a draw. They agree to play 3 more games. Calculate the probability that out of these 3 games, 2 games end in, a draw.
Answer:
Total number of games played \( = 12 \).
Number of games Mohan wins \( = 6 \).
Number of games Sohan wins \( = 4 \).
Number of games that end in a draw \( = 2 \).
Probability of a game ending in a draw \( p = \frac{\text{Number of draws}}{\text{Total games}} = \frac{2}{12} = \frac{1}{6} \).
Probability of a game not ending in a draw \( q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6} \).
They agree to play 3 more games, so the number of trials \( n = 3 \).
Let \( X \) be the random variable denoting the number of games that end in a draw out of these 3 games. All trials are independent.
This is a binomial variate with parameters \( n = 3 \), \( p = \frac{1}{6} \).
The probability mass function is \( P(X=r) = ^n C_r p^r q^{n-r} = ^3 C_r (\frac{1}{6})^r (\frac{5}{6})^{3-r} \).
We need to calculate the probability that 2 games end in a draw, i.e., \( P(X = 2) \).
\( P(X = 2) = ^3 C_2 (\frac{1}{6})^2 (\frac{5}{6})^{3-2} \)
\( P(X = 2) = 3 \times (\frac{1}{36}) \times (\frac{5}{6}) \)
\( P(X = 2) = \frac{15}{216} = \frac{5}{72} \)
The probability of 2 draws in 3 games is \( \frac{5}{72} \). This calculation helps in understanding future game outcomes.
In simple words: We first found the chance of a game ending in a draw based on past results. Then, using that chance and the number of new games, we calculated the probability that exactly two of those new games would end in a draw.

🎯 Exam Tip: Ensure that the probability of success \(p\) is correctly defined from the given data. Also, clearly state the binomial parameters (\(n, p\)) before applying the probability formula.

 

Question 3. On an average, out of 12 games of chess played by A and B, A wins 6, B wins 4 and 2 games end in a tie. A and B play a tournament of 3 games. Calculate the probability that (i) at least 2 games end in a tie. (ii) A and B win alternate games, no games end in a tie.
Answer:
Total number of games played \( = 12 \).
Number of games A wins \( = 6 \).
Number of games B wins \( = 4 \).
Number of games ending in a tie \( = 2 \).
A and B play a tournament of 3 games, so \( n = 3 \).

Let \( p \) be the probability of a game ending in a tie.
\( p = \frac{\text{Number of ties}}{\text{Total games}} = \frac{2}{12} = \frac{1}{6} \).
Let \( q \) be the probability of a game not ending in a tie.
\( q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6} \).
Let \( X \) be the random variable denoting the number of tie games out of 3. This is a binomial variate with parameters \( n = 3 \), \( p = \frac{1}{6} \).

(i) Probability that at least 2 games end in a tie, \( P(X \ge 2) \).
\( P(X \ge 2) = P(X = 2) + P(X = 3) \)
Using the binomial probability formula \( P(X=r) = ^n C_r p^r q^{n-r} \):
\( P(X = 2) = ^3 C_2 (\frac{1}{6})^2 (\frac{5}{6})^{3-2} = 3 \times \frac{1}{36} \times \frac{5}{6} = \frac{15}{216} \)
\( P(X = 3) = ^3 C_3 (\frac{1}{6})^3 (\frac{5}{6})^{3-3} = 1 \times \frac{1}{216} \times 1 = \frac{1}{216} \)
\( P(X \ge 2) = \frac{15}{216} + \frac{1}{216} = \frac{16}{216} = \frac{2}{27} \)

(ii) Probability that A and B win alternate games, with no games ending in a tie.
First, find the individual probabilities of A winning and B winning:
Probability of A winning \( P(A) = \frac{6}{12} = \frac{1}{2} \).
Probability of B winning \( P(B) = \frac{4}{12} = \frac{1}{3} \).
Since no games end in a tie, these are the only two possible outcomes for each game.
There are two possible alternating sequences for 3 games:
1. A wins, B wins, A wins (ABA): \( P(A) \times P(B) \times P(A) = \frac{1}{2} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{12} \)
2. B wins, A wins, B wins (BAB): \( P(B) \times P(A) \times P(B) = \frac{1}{3} \times \frac{1}{2} \times \frac{1}{3} = \frac{1}{18} \)
The total probability for this condition is the sum of these two sequence probabilities:
Total probability \( = \frac{1}{12} + \frac{1}{18} \)
To add these fractions, find a common denominator, which is 36:
Total probability \( = \frac{3}{36} + \frac{2}{36} = \frac{5}{36} \). This is a composite probability based on sequential events.
In simple words: First, we calculated the chance of a tie, and then for part (i), we added the chances of getting exactly two ties or exactly three ties. For part (ii), we looked at the chances of A and B winning back and forth without any ties, considering the two possible starting players, and then added those probabilities together.

🎯 Exam Tip: For "alternate games" scenarios, list all possible valid sequences. Calculate the probability of each sequence and then sum them up, assuming independence between games.

 

Question 4. The probability density function y of a continuous variable x is given by \( y = \frac{x}{k} \), \( 0 < x < 2 \) and \( y = 0 \) for all other values of x. Calculate the value of k and the probability that \( x<1 \).
Answer: For a continuous random variable, the total probability must always be equal to 1. We are given the probability density function \( y = \frac{x}{k} \) for \( 0 < x < 2 \). To find \( k \), we integrate this function from 0 to 2 and set the result to 1.
\[ \int_{0}^{2} \frac{x}{k} dx = 1 \] \[ \implies \frac{1}{k} \left[ \frac{x^2}{2} \right]_0^2 = 1 \] \[ \implies \frac{1}{k} \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = 1 \] \[ \implies \frac{2}{k} = 1 \] \[ \implies k = 2 \] Now that we have \( k=2 \), the probability density function is \( y = \frac{x}{2} \) for \( 0 < x < 2 \). We need to find the probability that \( x < 1 \). For this, we integrate the function from 0 to 1.
\[ P(x < 1) = \int_{0}^{1} \frac{x}{2} dx \] \[ \implies P(x < 1) = \frac{1}{2} \left[ \frac{x^2}{2} \right]_0^1 \] \[ \implies P(x < 1) = \frac{1}{2} \left( \frac{1^2}{2} - \frac{0^2}{2} \right) \] \[ \implies P(x < 1) = \frac{1}{2} \times \frac{1}{2} \] \[ \implies P(x < 1) = \frac{1}{4} \]In simple words: First, we use the rule that all probabilities add up to 1 over the whole range. We do this by integrating the given function. This helps us find the hidden number \( k \), which turns out to be 2. Next, we use this new function to find the probability that \( x \) is smaller than 1. We integrate the function again, but this time only from 0 to 1. The chance for \( x < 1 \) is \( \frac{1}{4} \).

🎯 Exam Tip: Remember that for a probability density function, the total area under the curve must be 1, which is found by integrating over the entire domain. Also, probabilities for specific ranges are found by integrating over those ranges.

 

Question 5. A pair of dice is thrown 5 times. If getting a doublet is considered a success, find the probability of 2 successes.
Answer: When two dice are rolled, there are \( 6 \times 6 = 36 \) possible outcomes. A "doublet" means both dice show the same number (e.g., (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)). There are 6 such doublets. So, the probability of getting a doublet in one throw (our success, \( p \)) is \( \frac{6}{36} = \frac{1}{6} \). The probability of not getting a doublet (our failure, \( q \)) is \( 1 - p = 1 - \frac{1}{6} = \frac{5}{6} \).
We throw the pair of dice 5 times, so the number of trials \( n=5 \). We want to find the probability of getting exactly 2 successes (2 doublets). This is a binomial distribution problem.
The formula for binomial probability is \( P(X=r) = { }^n C_r p^r q^{n-r} \).
For \( r=2 \), \( n=5 \), \( p=\frac{1}{6} \), \( q=\frac{5}{6} \):
\[ P(X=2) = { }^5 C_2 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{5-2} \] \[ \implies P(X=2) = { }^5 C_2 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^3 \] \[ \implies P(X=2) = \frac{5 \times 4}{2 \times 1} \times \frac{1}{36} \times \frac{125}{216} \] \[ \implies P(X=2) = 10 \times \frac{125}{36 \times 216} \] \[ \implies P(X=2) = \frac{1250}{7776} \] \[ \implies P(X=2) = \frac{625}{3888} \]In simple words: When you roll two dice, there are 36 possible results. Getting the same number on both dice (a doublet) happens 6 times. So, the chance of a doublet is \( \frac{6}{36} = \frac{1}{6} \). This is our success rate. The chance of not getting a doublet is \( \frac{5}{6} \). We roll the dice 5 times. We want to know the chance of getting exactly 2 doublets. Using the special binomial formula, this chance works out to be \( \frac{625}{3888} \).

🎯 Exam Tip: Clearly identify \( n, p \), and \( q \) from the problem statement. Remember that \( p \) is the probability of success for a single trial, and \( q = 1-p \).

 

Question 6. Assume, that on an average, 1 telephone out of 10 is busy, six telephone numbers are randomly selected and called. Find the probability that four of them will be busy.
Answer: On average, 1 out of 10 telephones is busy. So, the probability that a randomly selected telephone number is busy (our success, \( p \)) is \( \frac{1}{10} \). The probability that a telephone number is not busy (our failure, \( q \)) is \( 1 - p = 1 - \frac{1}{10} = \frac{9}{10} \).
We are selecting and calling six telephone numbers, so the number of trials \( n=6 \). We want to find the probability that exactly four of these six numbers will be busy. This is a binomial distribution problem.
The formula for binomial probability is \( P(X=r) = { }^n C_r p^r q^{n-r} \).
For \( r=4 \), \( n=6 \), \( p=\frac{1}{10} \), \( q=\frac{9}{10} \):
\[ P(X=4) = { }^6 C_4 \left(\frac{1}{10}\right)^4 \left(\frac{9}{10}\right)^{6-4} \] \[ \implies P(X=4) = { }^6 C_4 \left(\frac{1}{10}\right)^4 \left(\frac{9}{10}\right)^2 \] \[ \implies P(X=4) = \frac{6 \times 5}{2 \times 1} \times \frac{1}{10^4} \times \frac{9^2}{10^2} \] \[ \implies P(X=4) = 15 \times \frac{1}{10000} \times \frac{81}{100} \] \[ \implies P(X=4) = \frac{15 \times 81}{1000000} \] \[ \implies P(X=4) = \frac{1215}{1000000} \] \[ \implies P(X=4) = 0.001215 \]In simple words: If 1 out of 10 phones is busy, then the chance a phone is busy is \( \frac{1}{10} \), and not busy is \( \frac{9}{10} \). We call 6 phones. We want to find the chance that exactly 4 of these 6 phones are busy. Using the binomial formula, we calculate this probability to be 0.001215. It's a small chance, showing that finding four busy lines out of six is quite rare.

🎯 Exam Tip: Pay attention to the distinction between "at least," "at most," and "exactly" in probability questions, as they determine which terms of the binomial sum you need to include.

 

Question 7. (i) Comment on the following : For a binomial distribution, mean = 7 and variance = 11. (ii) Can the mean of a binomial distribution be less than variance? Bring out the fallacy, if any, in the statement. “The mean of a binomial distribution is 15 and its standard deviation is 5."
Answer:
(i) For a binomial distribution, the mean is given by \( np \) and the variance by \( npq \).
Given mean \( np = 7 \) (Equation 1)
Given variance \( npq = 11 \) (Equation 2)
If we divide Equation 2 by Equation 1:
\[ \frac{npq}{np} = \frac{11}{7} \] \[ \implies q = \frac{11}{7} \] However, \( q \) represents a probability, which must always be between 0 and 1 (i.e., \( 0 < q < 1 \)). Since \( \frac{11}{7} \) is greater than 1, it is an impossible value for a probability. Therefore, a binomial distribution with a mean of 7 and a variance of 11 cannot exist.
(ii) For a binomial distribution, we know Mean \( = np \) and Variance \( = npq \).
The difference between the mean and variance is \( \text{Mean} - \text{Variance} = np - npq = np(1-q) \).
Since \( 1-q = p \) (as \( p+q=1 \)), the difference becomes \( np \cdot p = np^2 \).
Given that \( n \) is a natural number (number of trials) and \( p \) is a probability of success, both \( n \) and \( p \) are positive. Therefore, \( np^2 \) must always be positive (\( np^2 > 0 \)).
This implies that \( \text{Mean} - \text{Variance} > 0 \), which means \( \text{Mean} > \text{Variance} \).
Thus, the mean of a binomial distribution can never be less than its variance.
Now, let's analyze the given statement: "The mean of a binomial distribution is 15 and its standard deviation is 5."
Given mean \( np = 15 \) (Equation 1)
Given standard deviation \( \sigma = 5 \). The variance is \( \sigma^2 \). So, \( npq = 5^2 = 25 \) (Equation 2)
If we divide Equation 2 by Equation 1:
\[ \frac{npq}{np} = \frac{25}{15} \] \[ \implies q = \frac{5}{3} \] Again, \( q \) is a probability and must be between 0 and 1. Since \( \frac{5}{3} \) is greater than 1, this value for \( q \) is impossible. Therefore, the statement is fallacious (incorrect).
In simple words:
(i) If a binomial distribution had a mean of 7 and a variance of 11, the calculation for \( q \) (the probability of failure) would be \( \frac{11}{7} \). But a probability cannot be bigger than 1. So, this distribution is not possible. For binomial distributions, the variance is always less than or equal to the mean.
(ii) For a binomial distribution, the mean is always greater than the variance. This is because the difference between them is \( np^2 \), which is always a positive number. So, the mean can never be smaller than the variance.
Regarding the statement about a mean of 15 and a standard deviation of 5: if the mean is 15, the variance is \( 5^2 = 25 \). If we calculate \( q \) from these values (\( \frac{\text{variance}}{\text{mean}} \)), we get \( \frac{25}{15} = \frac{5}{3} \). Again, this probability is greater than 1, which means such a statement cannot be true.

🎯 Exam Tip: Always check if calculated probabilities (\( p \) or \( q \)) fall within the valid range of 0 to 1. Also, remember the fundamental relationship: for a binomial distribution, mean \( \ge \) variance.

 

Question 8. If the random variable X follows binomial distribution with mean 3 and variance \( \frac{3}{2} \), find \( P(X < 5) \).
Answer: We are given the mean of the binomial distribution as 3 and the variance as \( \frac{3}{2} \).
Mean \( np = 3 \) (Equation 1)
Variance \( npq = \frac{3}{2} \) (Equation 2)
To find \( q \), we divide Equation 2 by Equation 1:
\[ q = \frac{npq}{np} = \frac{\frac{3}{2}}{3} = \frac{3}{2 \times 3} = \frac{1}{2} \] Now, we find \( p \) using \( p = 1-q \):
\[ p = 1 - \frac{1}{2} = \frac{1}{2} \] Substitute \( p = \frac{1}{2} \) back into Equation 1 to find \( n \):
\[ n \times \frac{1}{2} = 3 \] \[ \implies n = 6 \] So, the binomial distribution has parameters \( n=6 \) and \( p=\frac{1}{2} \). The probability mass function is \( P(X=r) = { }^n C_r p^r q^{n-r} \).
For \( n=6 \), \( p=\frac{1}{2} \), \( q=\frac{1}{2} \), the formula simplifies to:
\[ P(X=r) = { }^6 C_r \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{6-r} = { }^6 C_r \left(\frac{1}{2}\right)^6 = \frac{{ }^6 C_r}{64} \] We need to find \( P(X < 5) \). This means the probability of getting 0, 1, 2, 3, or 4 successes. It is easier to calculate this as \( 1 - P(X \ge 5) \), which is \( 1 - (P(X=5) + P(X=6)) \).
Calculate \( P(X=5) \):
\[ P(X=5) = { }^6 C_5 \left(\frac{1}{2}\right)^6 = 6 \times \frac{1}{64} = \frac{6}{64} \] Calculate \( P(X=6) \):
\[ P(X=6) = { }^6 C_6 \left(\frac{1}{2}\right)^6 = 1 \times \frac{1}{64} = \frac{1}{64} \] Now, we find \( P(X < 5) \):
\[ P(X < 5) = 1 - \left( P(X=5) + P(X=6) \right) \] \[ \implies P(X < 5) = 1 - \left( \frac{6}{64} + \frac{1}{64} \right) \] \[ \implies P(X < 5) = 1 - \frac{7}{64} \] \[ \implies P(X < 5) = \frac{64 - 7}{64} \] \[ \implies P(X < 5) = \frac{57}{64} \]In simple words: We start with the mean (3) and variance (\( \frac{3}{2} \)) to figure out the number of trials (\( n \)) and the probability of success (\( p \)) for our binomial distribution. We find that \( n=6 \) and \( p=\frac{1}{2} \). Now, we need the probability of having less than 5 successes. It's simpler to calculate the chance of having 5 or 6 successes and subtract that from 1. The chance of 5 successes is \( \frac{6}{64} \), and for 6 successes it's \( \frac{1}{64} \). Adding these up and subtracting from 1 gives \( 1 - \frac{7}{64} = \frac{57}{64} \).

🎯 Exam Tip: When asked for \( P(X < k) \) or \( P(X \ge k) \), consider using the complementary probability rule (\( P(A) = 1 - P(A') \)) if it simplifies calculations by reducing the number of terms to compute.

 

Question 9. The sum and product of the mean and variance of a binomial distribution are 24 and 128 respectively. Find the distribution. (NMOC)
Answer: Let \( M \) be the mean and \( V \) be the variance of the binomial distribution. We know \( M = np \) and \( V = npq \).
Given: Sum of mean and variance \( M + V = 24 \).
\[ np + npq = 24 \] \[ \implies np(1+q) = 24 \] (Equation 1)
Given: Product of mean and variance \( M \times V = 128 \).
\[ np \times npq = 128 \] \[ \implies n^2 p^2 q = 128 \] (Equation 2)
Square both sides of Equation 1:
\[ (np(1+q))^2 = 24^2 \] \[ \implies n^2 p^2 (1+q)^2 = 576 \] (Equation 3)
Divide Equation 3 by Equation 2:
\[ \frac{n^2 p^2 (1+q)^2}{n^2 p^2 q} = \frac{576}{128} \] \[ \implies \frac{(1+q)^2}{q} = \frac{9}{2} \] Now, solve for \( q \):
\[ 2(1+q)^2 = 9q \] \[ \implies 2(1 + 2q + q^2) = 9q \] \[ \implies 2 + 4q + 2q^2 = 9q \] \[ \implies 2q^2 - 5q + 2 = 0 \] This is a quadratic equation. We can factor it:
\[ (2q - 1)(q - 2) = 0 \] This gives two possible values for \( q \): \( 2q - 1 = 0 \implies q = \frac{1}{2} \) or \( q - 2 = 0 \implies q = 2 \).
Since \( q \) is a probability, its value must be between 0 and 1. So, \( q = 2 \) is not possible. Therefore, \( q = \frac{1}{2} \).
Now, find \( p \) using \( p = 1-q \):
\[ p = 1 - \frac{1}{2} = \frac{1}{2} \] Substitute \( p = \frac{1}{2} \) and \( q = \frac{1}{2} \) into Equation 1:
\[ n \times \frac{1}{2} (1 + \frac{1}{2}) = 24 \] \[ \implies n \times \frac{1}{2} \times \frac{3}{2} = 24 \] \[ \implies \frac{3n}{4} = 24 \] \[ \implies n = \frac{24 \times 4}{3} \] \[ \implies n = 8 \times 4 \] \[ \implies n = 32 \] The required binomial distribution is defined by its parameters \( n=32, p=\frac{1}{2}, q=\frac{1}{2} \). We can write it as \( \left(\frac{1}{2} + \frac{1}{2}\right)^{32} \).
In simple words: We are given that the sum of the mean and variance of a binomial distribution is 24, and their product is 128. We use these two facts to set up equations involving \( n, p \), and \( q \). By cleverly combining these equations, we solve for \( q \). We find \( q = \frac{1}{2} \) (as probability cannot be greater than 1). From \( q \), we find \( p = \frac{1}{2} \). Finally, we use these values to find \( n \), which turns out to be 32. So the distribution is \( (\frac{1}{2} + \frac{1}{2})^{32} \).

🎯 Exam Tip: When given relationships between mean and variance, use the formulas \( M=np \) and \( V=npq \) to form a system of equations. Solving these equations is often the quickest way to find \( n, p \), and \( q \).

 

Question 9. A die is tossed three times. Getting a ' 3 ' or a ' 5' is considered a success. Find the probability of at least two successes.
Answer: When a single die is tossed, there are 6 possible outcomes: \( \{1, 2, 3, 4, 5, 6\} \). Getting a '3' or a '5' means there are 2 favorable outcomes. So, the probability of success \( p = \frac{2}{6} = \frac{1}{3} \). The probability of failure \( q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \).
The die is tossed three times, so the number of trials \( n=3 \). We need to find the probability of at least two successes, which means \( P(X \ge 2) \). This is the sum of probabilities for 2 successes and 3 successes: \( P(X \ge 2) = P(X=2) + P(X=3) \).
Using the binomial probability formula \( P(X=r) = { }^n C_r p^r q^{n-r} \):
For \( P(X=2) \):
\[ P(X=2) = { }^3 C_2 \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^{3-2} \] \[ \implies P(X=2) = 3 \times \frac{1}{9} \times \frac{2}{3} = \frac{6}{27} \] For \( P(X=3) \):
\[ P(X=3) = { }^3 C_3 \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^{3-3} \] \[ \implies P(X=3) = 1 \times \frac{1}{27} \times 1 = \frac{1}{27} \] Adding these probabilities:
\[ P(X \ge 2) = \frac{6}{27} + \frac{1}{27} = \frac{7}{27} \]In simple words: When a die is rolled, getting a 3 or 5 is a success. The chance of success is \( \frac{2}{6} = \frac{1}{3} \), and the chance of not succeeding is \( \frac{2}{3} \). We roll the die 3 times. We want the chance of getting at least 2 successes, meaning either 2 successes or 3 successes. We use the binomial formula to find these. For 2 successes, the chance is \( \frac{6}{27} \). For 3 successes, it's \( \frac{1}{27} \). Adding them up, the total chance for at least 2 successes is \( \frac{7}{27} \).

🎯 Exam Tip: Ensure you correctly identify 'success' and 'failure' probabilities and the number of trials. The phrase "at least \( k \)" means \( k \) or more, so sum the probabilities for \( r=k, k+1, \dots, n \).

 

Question 10. A coin is tossed 5 times. What is the probability of getting at least three heads?
Answer: When a coin is tossed, the probability of getting a head (success, \( p \)) is \( \frac{1}{2} \). The probability of not getting a head (failure, \( q \)) is \( 1 - \frac{1}{2} = \frac{1}{2} \).
The coin is tossed 5 times, so the number of trials \( n=5 \). We need to find the probability of getting at least three heads, which means \( P(X \ge 3) \). This is the sum of probabilities for 3, 4, or 5 heads: \( P(X \ge 3) = P(X=3) + P(X=4) + P(X=5) \).
Using the binomial probability formula \( P(X=r) = { }^n C_r p^r q^{n-r} \). For \( n=5, p=\frac{1}{2}, q=\frac{1}{2} \), this simplifies to \( P(X=r) = { }^5 C_r \left(\frac{1}{2}\right)^5 = \frac{{ }^5 C_r}{32} \).
Calculate \( P(X=3) \):
\[ P(X=3) = { }^5 C_3 \left(\frac{1}{2}\right)^5 = 10 \times \frac{1}{32} = \frac{10}{32} \] Calculate \( P(X=4) \):
\[ P(X=4) = { }^5 C_4 \left(\frac{1}{2}\right)^5 = 5 \times \frac{1}{32} = \frac{5}{32} \] Calculate \( P(X=5) \):
\[ P(X=5) = { }^5 C_5 \left(\frac{1}{2}\right)^5 = 1 \times \frac{1}{32} = \frac{1}{32} \] Adding these probabilities:
\[ P(X \ge 3) = \frac{10}{32} + \frac{5}{32} + \frac{1}{32} = \frac{16}{32} = \frac{1}{2} \]In simple words: When you flip a coin, there's a \( \frac{1}{2} \) chance of getting a head (success) and a \( \frac{1}{2} \) chance of not getting a head (failure). We flip the coin 5 times. We want the chance of getting at least 3 heads. This means we calculate the chances of getting exactly 3, 4, or 5 heads and add them together. The chance for 3 heads is \( \frac{10}{32} \), for 4 heads it's \( \frac{5}{32} \), and for 5 heads it's \( \frac{1}{32} \). Adding these up gives a total probability of \( \frac{16}{32} \), which simplifies to \( \frac{1}{2} \).

🎯 Exam Tip: For coin toss problems, \( p \) and \( q \) are usually \( \frac{1}{2} \). This simplifies calculations, as \( p^r q^{n-r} \) becomes \( (\frac{1}{2})^n \).

 

Question 11. Four dice are thrown simultaneously. If the occurrence of an odd number in a single die is considered a success, find the probability of at most 2 successes.
Answer: When a single die is thrown, the outcomes are \( \{1, 2, 3, 4, 5, 6\} \). The odd numbers are \( \{1, 3, 5\} \). So, the probability of getting an odd number (success, \( p \)) is \( \frac{3}{6} = \frac{1}{2} \). The probability of not getting an odd number (failure, \( q \)) is \( 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \).
Four dice are thrown simultaneously, so the number of trials \( n=4 \). We need to find the probability of at most 2 successes, which means \( P(X \le 2) \). This is the sum of probabilities for 0, 1, or 2 successes: \( P(X \le 2) = P(X=0) + P(X=1) + P(X=2) \).
Using the binomial probability formula \( P(X=r) = { }^n C_r p^r q^{n-r} \). For \( n=4, p=\frac{1}{2}, q=\frac{1}{2} \), this simplifies to \( P(X=r) = { }^4 C_r \left(\frac{1}{2}\right)^4 = \frac{{ }^4 C_r}{16} \).
Calculate \( P(X=0) \):
\[ P(X=0) = { }^4 C_0 \left(\frac{1}{2}\right)^4 = 1 \times \frac{1}{16} = \frac{1}{16} \] Calculate \( P(X=1) \):
\[ P(X=1) = { }^4 C_1 \left(\frac{1}{2}\right)^4 = 4 \times \frac{1}{16} = \frac{4}{16} \] Calculate \( P(X=2) \):
\[ P(X=2) = { }^4 C_2 \left(\frac{1}{2}\right)^4 = 6 \times \frac{1}{16} = \frac{6}{16} \] Adding these probabilities:
\[ P(X \le 2) = \frac{1}{16} + \frac{4}{16} + \frac{6}{16} = \frac{11}{16} \]In simple words: When you roll a single die, the chance of getting an odd number (like 1, 3, or 5) is \( \frac{3}{6} = \frac{1}{2} \). This is our success rate. The chance of not getting an odd number is also \( \frac{1}{2} \). We roll four dice. We want to find the chance of getting at most 2 odd numbers, which means 0, 1, or 2 odd numbers. We calculate each chance: \( \frac{1}{16} \) for 0, \( \frac{4}{16} \) for 1, and \( \frac{6}{16} \) for 2. Adding them all up gives us a total chance of \( \frac{11}{16} \).

🎯 Exam Tip: The phrase "at most \( k \)" means \( k \) or fewer, so sum the probabilities for \( r=0, 1, \dots, k \). Listing the outcomes for success and failure first helps avoid errors.

 

Question 12. Assuming that on an average one tele- phone out of ten is busy, seven telephone numbers are randomly selected and called. Find the probability that three of them will be busy.
Answer: If, on average, one telephone out of ten is busy, then the probability of a telephone number being busy (our success, \( p \)) is \( \frac{1}{10} \). The probability of a telephone number not being busy (our failure, \( q \)) is \( 1 - p = 1 - \frac{1}{10} = \frac{9}{10} \).
Seven telephone numbers are randomly selected and called, so the number of trials \( n=7 \). We want to find the probability that exactly three of these numbers will be busy. This is a binomial distribution problem.
The formula for binomial probability is \( P(X=r) = { }^n C_r p^r q^{n-r} \).
For \( r=3 \), \( n=7 \), \( p=\frac{1}{10} \), \( q=\frac{9}{10} \):
\[ P(X=3) = { }^7 C_3 \left(\frac{1}{10}\right)^3 \left(\frac{9}{10}\right)^{7-3} \] \[ \implies P(X=3) = { }^7 C_3 \left(\frac{1}{10}\right)^3 \left(\frac{9}{10}\right)^4 \] \[ \implies P(X=3) = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{1}{1000} \times \frac{6561}{10000} \] \[ \implies P(X=3) = 35 \times \frac{6561}{10000000} \] \[ \implies P(X=3) = \frac{229635}{10000000} \] \[ \implies P(X=3) = 0.0229635 \]In simple words: If 1 out of 10 phones is usually busy, then the chance of a phone being busy is \( \frac{1}{10} \), and not busy is \( \frac{9}{10} \). We call 7 phone numbers. We want to find the exact chance that 3 of these 7 calls will reach a busy line. Using the binomial formula, we calculate this chance to be 0.0229635. This means it is quite an unlikely event.

🎯 Exam Tip: Write down all intermediate steps when calculating combinations and powers, especially for larger numbers, to minimize calculation errors.

 

Question 13. In a binomial distribution, the sum of its mean and the variance is 1.8 . Find the probability of two successes if the event was conducted 5 times.
Answer: Let \( M \) be the mean and \( V \) be the variance of the binomial distribution. We know \( M = np \) and \( V = npq \).
Given: Sum of mean and variance \( M + V = 1.8 \).
\[ np + npq = 1.8 \] \[ \implies np(1+q) = 1.8 \] We are given that the event was conducted 5 times, so \( n=5 \). Substituting \( n=5 \):
\[ 5p(1+q) = 1.8 \] We also know that \( q = 1-p \), so \( 1+q = 1+(1-p) = 2-p \). Substituting this:
\[ 5p(2-p) = 1.8 \] \[ \implies 10p - 5p^2 = 1.8 \] Rearranging into a quadratic equation:
\[ 5p^2 - 10p + 1.8 = 0 \] To remove the decimal, multiply by 10:
\[ 50p^2 - 100p + 18 = 0 \] Divide by 2:
\[ 25p^2 - 50p + 9 = 0 \] Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ p = \frac{50 \pm \sqrt{(-50)^2 - 4(25)(9)}}{2(25)} \] \[ p = \frac{50 \pm \sqrt{2500 - 900}}{50} \] \[ p = \frac{50 \pm \sqrt{1600}}{50} \] \[ p = \frac{50 \pm 40}{50} \] This gives two possible values for \( p \):
\( p_1 = \frac{50 + 40}{50} = \frac{90}{50} = \frac{9}{5} \)
\( p_2 = \frac{50 - 40}{50} = \frac{10}{50} = \frac{1}{5} \)
Since \( p \) is a probability, it must be between 0 and 1. So, \( p = \frac{9}{5} \) is not a valid value. Therefore, \( p = \frac{1}{5} \).
Now, find \( q \):
\[ q = 1 - p = 1 - \frac{1}{5} = \frac{4}{5} \] We have the parameters: \( n=5, p=\frac{1}{5}, q=\frac{4}{5} \). We need to find the probability of two successes (\( X=2 \)).
Using the binomial probability formula \( P(X=r) = { }^n C_r p^r q^{n-r} \):
\[ P(X=2) = { }^5 C_2 \left(\frac{1}{5}\right)^2 \left(\frac{4}{5}\right)^{5-2} \] \[ \implies P(X=2) = { }^5 C_2 \left(\frac{1}{5}\right)^2 \left(\frac{4}{5}\right)^3 \] \[ \implies P(X=2) = 10 \times \frac{1}{25} \times \frac{64}{125} \] \[ \implies P(X=2) = \frac{10 \times 64}{25 \times 125} = \frac{640}{3125} \] \[ \implies P(X=2) = \frac{128}{625} \]In simple words: We are told that the sum of the mean and variance of a binomial distribution is 1.8, and the event happened 5 times (\( n=5 \)). We use this information to set up an equation and solve for \( p \), the probability of success. After solving a quadratic equation, we find \( p = \frac{1}{5} \) (because probability must be between 0 and 1). This means \( q \) (probability of failure) is \( \frac{4}{5} \). Now, with \( n=5, p=\frac{1}{5}, q=\frac{4}{5} \), we calculate the probability of getting exactly two successes using the binomial formula. The chance is \( \frac{128}{625} \).

🎯 Exam Tip: Be careful with algebraic manipulations, especially when solving quadratic equations for \( p \) or \( q \). Always verify that your resulting probabilities are valid (between 0 and 1).

 

Question 14. Eight coins are thrown simultaneously. (i) Show that the probability of getting at least 6 heads is \( \frac{37}{256} \). (ii) What is the probability of getting at least 3 heads?
Answer: When a coin is thrown, the probability of getting a head (success, \( p \)) is \( \frac{1}{2} \). The probability of not getting a head (failure, \( q \)) is \( 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \).
Eight coins are thrown simultaneously, so the number of trials \( n=8 \). This is a binomial distribution problem.
The general formula for binomial probability is \( P(X=r) = { }^n C_r p^r q^{n-r} \).
For \( n=8 \), \( p=\frac{1}{2} \), \( q=\frac{1}{2} \), the formula simplifies to:
\[ P(X=r) = { }^8 C_r \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{8-r} = { }^8 C_r \left(\frac{1}{2}\right)^8 = \frac{{ }^8 C_r}{256} \] (i) We need to show that the probability of getting at least 6 heads is \( \frac{37}{256} \). "At least 6 heads" means \( P(X \ge 6) = P(X=6) + P(X=7) + P(X=8) \).
Calculate \( P(X=6) \):
\[ P(X=6) = { }^8 C_6 \left(\frac{1}{2}\right)^8 = 28 \times \frac{1}{256} = \frac{28}{256} \] Calculate \( P(X=7) \):
\[ P(X=7) = { }^8 C_7 \left(\frac{1}{2}\right)^8 = 8 \times \frac{1}{256} = \frac{8}{256} \] Calculate \( P(X=8) \):
\[ P(X=8) = { }^8 C_8 \left(\frac{1}{2}\right)^8 = 1 \times \frac{1}{256} = \frac{1}{256} \] Adding these probabilities:
\[ P(X \ge 6) = \frac{28}{256} + \frac{8}{256} + \frac{1}{256} = \frac{37}{256} \] This matches the given value.
(ii) We need to find the probability of getting at least 3 heads, which means \( P(X \ge 3) \). It is simpler to calculate this using the complementary probability rule: \( P(X \ge 3) = 1 - P(X < 3) = 1 - (P(X=0) + P(X=1) + P(X=2)) \).
Calculate \( P(X=0) \):
\[ P(X=0) = { }^8 C_0 \left(\frac{1}{2}\right)^8 = 1 \times \frac{1}{256} = \frac{1}{256} \] Calculate \( P(X=1) \):
\[ P(X=1) = { }^8 C_1 \left(\frac{1}{2}\right)^8 = 8 \times \frac{1}{256} = \frac{8}{256} \] Calculate \( P(X=2) \):
\[ P(X=2) = { }^8 C_2 \left(\frac{1}{2}\right)^8 = 28 \times \frac{1}{256} = \frac{28}{256} \] Adding these probabilities for \( P(X < 3) \):
\[ P(X < 3) = \frac{1}{256} + \frac{8}{256} + \frac{28}{256} = \frac{37}{256} \] Now, calculate \( P(X \ge 3) \):
\[ P(X \ge 3) = 1 - \frac{37}{256} \] \[ \implies P(X \ge 3) = \frac{256 - 37}{256} \] \[ \implies P(X \ge 3) = \frac{219}{256} \]In simple words: When flipping 8 coins, the chance of getting a head is \( \frac{1}{2} \), and a tail is also \( \frac{1}{2} \). We use the binomial formula for \( n=8 \).
(i) To check if the probability of at least 6 heads is \( \frac{37}{256} \), we add up the chances of getting exactly 6, 7, and 8 heads. These are \( \frac{28}{256}, \frac{8}{256} \), and \( \frac{1}{256} \) respectively. Their sum is indeed \( \frac{37}{256} \).
(ii) To find the chance of at least 3 heads, it's easier to find the chance of getting fewer than 3 heads (0, 1, or 2 heads) and subtract that from 1. The chances for 0, 1, and 2 heads are \( \frac{1}{256}, \frac{8}{256} \), and \( \frac{28}{256} \). Their sum is \( \frac{37}{256} \). So, \( 1 - \frac{37}{256} = \frac{219}{256} \) is the probability of at least 3 heads.

🎯 Exam Tip: When asked to "show that" a probability is a certain value, you must perform the full calculation and arrive at that value. The complementary rule is very useful for "at least" or "at most" questions, simplifying the work.

 

Question 15. The mean and variance of a binomial distribution are 4 and 2 respectively. Find the probability of at least 6 successes.
Answer: We are given the mean of the binomial distribution as 4 and the variance as 2.
Mean \( np = 4 \) (Equation 1)
Variance \( npq = 2 \) (Equation 2)
To find \( q \), we divide Equation 2 by Equation 1:
\[ q = \frac{npq}{np} = \frac{2}{4} = \frac{1}{2} \] Now, we find \( p \) using \( p = 1-q \):
\[ p = 1 - \frac{1}{2} = \frac{1}{2} \] Substitute \( p = \frac{1}{2} \) back into Equation 1 to find \( n \):
\[ n \times \frac{1}{2} = 4 \] \[ \implies n = 8 \] So, the binomial distribution has parameters \( n=8 \) and \( p=\frac{1}{2} \). The probability mass function is \( P(X=r) = { }^n C_r p^r q^{n-r} \).
For \( n=8 \), \( p=\frac{1}{2} \), \( q=\frac{1}{2} \), the formula simplifies to:
\[ P(X=r) = { }^8 C_r \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{8-r} = { }^8 C_r \left(\frac{1}{2}\right)^8 = \frac{{ }^8 C_r}{256} \] We need to find the probability of at least 6 successes, which means \( P(X \ge 6) \). This is the sum of probabilities for 6, 7, or 8 successes: \( P(X \ge 6) = P(X=6) + P(X=7) + P(X=8) \).
Calculate \( P(X=6) \):
\[ P(X=6) = { }^8 C_6 \left(\frac{1}{2}\right)^8 = 28 \times \frac{1}{256} = \frac{28}{256} \] Calculate \( P(X=7) \):
\[ P(X=7) = { }^8 C_7 \left(\frac{1}{2}\right)^8 = 8 \times \frac{1}{256} = \frac{8}{256} \] Calculate \( P(X=8) \):
\[ P(X=8) = { }^8 C_8 \left(\frac{1}{2}\right)^8 = 1 \times \frac{1}{256} = \frac{1}{256} \] Adding these probabilities:
\[ P(X \ge 6) = \frac{28}{256} + \frac{8}{256} + \frac{1}{256} = \frac{37}{256} \]In simple words: First, we use the given mean (4) and variance (2) to find the binomial distribution's basic numbers: \( n \) (number of trials) and \( p \) (probability of success). By dividing variance by mean, we get \( q = \frac{1}{2} \), so \( p \) is also \( \frac{1}{2} \). Then, using the mean, we find \( n=8 \). So, we have 8 trials with a \( \frac{1}{2} \) chance of success each time. Now, we want the chance of getting at least 6 successes. This means we add the chances of getting exactly 6, 7, or 8 successes. After calculating these, the total chance is \( \frac{37}{256} \).

🎯 Exam Tip: Always make sure to find all the parameters (\( n, p, q \)) of the binomial distribution first before attempting to calculate specific probabilities.

 

Question 16. Five bad eggs aretaixed with 10 good ones. If three eggs are drawn one by one with replacement, find the probability distribution of the number of good eggs drawn.
Answer: We have 5 bad eggs and 10 good eggs, making a total of \( 5+10=15 \) eggs. When an egg is drawn, the probability of it being a good egg (our success, \( p \)) is \( \frac{\text{Number of good eggs}}{\text{Total number of eggs}} = \frac{10}{15} = \frac{2}{3} \). The probability of it being a bad egg (our failure, \( q \)) is \( 1 - p = 1 - \frac{2}{3} = \frac{1}{3} \). Three eggs are drawn one by one with replacement, so the number of trials \( n=3 \). Let \( X \) be the random variable representing the number of good eggs drawn. \( X \) can take values 0, 1, 2, or 3. This is a binomial distribution problem.
The general formula for binomial probability is \( P(X=r) = { }^n C_r p^r q^{n-r} \).
For \( n=3 \), \( p=\frac{2}{3} \), \( q=\frac{1}{3} \), the formula is \( P(X=r) = { }^3 C_r \left(\frac{2}{3}\right)^r \left(\frac{1}{3}\right)^{3-r} \).
Calculate \( P(X=0) \): (0 good eggs, 3 bad eggs)
\[ P(X=0) = { }^3 C_0 \left(\frac{2}{3}\right)^0 \left(\frac{1}{3}\right)^3 = 1 \times 1 \times \frac{1}{27} = \frac{1}{27} \] Calculate \( P(X=1) \): (1 good egg, 2 bad eggs)
\[ P(X=1) = { }^3 C_1 \left(\frac{2}{3}\right)^1 \left(\frac{1}{3}\right)^2 = 3 \times \frac{2}{3} \times \frac{1}{9} = \frac{6}{27} = \frac{2}{9} \] Calculate \( P(X=2) \): (2 good eggs, 1 bad egg)
\[ P(X=2) = { }^3 C_2 \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^1 = 3 \times \frac{4}{9} \times \frac{1}{3} = \frac{12}{27} = \frac{4}{9} \] Calculate \( P(X=3) \): (3 good eggs, 0 bad eggs)
\[ P(X=3) = { }^3 C_3 \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^0 = 1 \times \frac{8}{27} \times 1 = \frac{8}{27} \] The probability distribution of \( X \) is shown in the table below:

In simple words: We have 10 good eggs and 5 bad eggs, totaling 15 eggs. The chance of picking a good egg is \( \frac{10}{15} = \frac{2}{3} \), and a bad egg is \( \frac{1}{3} \). We draw 3 eggs one by one, putting each back after drawing, which keeps the chances the same for each draw. We want to find the chances of getting 0, 1, 2, or 3 good eggs. Using the binomial formula: for 0 good eggs, it's \( \frac{1}{27} \); for 1, it's \( \frac{2}{9} \); for 2, it's \( \frac{4}{9} \); and for 3, it's \( \frac{8}{27} \). These results form the probability distribution.

🎯 Exam Tip: Drawing "with replacement" means trials are independent, making it a binomial distribution. Drawing "without replacement" indicates a hypergeometric distribution, which uses combinations for joint selections.

 

Question 17. A box contains 4 red and 5 black marbles. Find the probability distribution of the red marbles in a random draw of three marbles. Also, find the mean and standard deviation of the distribution.
Answer: We have 4 red marbles and 5 black marbles, making a total of \( 4+5=9 \) marbles. Three marbles are drawn randomly without replacement.
Let \( X \) be the random variable representing the number of red marbles drawn. \( X \) can take values 0, 1, 2, or 3 (since we draw 3 marbles and there are 4 red ones available).
The total number of ways to draw 3 marbles from 9 is \( { }^9 C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \).
Calculate \( P(X=0) \): (0 red marbles and 3 black marbles)
This means selecting 0 from 4 red marbles and 3 from 5 black marbles.
\[ P(X=0) = \frac{{ }^4 C_0 \times { }^5 C_3}{{ }^9 C_3} = \frac{1 \times 10}{84} = \frac{10}{84} = \frac{5}{42} \] Calculate \( P(X=1) \): (1 red marble and 2 black marbles)
This means selecting 1 from 4 red marbles and 2 from 5 black marbles.
\[ P(X=1) = \frac{{ }^4 C_1 \times { }^5 C_2}{{ }^9 C_3} = \frac{4 \times 10}{84} = \frac{40}{84} = \frac{10}{21} \] Calculate \( P(X=2) \): (2 red marbles and 1 black marble)
This means selecting 2 from 4 red marbles and 1 from 5 black marbles.
\[ P(X=2) = \frac{{ }^4 C_2 \times { }^5 C_1}{{ }^9 C_3} = \frac{6 \times 5}{84} = \frac{30}{84} = \frac{5}{14} \] Calculate \( P(X=3) \): (3 red marbles and 0 black marbles)
This means selecting 3 from 4 red marbles and 0 from 5 black marbles.
\[ P(X=3) = \frac{{ }^4 C_3 \times { }^5 C_0}{{ }^9 C_3} = \frac{4 \times 1}{84} = \frac{4}{84} = \frac{1}{21} \] The probability distribution of \( X \) is given in the table below:

Now, we find the mean \( (\mu) \) and standard deviation \( (\sigma) \) of this distribution. For a discrete probability distribution: Mean \( \mu = \sum X_i P(X_i) \) Variance \( \sigma^2 = \sum X_i^2 P(X_i) - \mu^2 \) Let's compute the necessary sums using a table:Calculate Mean \( \mu \):
\[ \mu = 0 + \frac{10}{21} + \frac{5}{7} + \frac{1}{7} \] \[ \mu = \frac{10}{21} + \frac{15}{21} + \frac{3}{21} \] \[ \implies \mu = \frac{10+15+3}{21} = \frac{28}{21} = \frac{4}{3} \] Calculate \( \sum X_i^2 P(X_i) \):
\[ \sum X_i^2 P(X_i) = 0 + \frac{10}{21} + \frac{10}{7} + \frac{3}{7} \] \[ \sum X_i^2 P(X_i) = \frac{10}{21} + \frac{30}{21} + \frac{9}{21} \] \[ \implies \sum X_i^2 P(X_i) = \frac{10+30+9}{21} = \frac{49}{21} = \frac{7}{3} \] Calculate Variance \( \sigma^2 \):
\[ \sigma^2 = \sum X_i^2 P(X_i) - \mu^2 \] \[ \implies \sigma^2 = \frac{7}{3} - \left(\frac{4}{3}\right)^2 \] \[ \implies \sigma^2 = \frac{7}{3} - \frac{16}{9} \] \[ \implies \sigma^2 = \frac{21}{9} - \frac{16}{9} = \frac{5}{9} \] Calculate Standard Deviation \( \sigma \):
\[ \sigma = \sqrt{\sigma^2} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]In simple words: We have a total of 9 marbles, with 4 red and 5 black. We pick 3 marbles without putting any back. Let \( X \) be the number of red marbles we pick (it can be 0, 1, 2, or 3). We find the probability for each value of \( X \) by calculating combinations. For example, the chance of picking 0 red marbles is \( \frac{5}{42} \), 1 red marble is \( \frac{10}{21} \), and so on. These probabilities make up the distribution.
Then, we find the mean (average) number of red marbles we expect to pick, which is \( \frac{4}{3} \). We also find the variance, which tells us how spread out the numbers are, and it is \( \frac{5}{9} \). Finally, the standard deviation, which is the square root of the variance, is \( \frac{\sqrt{5}}{3} \).

🎯 Exam Tip: For problems involving drawing without replacement, always use combinations (\( { }^n C_r \)) to calculate probabilities. Remember to calculate mean and variance using the respective formulas for discrete probability distributions.

 

Question 18. The probability that a bulb produced by a factory will fuse in 100 days of use is 0.05 . Find the probability that out of 5 such bulbs after 100 days of use: (i) None (ii) Not more than one (iii) More than one (iv) At least one will fuse.
Answer: The probability that a bulb will fuse (our success, \( p \)) is 0.05. This can be written as \( \frac{5}{100} = \frac{1}{20} \). The probability that a bulb will not fuse (our failure, \( q \)) is \( 1 - 0.05 = 0.95 \), which is \( \frac{95}{100} = \frac{19}{20} \).
We are considering 5 bulbs, so the number of trials \( n=5 \). Let \( X \) be the number of bulbs that fuse. This situation follows a binomial distribution.
The general formula for binomial probability is \( P(X=r) = { }^n C_r p^r q^{n-r} = { }^5 C_r \left(\frac{1}{20}\right)^r \left(\frac{19}{20}\right)^{5-r} \).
(i) Probability that none of the bulbs will fuse: \( P(X=0) \).
\[ P(X=0) = { }^5 C_0 \left(\frac{1}{20}\right)^0 \left(\frac{19}{20}\right)^{5-0} = 1 \times 1 \times \left(\frac{19}{20}\right)^5 = \left(\frac{19}{20}\right)^5 \] (ii) Probability that not more than one bulb will fuse: \( P(X \le 1) \). This means \( P(X=0) + P(X=1) \).
We already have \( P(X=0) = \left(\frac{19}{20}\right)^5 \).
Now calculate \( P(X=1) \):
\[ P(X=1) = { }^5 C_1 \left(\frac{1}{20}\right)^1 \left(\frac{19}{20}\right)^{5-1} = 5 \times \frac{1}{20} \times \left(\frac{19}{20}\right)^4 \] \[ \implies P(X=1) = \frac{5}{20} \left(\frac{19}{20}\right)^4 = \frac{1}{4} \left(\frac{19}{20}\right)^4 \] Adding \( P(X=0) \) and \( P(X=1) \):
\[ P(X \le 1) = \left(\frac{19}{20}\right)^5 + \frac{1}{4} \left(\frac{19}{20}\right)^4 \] Factor out \( \left(\frac{19}{20}\right)^4 \):
\[ P(X \le 1) = \left(\frac{19}{20}\right)^4 \left[ \frac{19}{20} + \frac{1}{4} \right] \] \[ \implies P(X \le 1) = \left(\frac{19}{20}\right)^4 \left[ \frac{19}{20} + \frac{5}{20} \right] \] \[ \implies P(X \le 1) = \left(\frac{19}{20}\right)^4 \left[ \frac{24}{20} \right] = \frac{6}{5} \left(\frac{19}{20}\right)^4 \] (iii) Probability that more than one bulb will fuse: \( P(X > 1) \). This is the complement of "not more than one," so \( P(X > 1) = 1 - P(X \le 1) \).
\[ P(X > 1) = 1 - \frac{6}{5} \left(\frac{19}{20}\right)^4 \] (iv) Probability that at least one bulb will fuse: \( P(X \ge 1) \). This is the complement of "none will fuse," so \( P(X \ge 1) = 1 - P(X=0) \).
\[ P(X \ge 1) = 1 - \left(\frac{19}{20}\right)^5 \]In simple words: The chance a bulb fuses is 0.05 (\( \frac{1}{20} \)), and the chance it doesn't fuse is 0.95 (\( \frac{19}{20} \)). We look at 5 bulbs. This is a binomial problem.
(i) For no bulbs to fuse, the probability is \( (\frac{19}{20})^5 \).
(ii) For not more than one bulb to fuse, we add the chances of 0 bulbs fusing and 1 bulb fusing. This equals \( \frac{6}{5} (\frac{19}{20})^4 \).
(iii) For more than one bulb to fuse, we subtract the result from part (ii) from 1, because it's the opposite event. So, \( 1 - \frac{6}{5} (\frac{19}{20})^4 \).
(iv) For at least one bulb to fuse, we subtract the chance of zero bulbs fusing from 1. So, \( 1 - (\frac{19}{20})^5 \).

🎯 Exam Tip: Carefully read the wording "none," "not more than one," "more than one," and "at least one" as they dictate which combinations of binomial probabilities you need to sum or subtract from 1.

 

Question 19. If the sum and product of the mean and variance of a Binomial distribution are 1.8 and 0.8 respectively, find the probability distribution and the probability of at least one success.
Answer: Let \( M \) be the mean and \( V \) be the variance of the binomial distribution. We know \( M = np \) and \( V = npq \).
Given: Sum of mean and variance \( M + V = 1.8 \).
\[ np + npq = 1.8 \] \[ \implies np(1+q) = 1.8 \] (Equation 1)
Given: Product of mean and variance \( M \times V = 0.8 \).
\[ np \times npq = 0.8 \] \[ \implies n^2 p^2 q = 0.8 \] (Equation 2)
Square both sides of Equation 1:
\[ (np(1+q))^2 = (1.8)^2 \] \[ \implies n^2 p^2 (1+q)^2 = 3.24 \] (Equation 3)
Divide Equation 3 by Equation 2:
\[ \frac{n^2 p^2 (1+q)^2}{n^2 p^2 q} = \frac{3.24}{0.8} \] \[ \implies \frac{(1+q)^2}{q} = \frac{324}{80} \] \[ \implies \frac{(1+q)^2}{q} = \frac{81}{20} \] Now, solve for \( q \):
\[ 20(1+q)^2 = 81q \] \[ \implies 20(1 + 2q + q^2) = 81q \] \[ \implies 20 + 40q + 20q^2 = 81q \] \[ \implies 20q^2 - 41q + 20 = 0 \] This is a quadratic equation. We can solve it using the quadratic formula \( q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ q = \frac{41 \pm \sqrt{(-41)^2 - 4(20)(20)}}{2(20)} \] \[ q = \frac{41 \pm \sqrt{1681 - 1600}}{40} \] \[ q = \frac{41 \pm \sqrt{81}}{40} \] \[ q = \frac{41 \pm 9}{40} \] This gives two possible values for \( q \):
\( q_1 = \frac{41 + 9}{40} = \frac{50}{40} = \frac{5}{4} \)
\( q_2 = \frac{41 - 9}{40} = \frac{32}{40} = \frac{4}{5} \)
Since \( q \) is a probability, its value must be between 0 and 1. So, \( q = \frac{5}{4} \) is not possible. Therefore, \( q = \frac{4}{5} \).
Now, find \( p \) using \( p = 1-q \):
\[ p = 1 - \frac{4}{5} = \frac{1}{5} \] Substitute \( p = \frac{1}{5} \) and \( q = \frac{4}{5} \) into Equation 1:
\[ n \times \frac{1}{5} (1 + \frac{4}{5}) = 1.8 \] \[ \implies n \times \frac{1}{5} \times \frac{9}{5} = 1.8 \] \[ \implies \frac{9n}{25} = 1.8 \] \[ \implies 9n = 1.8 \times 25 \] \[ \implies 9n = 45 \] \[ \implies n = \frac{45}{9} \] \[ \implies n = 5 \] The probability distribution is defined by its parameters \( n=5, p=\frac{1}{5}, q=\frac{4}{5} \). We can write it as \( \left(\frac{4}{5} + \frac{1}{5}\right)^5 \).
Next, we find the probability of at least one success: \( P(X \ge 1) \). It is simpler to calculate this using the complementary probability rule: \( P(X \ge 1) = 1 - P(X=0) \).
Calculate \( P(X=0) \):
\[ P(X=0) = { }^n C_0 p^0 q^{n-0} = { }^5 C_0 \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^5 \] \[ \implies P(X=0) = 1 \times 1 \times \left(\frac{4}{5}\right)^5 = \left(\frac{4}{5}\right)^5 \] Now, calculate \( P(X \ge 1) \):
\[ P(X \ge 1) = 1 - P(X=0) \] \[ \implies P(X \ge 1) = 1 - \left(\frac{4}{5}\right)^5 \]In simple words: We are given the sum (1.8) and product (0.8) of the mean and variance of a binomial distribution. We use these two pieces of information to create mathematical equations and solve for \( n, p \), and \( q \). After solving, we find that \( n=5 \), \( p=\frac{1}{5} \), and \( q=\frac{4}{5} \). This defines our distribution. Then, we need to find the chance of at least one success. This is easily found by taking 1 minus the chance of having zero successes. The chance of zero successes is \( (\frac{4}{5})^5 \). So, the chance of at least one success is \( 1 - (\frac{4}{5})^5 \).

🎯 Exam Tip: Problems that provide mean and variance require solving a system of equations to determine \( n, p, \) and \( q \). Always double-check that derived probabilities (\( p, q \)) are within the \( [0, 1] \) range.

 

Question 20. On dialling certain telephone numbers, assume that on an average one telephone number out of five is busy. Ten telephone numbers are randomly selected and dialled. Find the probability that at least three of them will be busy.
Answer: Let \( p \) be the probability that a telephone number is busy. We are given that \( p = \frac{1}{5} \).
So, \( q \), the probability that a telephone number is not busy, is \( 1 - p = 1 - \frac{1}{5} = \frac{4}{5} \).
Here, \( n = 10 \) is the number of telephone numbers selected. Let \( X \) be the random variable representing the number of busy telephone numbers out of these 10.
This is a binomial distribution with parameters \( n = 10 \) and \( p = \frac{1}{5} \).
The probability mass function is \( P(X=r) = {}^{10}C_r p^r q^{n-r} = {}^{10}C_r \left(\frac{1}{5}\right)^r \left(\frac{4}{5}\right)^{10-r} \).
We need to find the probability that at least three numbers will be busy, which is \( P(X \ge 3) \).
This can be found as \( 1 - P(X < 3) \), which means \( 1 - [P(X=0) + P(X=1) + P(X=2)] \).
\( P(X=0) = {}^{10}C_0 \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^{10} = 1 \cdot 1 \cdot \left(\frac{4}{5}\right)^{10} = \left(\frac{4}{5}\right)^{10} \).
\( P(X=1) = {}^{10}C_1 \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^9 = 10 \cdot \frac{1}{5} \cdot \left(\frac{4}{5}\right)^9 = 2 \left(\frac{4}{5}\right)^9 \).
\( P(X=2) = {}^{10}C_2 \left(\frac{1}{5}\right)^2 \left(\frac{4}{5}\right)^8 = \frac{10 \times 9}{2 \times 1} \cdot \frac{1}{25} \cdot \left(\frac{4}{5}\right)^8 = 45 \cdot \frac{1}{25} \cdot \left(\frac{4}{5}\right)^8 = \frac{9}{5} \left(\frac{4}{5}\right)^8 \).
Now, let's substitute these values back:
\( P(X \ge 3) = 1 - \left[ \left(\frac{4}{5}\right)^{10} + 2 \left(\frac{4}{5}\right)^9 + \frac{9}{5} \left(\frac{4}{5}\right)^8 \right] \).
\( = 1 - \left(\frac{4}{5}\right)^8 \left[ \left(\frac{4}{5}\right)^2 + 2 \left(\frac{4}{5}\right) + \frac{9}{5} \right] \)
\( = 1 - \left(\frac{4}{5}\right)^8 \left[ \frac{16}{25} + \frac{8}{5} + \frac{9}{5} \right] \)
\( = 1 - \left(\frac{4}{5}\right)^8 \left[ \frac{16}{25} + \frac{40}{25} + \frac{45}{25} \right] \)
\( = 1 - \left(\frac{4}{5}\right)^8 \left[ \frac{16+40+45}{25} \right] \)
\( = 1 - \left(\frac{4}{5}\right)^8 \left[ \frac{101}{25} \right] \)
\( = 1 - \frac{101 \times 4^8}{25 \times 5^8} = 1 - \frac{101 \times 65536}{25 \times 390625} \)
\( = 1 - \frac{6619136}{9765625} = \frac{9765625 - 6619136}{9765625} = \frac{3146489}{9765625} \approx 0.3221 \).
In simple words: First, we figure out the chance of a phone being busy (p) and not busy (q). Then, we use the binomial distribution formula to calculate the chance that 0, 1, or 2 phones are busy out of 10. Finally, we subtract this from 1 to find the chance that at least 3 phones are busy. Binomial distributions are very common for a fixed number of trials with two outcomes.

🎯 Exam Tip: When asked for "at least" a certain number of successes, it's often easier to calculate the probability of "less than" that number and subtract from 1. This avoids summing many individual probabilities. Always ensure your \( p \) and \( q \) probabilities add up to 1.

 

Question 21. If the occurrence of an odd number is a single die is considered a success, find the probability of maximum successes.
Answer: Let \( p \) be the probability of success, which is getting an odd number when a single die is thrown. The odd numbers on a die are 1, 3, 5. There are 3 odd numbers out of 6 total possibilities.
So, \( p = \frac{3}{6} = \frac{1}{2} \).
The probability of failure, \( q \), is \( 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \).
The problem statement "find the probability of maximum successes" is interpreted here as the maximum possible number of successes given the implied number of trials in the solution, which calculates \( P(X \le 3) \) for \( n=5 \) throws.
So, let \( n=5 \) be the number of trials (throws of the die). Let \( X \) be the random variable representing the number of successes (odd numbers) in 5 throws.
This is a binomial variate with parameters \( n=5 \) and \( p=\frac{1}{2} \).
The probability mass function is \( P(X=r) = {}^nC_r p^r q^{n-r} = {}^5C_r \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{5-r} = {}^5C_r \left(\frac{1}{2}\right)^5 \).
Assuming "probability of maximum successes" refers to the probability of getting "at most 3 successes", we calculate \( P(X \le 3) \).
\( P(X \le 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) \)
\( P(X=0) = {}^5C_0 \left(\frac{1}{2}\right)^5 = 1 \cdot \frac{1}{32} = \frac{1}{32} \).
\( P(X=1) = {}^5C_1 \left(\frac{1}{2}\right)^5 = 5 \cdot \frac{1}{32} = \frac{5}{32} \).
\( P(X=2) = {}^5C_2 \left(\frac{1}{2}\right)^5 = \frac{5 \times 4}{2 \times 1} \cdot \frac{1}{32} = 10 \cdot \frac{1}{32} = \frac{10}{32} \).
\( P(X=3) = {}^5C_3 \left(\frac{1}{2}\right)^5 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} \cdot \frac{1}{32} = 10 \cdot \frac{1}{32} = \frac{10}{32} \).
So, \( P(X \le 3) = \frac{1}{32} + \frac{5}{32} + \frac{10}{32} + \frac{10}{32} = \frac{1+5+10+10}{32} = \frac{26}{32} = \frac{13}{16} \).
In simple words: When you roll a die, the chance of getting an odd number (like 1, 3, or 5) is one-half. If you roll the die five times, we want to find the total chance of getting up to three odd numbers. We calculate the chance of getting 0, 1, 2, or 3 odd numbers separately and add them up. This sum tells us the probability of having "at most 3 successes".

🎯 Exam Tip: Always define your success probability \( p \) and the number of trials \( n \) clearly. For binomial distributions, remember that \( {}^nC_r = {}^nC_{n-r} \), which can simplify calculations for symmetric probabilities like \( p=0.5 \).

 

Question 22. The difference between mean and variance of a binomial distribution is 1 and the difference of their squares is 11. Find the distribution.
Answer: Let the mean of the binomial distribution be \( \mu = np \) and the variance be \( \sigma^2 = npq \).
According to the given conditions:
1. The difference between mean and variance is 1:
\( np - npq = 1 \)
\( \implies np(1 - q) = 1 \)
Since \( 1 - q = p \), we have
\( np \cdot p = 1 \)
\( \implies np^2 = 1 \) .........(1)
2. The difference of their squares is 11:
\( (np)^2 - (npq)^2 = 11 \)
\( \implies (np)^2 - (npq)^2 = 11 \)
\( \implies n^2p^2 - n^2p^2q^2 = 11 \)
\( \implies n^2p^2(1 - q^2) = 11 \)
Since \( 1 - q^2 = (1-q)(1+q) = p(1+q) \), we have
\( n^2p^2 \cdot p(1+q) = 11 \)
\( \implies n^2p^3(1+q) = 11 \) .........(2)

From equation (1), \( n = \frac{1}{p^2} \). Substitute this into equation (2):
\( \left(\frac{1}{p^2}\right)^2 p^3 (1+q) = 11 \)
\( \implies \frac{1}{p^4} p^3 (1+q) = 11 \)
\( \implies \frac{1}{p} (1+q) = 11 \)
\( \implies 1+q = 11p \)
Since \( p+q=1 \), we have \( q=1-p \). Substitute this into the equation:
\( 1 + (1-p) = 11p \)
\( \implies 2 - p = 11p \)
\( \implies 2 = 12p \)
\( \implies p = \frac{2}{12} = \frac{1}{6} \).
Now find \( q \):
\( q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6} \).
Now find \( n \) using \( np^2 = 1 \):
\( n \left(\frac{1}{6}\right)^2 = 1 \)
\( n \cdot \frac{1}{36} = 1 \)
\( \implies n = 36 \).
Thus, the required binomial distribution is \( (q+p)^n \), which is \( \left(\frac{5}{6} + \frac{1}{6}\right)^{36} \).
In simple words: We are given two conditions involving the mean (average) and variance (spread) of a binomial distribution. We write these conditions as equations using \( n \), \( p \), and \( q \). By solving these equations, we find the values for \( n \) (number of trials), \( p \) (probability of success), and \( q \) (probability of failure). This gives us the full binomial distribution. Remember, \( np \) is the mean and \( npq \) is the variance, and \( p+q \) always equals 1.

🎯 Exam Tip: For problems involving mean and variance of binomial distributions, always start by writing down the formulas for mean \( (np) \) and variance \( (npq) \). Use the relation \( q = 1-p \) to reduce the number of variables and simplify the equations. Squaring equations can be a useful algebraic technique here.