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

 

Question 1. In a single throw of a die, if X denotes the number on its upper face, find the mean of X.
Answer: Here, \( X \) denotes the number on the upper face when a die is thrown. The possible outcomes are 1, 2, 3, 4, 5, and 6. Each outcome has an equal probability.
So, \( P(X = 1) = P(X = 2) = P(X = 3) = P(X = 4) = P(X = 5) = P(X = 6) = \frac{1}{6} \).
The mean, denoted by \( \mu \), is calculated as the sum of each outcome multiplied by its probability:
\( \mu = \sum_{i=1}^6 X_i P(X_i) \)
\( \mu = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + 3 \times \frac{1}{6} + 4 \times \frac{1}{6} + 5 \times \frac{1}{6} + 6 \times \frac{1}{6} \)
\( \mu = \frac{1}{6} (1 + 2 + 3 + 4 + 5 + 6) \)
\( \mu = \frac{1}{6} (21) \)
\( \mu = \frac{21}{6} \)
\( \mu = \frac{7}{2} \)
\( \mu = 3.5 \)
In simple words: When you roll a die, each number from 1 to 6 has an equal chance of showing up. To find the average (mean) value you expect, you add up all the numbers (1 to 6) and divide by how many there are (6), which gives 3.5. This is like the average score you would get over many rolls.

🎯 Exam Tip: Remember that for a fair die, each face (1 to 6) has a probability of \( \frac{1}{6} \). The mean of a discrete random variable is found by summing the product of each possible value and its probability.

 

Question 2. A salesman wants to know the average number of units he sells per sales call. He checks his past sales records and comes up with the following probabilities.

The sum of \( P_i X_i \) gives the mean:
\( \sum P_i X_i = 0 + 0.20 + 0.20 + 0.15 + 1.20 + 1.00 = 2.75 \)
Therefore, the mean number of units the salesman sells per sales call is \( \mu = 2.75 \).
In simple words: To find the average sales, we multiply each possible number of sales (like 0, 1, 2) by how often it happens (its probability). Then, we add all those results together. This gives us 2.75 units, meaning on average, the salesman sells about 2.75 units per call.

🎯 Exam Tip: When calculating the mean of a discrete probability distribution, always set up a clear table to multiply each value of X by its probability P(X), and then sum the results. This structured approach helps prevent errors.

 

Question 3. Find \( \mu \) and variance for the following probability distribution.
(i)

(ii)(iii)(iv)
Answer:
(i) We need to find the mean (\( \mu \)) and variance (\( \sigma^2 \)) for the given distribution.
First, we create a table to calculate \( X P(X) \) and \( X^2 P(X) \).Now, we sum the columns:
\( \sum X P(X) = 0.8 + 3.0 = 3.8 \)
\( \sum X^2 P(X) = 1.6 + 1.5 = 3.1 \) (The source calculation of \( \sum X^2 P(X) = 16.6 \) appears to be a typo, \(1.6 + 1.5 = 3.1\). I will proceed with 3.1 for the sum, and show the variance calculation using 3.1. However, the next line in the source uses 16.6, which leads to 2.16. Let's re-evaluate: \( 2^2 \times 0.4 = 4 \times 0.4 = 1.6 \). \( 5^2 \times 0.6 = 25 \times 0.6 = 15.0 \). So \( \sum X^2 P(X) = 1.6 + 15.0 = 16.6 \). The table in the source had a typo for \(X^2 P(X)\) for X=5. It should be 15.0, not 1.5. I will correct the table to `15.0` to match the sum `16.6` given in the source).
The mean is \( \mu = \sum X P(X) = 3.8 \).
The variance is \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
\( \implies \sigma^2 = 16.6 - (3.8)^2 \)
\( \implies \sigma^2 = 16.6 - 14.44 \)
\( \implies \sigma^2 = 2.16 \)
(ii) We follow the same steps to find the mean and variance.
First, we calculate \( X P(X) \) and \( X^2 P(X) \).The mean is \( \mu = \sum X P(X) = 2.0 \).
The variance is \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
\( \implies \sigma^2 = 5 - (2)^2 \)
\( \implies \sigma^2 = 5 - 4 \)
\( \implies \sigma^2 = 1 \)
(iii) For this part, we again calculate the mean and variance.
First, we create a table for \( X P(X) \) and \( X^2 P(X) \).The mean is \( \mu = \sum X P(X) = -0.4 \).
The variance is \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
\( \implies \sigma^2 = 1.6 - (-0.4)^2 \)
\( \implies \sigma^2 = 1.6 - 0.16 \)
\( \implies \sigma^2 = 1.44 \)
(iv) We need to find the mean and variance in terms of \( p \).
First, we calculate \( X P(X) \) and \( X^2 P(X) \).The mean is \( \mu = \sum X P(X) = 4 - 3p \).
The variance is \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
\( \implies \sigma^2 = (16 - 19p) - (4 - 3p)^2 \)
\( \implies \sigma^2 = 16 - 19p - (16 - 24p + 9p^2) \)
\( \implies \sigma^2 = 16 - 19p - 16 + 24p - 9p^2 \)
\( \implies \sigma^2 = 5p - 9p^2 \)
In simple words: For each probability distribution given, we first make a table to help us calculate. We multiply each value of X by its chance of happening (P(X)) to get X P(X), and also by X squared to get X^2 P(X). Adding up the X P(X) column gives us the average (mean). Then, we use a special formula: add up all the X^2 P(X) values and subtract the mean squared. This gives us the variance, which tells us how spread out the numbers are.

🎯 Exam Tip: Always double-check your calculations for \( X P(X) \) and \( X^2 P(X) \) in the table before summing. A common mistake is arithmetic errors, especially with negative numbers or squaring terms involving variables like \( p \).

 

Question 4. A die is tossed thrice. If getting 'four' is considered a success, find the mean and variance of probability distribution of the number of successes.
Answer: When a die is tossed, the sample space is \( S = \{1, 2, 3, 4, 5, 6\} \).
If getting a 'four' is a success, then the probability of success \( p \) is the probability of getting 4.
\( p = P(\text{getting 4}) = \frac{1}{6} \).
The probability of failure \( q \) is \( 1 - p \).
\( q = 1 - \frac{1}{6} = \frac{5}{6} \).
Let \( X \) be the random variable representing the number of successes when the die is tossed thrice. So, \( X \) can take values 0, 1, 2, or 3.
Now we calculate the probability for each value of \( X \):
\( P(X = 0) = P(\text{no success in 3 tosses}) = qqq \)
\( \implies P(X = 0) = \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{125}{216} \)
\( P(X = 1) = P(\text{1 success in 3 tosses}) = pqq + qpq + qqp \)
\( \implies P(X = 1) = \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} + \frac{5}{6} \times \frac{1}{6} \times \frac{5}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} \)
\( \implies P(X = 1) = 3 \times \left( \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \right) = 3 \times \frac{25}{216} = \frac{75}{216} \)
\( P(X = 2) = P(\text{2 successes in 3 tosses}) = ppq + pqp + qpp \)
\( \implies P(X = 2) = \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} + \frac{1}{6} \times \frac{5}{6} \times \frac{1}{6} + \frac{5}{6} \times \frac{1}{6} \times \frac{1}{6} \)
\( \implies P(X = 2) = 3 \times \left( \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} \right) = 3 \times \frac{5}{216} = \frac{15}{216} \)
\( P(X = 3) = P(\text{all 3 successes}) = ppp \)
\( \implies P(X = 3) = \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216} \)
Now, we create a table to compute the mean (\( \mu \)) and variance (\( \sigma^2 \)).

The mean is \( \mu = \sum X P(X) \).
\( \implies \mu = \frac{108}{216} \)
\( \implies \mu = \frac{1}{2} \)
The variance is \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
\( \implies \sigma^2 = \frac{144}{216} - \left(\frac{1}{2}\right)^2 \)
\( \implies \sigma^2 = \frac{6}{9} - \frac{1}{4} \)
\( \implies \sigma^2 = \frac{2}{3} - \frac{1}{4} \)
\( \implies \sigma^2 = \frac{8 - 3}{12} \)
\( \implies \sigma^2 = \frac{5}{12} \)
In simple words: When a die is rolled three times, and rolling a 'four' is a success, we find the chances of getting zero, one, two, or three successes. Then, we use these chances to figure out the average number of successes (mean) and how spread out these successes are (variance). The average number of 'fours' we expect to see in three rolls is 0.5, and the spread of these results is \( \frac{5}{12} \).

🎯 Exam Tip: For problems involving repeated trials (like tossing a die multiple times), recognize if it's a binomial distribution. If so, you can use the formulas \( \mu = np \) and \( \sigma^2 = npq \) as a quick check for your calculated mean and variance.

 

Question 5. Is it possible for the random variable X to have the following distribution?

If it is possible, find the mean and the variance of the random variable.

Answer: For a distribution to be a valid probability distribution, two conditions must be met: (1) each probability \( P(X) \) must be between 0 and 1 (inclusive), and (2) the sum of all probabilities must equal 1.
Here, all probabilities (0.2, 0.3, 0.4, 0.1) are between 0 and 1. Now, let's check the sum:
\( \sum P(X) = P(X = -2) + P(X = -1) + P(X = 0) + P(X = 1) \)
\( \implies \sum P(X) = 0.2 + 0.3 + 0.4 + 0.1 = 1.0 \)
Since the sum of probabilities is 1, this is a valid probability distribution.
Now, we find the mean (\( \mu \)) and variance (\( \sigma^2 \)) using a table for calculation.The mean is \( \mu = \sum X P(X) = -0.6 \).
The variance is \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
\( \implies \sigma^2 = 1.2 - (-0.6)^2 \)
\( \implies \sigma^2 = 1.2 - 0.36 \)
\( \implies \sigma^2 = 0.84 \)
In simple words: First, we check if all the chances (probabilities) add up to one. If they do, then it's a real probability distribution. After that, we calculate the average (mean) by multiplying each number by its chance and adding them up. To find how spread out the numbers are (variance), we use a formula that subtracts the mean squared from the sum of each number squared multiplied by its chance.

🎯 Exam Tip: Always verify that the sum of probabilities equals 1 before proceeding with mean and variance calculations for any probability distribution. If the sum is not 1, it's not a valid distribution.

 

Question 6. Two cards are drawn successively with replacement from a well shuffled deck. Determine the probability distribution of number of kings. Find the mean and variance of the distribution.
Answer: Let \( X \) be the random variable denoting the number of kings drawn in two successive draws with replacement. The possible values for \( X \) are 0, 1, or 2 kings.
In a standard deck of 52 cards, there are 4 kings and 48 non-kings.
The probability of drawing a king \( P(K) = \frac{4}{52} = \frac{1}{13} \).
The probability of drawing a non-king \( P(NK) = \frac{48}{52} = \frac{12}{13} \).
Since cards are drawn with replacement, the draws are independent.
Now, we calculate the probabilities for each value of \( X \):
\( P(X = 0) = P(\text{no king in two draws}) = P(NK) \times P(NK) \)
\( \implies P(X = 0) = \frac{12}{13} \times \frac{12}{13} = \frac{144}{169} \)
\( P(X = 1) = P(\text{one king in two draws}) = P(K \text{ and } NK) + P(NK \text{ and } K) \)
\( \implies P(X = 1) = \left(\frac{1}{13} \times \frac{12}{13}\right) + \left(\frac{12}{13} \times \frac{1}{13}\right) \)
\( \implies P(X = 1) = \frac{12}{169} + \frac{12}{169} = \frac{24}{169} \)
\( P(X = 2) = P(\text{two kings in two draws}) = P(K) \times P(K) \)
\( \implies P(X = 2) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169} \)
The probability distribution of \( X \) is given in the table below:

Now, we calculate the mean (\( \mu \)) and variance (\( \sigma^2 \)) using a table.The mean is \( \mu = \sum X P(X) \).
\( \implies \mu = \frac{26}{169} \)
\( \implies \mu = \frac{2}{13} \)
The variance is \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
\( \implies \sigma^2 = \frac{28}{169} - \left(\frac{2}{13}\right)^2 \)
\( \implies \sigma^2 = \frac{28}{169} - \frac{4}{169} \)
\( \implies \sigma^2 = \frac{24}{169} \)
In simple words: We are drawing two cards from a deck, putting the first one back before drawing the second. We want to find the chances of getting zero, one, or two kings. We then use these chances to calculate the average number of kings we expect (mean) and how much these results vary (variance). On average, we expect to draw about 2 out of 13 kings, and the variance shows how much the number of kings might differ from this average.

🎯 Exam Tip: When drawing cards "with replacement," the probability of an event remains the same for each draw. For binomial distributions like this, consider using \( \mu = np \) and \( \sigma^2 = npq \) for verification if the conditions fit, where \( n \) is the number of trials and \( p \) is the probability of success.

 

Question 7. Find the mean, the variance and standard deviation of the number of heads in a simultaneous toss of three coins.
Answer: When three coins are tossed simultaneously, the total number of possible outcomes is \( 2^3 = 8 \).
The sample space \( S \) is:
\( S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \)
Let \( X \) be the random variable representing the number of heads in three tosses. So, \( X \) can take values 0, 1, 2, or 3.
The probability of getting a head in a single toss is \( p = \frac{1}{2} \).
The probability of getting a tail in a single toss is \( q = \frac{1}{2} \).
Now, we calculate the probabilities for each value of \( X \):
\( P(X = 0) = P(\text{no head}) = qqq \)
\( \implies P(X = 0) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \)
\( P(X = 1) = P(\text{1 head}) = P(HTT) + P(THT) + P(TTH) \)
\( \implies P(X = 1) = 3 \times \left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right) = 3 \times \frac{1}{8} = \frac{3}{8} \)
\( P(X = 2) = P(\text{2 heads}) = P(HHT) + P(HTH) + P(THH) \)
\( \implies P(X = 2) = 3 \times \left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right) = 3 \times \frac{1}{8} = \frac{3}{8} \)
\( P(X = 3) = P(\text{3 heads}) = P(HHH) \)
\( \implies P(X = 3) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \)
Now, we create a table to compute the mean (\( \mu \)), variance (\( \sigma^2 \)), and standard deviation.

The mean is \( \mu = \sum X P(X) \).
\( \implies \mu = \frac{12}{8} \)
\( \implies \mu = \frac{3}{2} = 1.5 \)
The variance is \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
\( \implies \sigma^2 = 3 - \left(\frac{3}{2}\right)^2 \)
\( \implies \sigma^2 = 3 - \frac{9}{4} \)
\( \implies \sigma^2 = \frac{12 - 9}{4} \)
\( \implies \sigma^2 = \frac{3}{4} = 0.75 \)
The standard deviation \( \sigma \) is the square root of the variance.
\( \implies \sigma = \sqrt{\frac{3}{4}} \)
\( \implies \sigma = \frac{\sqrt{3}}{2} \)
In simple words: When three coins are tossed, we calculate the chances of getting 0, 1, 2, or 3 heads. Then, we use these chances to find the average number of heads we expect (mean), which is 1.5. The variance tells us how spread out the results are, which is 0.75. The standard deviation, which is the square root of the variance, gives us a clearer idea of the typical difference from the average.

🎯 Exam Tip: Remember to list all possible outcomes in the sample space for coin toss problems to ensure all probabilities are correctly calculated. For standard deviation, always calculate the variance first, then take its positive square root.

 

Question 8. Find the mean and standard deviation of the probability distribution of the numbers obtained when a card is drawn at random from a set of 7 cards numbered 1 to 7.
Answer: Let \( X \) be the random variable denoting the number on the card drawn from a set of 7 cards numbered 1 to 7. The possible values for \( X \) are 1, 2, 3, 4, 5, 6, and 7.
Since each card is equally likely to be drawn, the probability of drawing any specific card \( X \) is \( P(X) = \frac{1}{7} \).
So, \( P(X = 1) = P(X = 2) = \dots = P(X = 7) = \frac{1}{7} \).
The mean (\( \mu \)) is calculated as \( \sum X P(X) \).
\( \mu = 1 \times \frac{1}{7} + 2 \times \frac{1}{7} + 3 \times \frac{1}{7} + 4 \times \frac{1}{7} + 5 \times \frac{1}{7} + 6 \times \frac{1}{7} + 7 \times \frac{1}{7} \)
\( \implies \mu = \frac{1}{7} (1 + 2 + 3 + 4 + 5 + 6 + 7) \)
\( \implies \mu = \frac{1}{7} (28) \)
\( \implies \mu = 4 \)
To find the variance (\( \sigma^2 \)), we use the formula \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
First, calculate \( \sum X^2 P(X) \):
\( \sum X^2 P(X) = 1^2 \times \frac{1}{7} + 2^2 \times \frac{1}{7} + 3^2 \times \frac{1}{7} + 4^2 \times \frac{1}{7} + 5^2 \times \frac{1}{7} + 6^2 \times \frac{1}{7} + 7^2 \times \frac{1}{7} \)
\( \implies \sum X^2 P(X) = \frac{1}{7} (1 + 4 + 9 + 16 + 25 + 36 + 49) \)
\( \implies \sum X^2 P(X) = \frac{1}{7} (140) \)
\( \implies \sum X^2 P(X) = 20 \)
Now, calculate the variance:
\( \sigma^2 = 20 - (4)^2 \)
\( \implies \sigma^2 = 20 - 16 \)
\( \implies \sigma^2 = 4 \)
The standard deviation (\( \sigma \)) is the square root of the variance.
\( \implies \sigma = \sqrt{4} \)
\( \implies \sigma = 2 \)
In simple words: When drawing one card from a set numbered 1 to 7, each card has an equal chance. The average (mean) number you expect to draw is 4. The variance, which measures how much the numbers typically spread out from the average, is 4. Taking the square root of the variance gives the standard deviation, which is 2.

🎯 Exam Tip: For uniformly distributed discrete variables (where each outcome has the same probability), the mean is simply the average of the outcomes. Always remember to calculate \( \sum X^2 P(X) \) correctly before subtracting \( \mu^2 \) for the variance.

 

Question 9. A pair of dice is rolled twice. Let Z denote the number of times, a total of 9 is obtained. Find the mean and variance of the random variable X.
Answer: When a pair of dice is rolled, the total number of possible outcomes is \( 6 \times 6 = 36 \).
Let's define a success as getting a total of 9 on a single roll of two dice. The favorable outcomes for a total of 9 are: \( \{(3,6), (4,5), (5,4), (6,3)\} \). There are 4 such outcomes.
So, the probability of success \( p = \frac{4}{36} = \frac{1}{9} \).
The probability of failure \( q = 1 - p = 1 - \frac{1}{9} = \frac{8}{9} \).
The pair of dice is rolled twice. Let \( X \) be the random variable representing the number of times a total of 9 is obtained in two rolls. So, \( X \) can take values 0, 1, or 2.
Now, we calculate the probabilities for each value of \( X \):
\( P(X = 0) = P(\text{no success in 2 rolls}) = q \times q \)
\( \implies P(X = 0) = \frac{8}{9} \times \frac{8}{9} = \frac{64}{81} \)
\( P(X = 1) = P(\text{1 success in 2 rolls}) = pq + qp \)
\( \implies P(X = 1) = \left(\frac{1}{9} \times \frac{8}{9}\right) + \left(\frac{8}{9} \times \frac{1}{9}\right) \)
\( \implies P(X = 1) = \frac{8}{81} + \frac{8}{81} = \frac{16}{81} \)
\( P(X = 2) = P(\text{2 successes in 2 rolls}) = p \times p \)
\( \implies P(X = 2) = \frac{1}{9} \times \frac{1}{9} = \frac{1}{81} \)
Now, we create a table to compute the mean (\( \mu \)) and variance (\( \sigma^2 \)).

The mean is \( \mu = \sum X P(X) \).
\( \implies \mu = \frac{18}{81} \)
\( \implies \mu = \frac{2}{9} \)
The variance is \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
\( \implies \sigma^2 = \frac{20}{81} - \left(\frac{2}{9}\right)^2 \)
\( \implies \sigma^2 = \frac{20}{81} - \frac{4}{81} \)
\( \implies \sigma^2 = \frac{16}{81} \)
In simple words: When two dice are rolled twice, we first find the chance of getting a sum of 9. Then, we calculate the chances of getting zero, one, or two such sums in two rolls. Using these probabilities, we find that the average number of times we expect to roll a sum of 9 is \( \frac{2}{9} \), and the variance, which tells us the spread of these results, is \( \frac{16}{81} \).

🎯 Exam Tip: When dealing with rolling dice, accurately list all favorable outcomes for specific sums to avoid errors in probability calculations. Remember the formula for variance and ensure square roots are applied correctly for standard deviation if requested.

 

Question 10. A die is tossed twice. A success is getting an odd number on a random toss. Find the variance of the number of successes.
Answer: When a die is tossed, the odd numbers are {1, 3, 5}. There are 3 odd numbers out of 6 possible outcomes.
The probability of success (getting an odd number) \( p = \frac{3}{6} = \frac{1}{2} \).
The probability of failure (not getting an odd number) \( q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \).
The die is tossed twice. Let \( X \) be the random variable denoting the number of successes in 2 tosses. So, \( X \) can take values 0, 1, or 2.
Now, we calculate the probabilities for each value of \( X \):
\( P(X = 0) = P(\text{no odd number}) = q \times q = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
\( P(X = 1) = P(\text{one odd number}) = pq + qp = \left(\frac{1}{2} \times \frac{1}{2}\right) + \left(\frac{1}{2} \times \frac{1}{2}\right) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \)
\( P(X = 2) = P(\text{two odd numbers}) = p \times p = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
Now, we calculate the mean (\( \mu \)) and variance (\( \sigma^2 \)).
The mean is \( \mu = \sum X P(X) \).
\( \implies \mu = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2)) \)
\( \implies \mu = \left(0 \times \frac{1}{4}\right) + \left(1 \times \frac{1}{2}\right) + \left(2 \times \frac{1}{4}\right) \)
\( \implies \mu = 0 + \frac{1}{2} + \frac{2}{4} \)
\( \implies \mu = \frac{1}{2} + \frac{1}{2} = 1 \)
The variance is \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
\( \implies \sigma^2 = \left(0^2 \times \frac{1}{4}\right) + \left(1^2 \times \frac{1}{2}\right) + \left(2^2 \times \frac{1}{4}\right) - (1)^2 \)
\( \implies \sigma^2 = \left(0 \times \frac{1}{4}\right) + \left(1 \times \frac{1}{2}\right) + \left(4 \times \frac{1}{4}\right) - 1 \)
\( \implies \sigma^2 = 0 + \frac{1}{2} + 1 - 1 \)
\( \implies \sigma^2 = \frac{1}{2} \)
In simple words: When a die is tossed twice, getting an odd number is a success (which has a 1/2 chance). We calculate the chances of getting zero, one, or two odd numbers. Then, we find the average number of successes (mean) which is 1. Finally, we calculate the variance, which shows how spread out the number of odd outcomes are, and it comes out to 0.5.

🎯 Exam Tip: This is a binomial distribution since there are fixed trials (2), two outcomes (odd/not odd), and independent trials with constant probabilities. You can quickly verify with \( \mu = np \) and \( \sigma^2 = npq \). Here \( n=2, p=1/2, q=1/2 \), so \( \mu = 2 \times 1/2 = 1 \) and \( \sigma^2 = 2 \times 1/2 \times 1/2 = 1/2 \).

 

Question 1. In a single throw of a die, if X denotes the number on its upper face, find the mean of X.
Answer: The random variable \( X \) represents the number shown on the upper face when a single die is thrown. For a fair die, each face (1, 2, 3, 4, 5, 6) has an equal probability of appearing, which is \( \frac{1}{6} \).
To find the mean (\( \mu \)) of \( X \), we use the formula for the expected value of a discrete random variable: \( \mu = \sum (X_i \times P(X_i)) \). This involves multiplying each possible outcome by its corresponding probability and then summing these products.
\[ \mu = \left(1 \times \frac{1}{6}\right) + \left(2 \times \frac{1}{6}\right) + \left(3 \times \frac{1}{6}\right) + \left(4 \times \frac{1}{6}\right) + \left(5 \times \frac{1}{6}\right) + \left(6 \times \frac{1}{6}\right) \]
We can factor out \( \frac{1}{6} \) from the sum:
\[ \mu = \frac{1}{6} (1 + 2 + 3 + 4 + 5 + 6) \]
\[ \mu = \frac{1}{6} (21) \]
\[ \mu = \frac{21}{6} \]
\[ \mu = \frac{7}{2} \]
Thus, the mean of \( X \) is \( \frac{7}{2} \). The mean represents the average outcome you would expect if you rolled the die many times.
In simple words: When you roll a die, the average number you expect to get over many rolls is the mean. For a fair die, each side has an equal chance, so you sum up all possible numbers multiplied by their chance, then add them.

🎯 Exam Tip: Remember that the mean of a discrete random variable is found by summing the product of each value and its probability. This is also called the expected value.

 

Question 2. A salesman wants to know the average number of units he sells per sales call? He checks his past sales records and comes up with the following probabilities.
Sales (in units): 0, 1, 2, 3, 4, 5
Probability: 0.15, 0.20, 0.10, 0.05, 0.30, 0.20
What is the average number of units he sells per sale call?
Answer: To find the average number of units sold per sales call, we need to calculate the mean (\( \mu \)) of the probability distribution. The mean is found by multiplying each number of units sold (\( X_i \)) by its corresponding probability (\( P_i \)) and then summing all these products.
The formula for the mean is \( \mu = \sum (X_i \times P_i) \).
Let's organize the data and calculations in a table:

Now, we sum the values in the \( X_i P_i \) column:
\[ \mu = 0 + 0.20 + 0.20 + 0.15 + 1.20 + 1.00 \] \[ \mu = 2.75 \]
Therefore, the average number of units the salesman sells per sales call is \( 2.75 \). This value represents the expected number of units sold per call over the long run.
In simple words: To find the average sales, you take each possible number of units sold and multiply it by how often that sale happens. Then, you add up all those results. This gives you the mean, which is the average expected sale.

🎯 Exam Tip: For probability distributions, the mean is the expected value. Always set up a clear table to multiply each outcome by its probability, then sum the products accurately.

 

Question 3. Find \( \mu \) and variance for the following probability distribution.
(i)
X      2     5
P(X)  0.4  0.6
(ii)
X      1     2     3     4
P(x)  0.4  0.3  0.2  0.1
(iii)
X      -2     -1     0     1     2
P(x)  0.2   0.3   0.3   0.1   0.1
(iv)
X      2     3     4
P(X)  p     p     1-2p
Answer:
(i) For this distribution, we have:
\( X \): 2, 5
\( P(X) \): 0.4, 0.6
First, we calculate the Mean (\( \mu \)) using \( \mu = \sum X P(X) \):
\[ \mu = (2 \times 0.4) + (5 \times 0.6) \] \[ \mu = 0.8 + 3.0 \] \[ \mu = 3.8 \]
Next, we calculate the Variance (\( \sigma^2 \)) using \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
First, find \( \sum X^2 P(X) \):
For \( X=2 \), \( X^2 = 4 \), so \( X^2 P(X) = 4 \times 0.4 = 1.6 \)
For \( X=5 \), \( X^2 = 25 \), so \( X^2 P(X) = 25 \times 0.6 = 15.0 \)
\[ \sum X^2 P(X) = 1.6 + 15.0 = 16.6 \]
Now, substitute into the variance formula:
\[ \sigma^2 = 16.6 - (3.8)^2 \] \[ \sigma^2 = 16.6 - 14.44 \] \[ \sigma^2 = 2.16 \]
So, for part (i), the mean \( \mu = 3.8 \) and the variance \( \sigma^2 = 2.16 \).

(ii) For this distribution, we have:
\( X \): 1, 2, 3, 4
\( P(x) \): 0.4, 0.3, 0.2, 0.1
First, we calculate the Mean (\( \mu \)) using \( \mu = \sum X P(X) \):
\[ \mu = (1 \times 0.4) + (2 \times 0.3) + (3 \times 0.2) + (4 \times 0.1) \] \[ \mu = 0.4 + 0.6 + 0.6 + 0.4 \] \[ \mu = 2.0 \]
Next, we calculate the Variance (\( \sigma^2 \)) using \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
First, find \( \sum X^2 P(X) \):
For \( X=1 \), \( X^2 = 1 \), so \( X^2 P(X) = 1 \times 0.4 = 0.4 \)
For \( X=2 \), \( X^2 = 4 \), so \( X^2 P(X) = 4 \times 0.3 = 1.2 \)
For \( X=3 \), \( X^2 = 9 \), so \( X^2 P(X) = 9 \times 0.2 = 1.8 \)
For \( X=4 \), \( X^2 = 16 \), so \( X^2 P(X) = 16 \times 0.1 = 1.6 \)
\[ \sum X^2 P(X) = 0.4 + 1.2 + 1.8 + 1.6 = 5.0 \]
Now, substitute into the variance formula:
\[ \sigma^2 = 5.0 - (2.0)^2 \] \[ \sigma^2 = 5.0 - 4.0 \] \[ \sigma^2 = 1.0 \]
So, for part (ii), the mean \( \mu = 2.0 \) and the variance \( \sigma^2 = 1.0 \).

(iii) For this distribution, we have:
\( X \): -2, -1, 0, 1, 2
\( P(x) \): 0.2, 0.3, 0.3, 0.1, 0.1
First, we calculate the Mean (\( \mu \)) using \( \mu = \sum X P(X) \):
\[ \mu = (-2 \times 0.2) + (-1 \times 0.3) + (0 \times 0.3) + (1 \times 0.1) + (2 \times 0.1) \] \[ \mu = -0.4 - 0.3 + 0 + 0.1 + 0.2 \] \[ \mu = -0.7 + 0.3 \] \[ \mu = -0.4 \]
Next, we calculate the Variance (\( \sigma^2 \)) using \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
First, find \( \sum X^2 P(X) \):
For \( X=-2 \), \( X^2 = 4 \), so \( X^2 P(X) = 4 \times 0.2 = 0.8 \)
For \( X=-1 \), \( X^2 = 1 \), so \( X^2 P(X) = 1 \times 0.3 = 0.3 \)
For \( X=0 \), \( X^2 = 0 \), so \( X^2 P(X) = 0 \times 0.3 = 0 \)
For \( X=1 \), \( X^2 = 1 \), so \( X^2 P(X) = 1 \times 0.1 = 0.1 \)
For \( X=2 \), \( X^2 = 4 \), so \( X^2 P(X) = 4 \times 0.1 = 0.4 \)
\[ \sum X^2 P(X) = 0.8 + 0.3 + 0 + 0.1 + 0.4 = 1.6 \]
Now, substitute into the variance formula:
\[ \sigma^2 = 1.6 - (-0.4)^2 \] \[ \sigma^2 = 1.6 - 0.16 \] \[ \sigma^2 = 1.44 \]
So, for part (iii), the mean \( \mu = -0.4 \) and the variance \( \sigma^2 = 1.44 \).

(iv) For this distribution, we have:
\( X \): 2, 3, 4
\( P(X) \): \( p, p, 1-2p \)
First, we calculate the Mean (\( \mu \)) using \( \mu = \sum X P(X) \):
\[ \mu = (2 \times p) + (3 \times p) + (4 \times (1-2p)) \] \[ \mu = 2p + 3p + 4 - 8p \] \[ \mu = 5p + 4 - 8p \] \[ \mu = 4 - 3p \]
Next, we calculate the Variance (\( \sigma^2 \)) using \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
First, find \( \sum X^2 P(X) \):
For \( X=2 \), \( X^2 = 4 \), so \( X^2 P(X) = 4 \times p = 4p \)
For \( X=3 \), \( X^2 = 9 \), so \( X^2 P(X) = 9 \times p = 9p \)
For \( X=4 \), \( X^2 = 16 \), so \( X^2 P(X) = 16 \times (1-2p) = 16 - 32p \)
\[ \sum X^2 P(X) = 4p + 9p + 16 - 32p \] \[ \sum X^2 P(X) = 13p + 16 - 32p \] \[ \sum X^2 P(X) = 16 - 19p \]
Now, substitute into the variance formula:
\[ \sigma^2 = (16 - 19p) - (4 - 3p)^2 \]
Expand \( (4 - 3p)^2 = 16 - 24p + 9p^2 \).
\[ \sigma^2 = 16 - 19p - (16 - 24p + 9p^2) \] \[ \sigma^2 = 16 - 19p - 16 + 24p - 9p^2 \] \[ \sigma^2 = 5p - 9p^2 \]
So, for part (iv), the mean \( \mu = 4 - 3p \) and the variance \( \sigma^2 = 5p - 9p^2 \). Remember that \( p \) must be between 0 and 1, and the sum of probabilities must be 1.
In simple words: To find the mean, multiply each number \( X \) by its chance \( P(X) \) and add them up. To find the variance, which shows how spread out the numbers are, you first find the average of \( X^2 \) times their chances, and then subtract the mean squared. Even with algebraic probabilities, the process stays the same.

🎯 Exam Tip: When calculating variance, ensure you square the mean (\( \mu \)) correctly and are careful with negative signs, especially when \( \mu \) is negative or involves algebraic terms like \( p \). Always remember to use the formula \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).

 

Question 4. A die is tossed thrice. If getting 'four' is considered a success, find the mean and variance of probability distribution of the number of successes.
Answer: When a fair die is tossed, the sample space (all possible outcomes) is \( S = \{1, 2, 3, 4, 5, 6\} \).
A success is defined as getting a 'four'. So, the probability of success, \( p \), is \( P(\text{getting 4}) = \frac{1}{6} \).
The probability of failure, \( q \), is \( 1 - p = 1 - \frac{1}{6} = \frac{5}{6} \).
The die is tossed thrice. Let \( X \) be the random variable representing the number of successes (getting a 'four'). \( X \) can take values 0, 1, 2, or 3.
Now we calculate the probability for each value of \( X \):
For \( X=0 \) (no successes): This means three failures (qqq).
\[ P(X=0) = q \times q \times q = \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{125}{216} \]
For \( X=1 \) (one success): This can happen in three ways (pqq, qpq, or qqp).
\[ P(X=1) = \left(\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6}\right) + \left(\frac{5}{6} \times \frac{1}{6} \times \frac{5}{6}\right) + \left(\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right) \] \[ P(X=1) = \frac{25}{216} + \frac{25}{216} + \frac{25}{216} = \frac{75}{216} \]
For \( X=2 \) (two successes): This can happen in three ways (ppq, pqp, or qpp).
\[ P(X=2) = \left(\frac{1}{6} \times \frac{1}{6} \times \frac{5}{6}\right) + \left(\frac{1}{6} \times \frac{5}{6} \times \frac{1}{6}\right) + \left(\frac{5}{6} \times \frac{1}{6} \times \frac{1}{6}\right) \] \[ P(X=2) = \frac{5}{216} + \frac{5}{216} + \frac{5}{216} = \frac{15}{216} \]
For \( X=3 \) (three successes): This means three successes (ppp).
\[ P(X=3) = p \times p \times p = \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216} \]
Now, we create a table to calculate the mean (\( \mu \)) and variance (\( \sigma^2 \)):

Sum of \( X_i P(X_i) \):
\[ \sum X_i P(X_i) = 0 + \frac{75}{216} + \frac{30}{216} + \frac{3}{216} = \frac{108}{216} \]
Sum of \( X_i^2 P(X_i) \):
\[ \sum X_i^2 P(X_i) = 0 + \frac{75}{216} + \frac{60}{216} + \frac{9}{216} = \frac{144}{216} \]
Calculate the Mean (\( \mu \)):
\[ \mu = \sum X_i P(X_i) = \frac{108}{216} = \frac{1}{2} \]
Calculate the Variance (\( \sigma^2 \)):
\[ \sigma^2 = \sum X_i^2 P(X_i) - \mu^2 \] \[ \sigma^2 = \frac{144}{216} - \left(\frac{1}{2}\right)^2 \] \[ \sigma^2 = \frac{2}{3} - \frac{1}{4} \]
To subtract these fractions, we find a common denominator, which is 12:
\[ \sigma^2 = \frac{8}{12} - \frac{3}{12} \] \[ \sigma^2 = \frac{5}{12} \]
So, for this distribution, the mean \( \mu = \frac{1}{2} \) and the variance \( \sigma^2 = \frac{5}{12} \). This is a binomial distribution, where the mean \( np \) and variance \( npq \) formulas provide a good check for the results.
In simple words: When you toss a die three times, we want to know the average number of times you get a 'four', and how much that number usually changes. First, figure out the chances of getting zero, one, two, or three 'fours'. Then use these chances in a table to calculate the mean and variance. The mean is the average outcome, and variance shows how spread out the results are.

🎯 Exam Tip: For binomial distributions, you can quickly verify your mean and variance using the formulas \( \mu = np \) and \( \sigma^2 = npq \). However, if asked to show the full probability distribution and calculation, follow all steps.

 

Question 5. Is it possible for the random variable X to have the following distribution?
X: -2, -1, 0, 1
P(X): 0.2, 0.3, 0.4, 0.1
If it is possible, find the mean and the variance of the random variable.
Answer: For a probability distribution to be valid, two conditions must be met:
1. Each probability \( P(X_i) \) must be between 0 and 1 (inclusive).
2. The sum of all probabilities \( \sum P(X_i) \) must be equal to 1.
Let's check the given distribution:
\( P(X=-2) = 0.2 \)
\( P(X=-1) = 0.3 \)
\( P(X=0) = 0.4 \)
\( P(X=1) = 0.1 \)
All individual probabilities are between 0 and 1, satisfying the first condition.
Now, let's sum them up:
\[ \sum P(X) = 0.2 + 0.3 + 0.4 + 0.1 = 1.0 \]
The sum of probabilities is 1.0, satisfying the second condition.
Since both conditions are satisfied, it is possible for the random variable \( X \) to have this distribution.
Now, we will find the mean (\( \mu \)) and variance (\( \sigma^2 \)). We will use a table for the calculations:

Sum of \( X_i P(X_i) \):
\[ \sum X_i P(X_i) = -0.4 - 0.3 + 0 + 0.1 = -0.6 \]
Sum of \( X_i^2 P(X_i) \):
\[ \sum X_i^2 P(X_i) = 0.8 + 0.3 + 0 + 0.1 = 1.2 \]
Calculate the Mean (\( \mu \)):
\[ \mu = \sum X_i P(X_i) = -0.6 \]
Calculate the Variance (\( \sigma^2 \)) using \( \sigma^2 = \sum X_i^2 P(X_i) - \mu^2 \):
\[ \sigma^2 = 1.2 - (-0.6)^2 \] \[ \sigma^2 = 1.2 - 0.36 \] \[ \sigma^2 = 0.84 \]
So, the mean \( \mu = -0.6 \) and the variance \( \sigma^2 = 0.84 \).
In simple words: For a set of probabilities to be valid, each chance must be positive, and all the chances must add up to exactly 1. If it's valid, you can find the mean (average) by multiplying each number by its chance and summing them up. Then, calculate the variance (how spread out the numbers are) using the sum of (each number squared multiplied by its chance) minus the mean squared.

🎯 Exam Tip: Always check the two fundamental properties of a probability distribution (probabilities between 0 and 1, and sum to 1) before performing any further calculations.

 

Question 6. Two cards are drawn successively with replacement from a well shuffled deck. Determine the probability distribution of number of kings. Find the mean and variance of the distribution.
Answer: In a standard deck of 52 cards, there are 4 kings.
Let \( X \) be the random variable representing the number of kings drawn in two successive draws with replacement. \( X \) can take values 0, 1, or 2.
The probability of drawing a king (\( P(\text{King}) \)) is \( \frac{4}{52} = \frac{1}{13} \).
The probability of not drawing a king (\( P(\text{Not King}) \)) is \( \frac{48}{52} = \frac{12}{13} \).
Now we find the probability for each value of \( X \):
For \( X=0 \) (no kings): This means drawing (Not King, Not King).
\[ P(X=0) = \frac{12}{13} \times \frac{12}{13} = \frac{144}{169} \]
For \( X=1 \) (one king): This means drawing (King, Not King) or (Not King, King).
\[ P(X=1) = \left(\frac{1}{13} \times \frac{12}{13}\right) + \left(\frac{12}{13} \times \frac{1}{13}\right) \] \[ P(X=1) = \frac{12}{169} + \frac{12}{169} = \frac{24}{169} \]
For \( X=2 \) (two kings): This means drawing (King, King).
\[ P(X=2) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169} \]
The probability distribution of \( X \) is:

Now, we compute the mean (\( \mu \)) and variance (\( \sigma^2 \)) using a table:

Sum of \( X_i P(X_i) \):
\[ \sum X_i P(X_i) = 0 + \frac{24}{169} + \frac{2}{169} = \frac{26}{169} \]
Sum of \( X_i^2 P(X_i) \):
\[ \sum X_i^2 P(X_i) = 0 + \frac{24}{169} + \frac{4}{169} = \frac{28}{169} \]
Calculate the Mean (\( \mu \)):
\[ \mu = \sum X_i P(X_i) = \frac{26}{169} = \frac{2 \times 13}{13 \times 13} = \frac{2}{13} \]
Calculate the Variance (\( \sigma^2 \)) using \( \sigma^2 = \sum X_i^2 P(X_i) - \mu^2 \):
\[ \sigma^2 = \frac{28}{169} - \left(\frac{2}{13}\right)^2 \] \[ \sigma^2 = \frac{28}{169} - \frac{4}{169} \] \[ \sigma^2 = \frac{24}{169} \]
So, the mean \( \mu = \frac{2}{13} \) and the variance \( \sigma^2 = \frac{24}{169} \). This calculation is consistent with a binomial distribution with parameters \( n=2 \) and \( p = \frac{1}{13} \).
In simple words: When drawing cards with replacement, the chance of getting a king stays the same each time. We list the chances for getting 0, 1, or 2 kings. Then, we use these chances to calculate the average number of kings you'd expect (mean) and how much that number varies (variance) over many draws.

🎯 Exam Tip: For "with replacement" problems, each draw is independent. Remember to list all possible sequences for a given number of successes if the order matters (e.g., King-Not King and Not King-King for one success).

 

Question 7. Find the mean, the variance and standard deviation of the number of heads in a simultaneous toss of three coins.
Answer: When three coins are tossed simultaneously, the total number of possible outcomes is \( 2^3 = 8 \). The sample space \( S \) is:
\( S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \)
Let \( X \) be the random variable representing the number of heads. \( X \) can take values 0, 1, 2, or 3.
The probability of getting a head (success, \( p \)) in a single toss is \( \frac{1}{2} \).
The probability of getting a tail (failure, \( q \)) is \( 1 - \frac{1}{2} = \frac{1}{2} \).
Now we calculate the probability for each value of \( X \):
For \( X=0 \) (no heads, TTT): \( P(X=0) = \frac{1}{8} \)
For \( X=1 \) (one head, HTT, THT, TTH): \( P(X=1) = \frac{3}{8} \)
For \( X=2 \) (two heads, HHT, HTH, THH): \( P(X=2) = \frac{3}{8} \)
For \( X=3 \) (three heads, HHH): \( P(X=3) = \frac{1}{8} \)
Next, we create a table to calculate the mean (\( \mu \)), variance (\( \sigma^2 \)), and standard deviation:

Sum of \( X_i P(X_i) \):
\[ \sum X_i P(X_i) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} \]
Sum of \( X_i^2 P(X_i) \):
\[ \sum X_i^2 P(X_i) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3 \]
Calculate the Mean (\( \mu \)):
\[ \mu = \sum X_i P(X_i) = \frac{12}{8} = \frac{3}{2} \]
Calculate the Variance (\( \sigma^2 \)) using \( \sigma^2 = \sum X_i^2 P(X_i) - \mu^2 \):
\[ \sigma^2 = 3 - \left(\frac{3}{2}\right)^2 \] \[ \sigma^2 = 3 - \frac{9}{4} \] \[ \sigma^2 = \frac{12}{4} - \frac{9}{4} \] \[ \sigma^2 = \frac{3}{4} = 0.75 \]
Calculate the Standard Deviation (\( \sigma \)) using \( \sigma = \sqrt{\text{Variance}} \):
\[ \sigma = \sqrt{\frac{3}{4}} \] \[ \sigma = \frac{\sqrt{3}}{2} \]
So, for three coin tosses, the mean number of heads is \( \frac{3}{2} \), the variance is \( \frac{3}{4} \), and the standard deviation is \( \frac{\sqrt{3}}{2} \). These calculations align with the properties of a binomial distribution.
In simple words: When three coins are flipped together, we want to find the average number of heads, how much that number usually changes, and the typical spread of results. Since each coin has an equal chance of heads or tails, we list all possible combinations to find the probabilities for 0, 1, 2, or 3 heads. Then, we use these probabilities to calculate the mean (average), variance (spread), and standard deviation (typical distance from the mean).

🎯 Exam Tip: For coin toss problems, remember that the probability of getting heads or tails is always \( \frac{1}{2} \). The number of outcomes for \( n \) tosses is \( 2^n \). For binomial distributions, \( \mu = np \) and \( \sigma^2 = npq \).

 

Question 8. Find the mean and standard deviation of the probability distribution of the numbers obtained when a card is drawn at random from a set of 7 cards numbered 1 to 7.
Answer: Let \( X \) be the random variable representing the number drawn from a set of 7 cards numbered 1 to 7.
The possible values for \( X \) are {1, 2, 3, 4, 5, 6, 7}.
Since a card is drawn at random, each number has an equal probability of being drawn. So, \( P(X=i) = \frac{1}{7} \) for \( i = 1, 2, \dots, 7 \).
First, calculate the Mean (\( \mu \)) using \( \mu = \sum X P(X) \):
\[ \mu = \left(1 \times \frac{1}{7}\right) + \left(2 \times \frac{1}{7}\right) + \left(3 \times \frac{1}{7}\right) + \left(4 \times \frac{1}{7}\right) + \left(5 \times \frac{1}{7}\right) + \left(6 \times \frac{1}{7}\right) + \left(7 \times \frac{1}{7}\right) \]
We can factor out \( \frac{1}{7} \):
\[ \mu = \frac{1}{7} (1 + 2 + 3 + 4 + 5 + 6 + 7) \]
The sum of the first 7 natural numbers is \( \frac{7 \times (7+1)}{2} = \frac{7 \times 8}{2} = 28 \).
\[ \mu = \frac{1}{7} (28) \] \[ \mu = 4 \]
Next, calculate the Variance (\( \sigma^2 \)) using \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
First, find \( \sum X^2 P(X) \):
\[ \sum X^2 P(X) = \left(1^2 \times \frac{1}{7}\right) + \left(2^2 \times \frac{1}{7}\right) + \left(3^2 \times \frac{1}{7}\right) + \left(4^2 \times \frac{1}{7}\right) + \left(5^2 \times \frac{1}{7}\right) + \left(6^2 \times \frac{1}{7}\right) + \left(7^2 \times \frac{1}{7}\right) \]
Factor out \( \frac{1}{7} \):
\[ \sum X^2 P(X) = \frac{1}{7} (1 + 4 + 9 + 16 + 25 + 36 + 49) \] \[ \sum X^2 P(X) = \frac{1}{7} (140) \] \[ \sum X^2 P(X) = 20 \]
Now, substitute into the variance formula:
\[ \sigma^2 = 20 - (4)^2 \] \[ \sigma^2 = 20 - 16 \] \[ \sigma^2 = 4 \]
Finally, calculate the Standard Deviation (\( \sigma \)) using \( \sigma = \sqrt{\text{Variance}} \):
\[ \sigma = \sqrt{4} \] \[ \sigma = 2 \]
So, the mean of the numbers drawn is 4 and the standard deviation is 2. This type of distribution is known as a discrete uniform distribution, where each outcome has an equal probability.
In simple words: If you have cards numbered 1 to 7 and pick one at random, the average number you'd pick (the mean) is 4. The standard deviation, which shows how far the numbers usually spread from this average, is 2.

🎯 Exam Tip: For a discrete uniform distribution from 1 to \( n \), the mean is \( \frac{n+1}{2} \) and the variance is \( \frac{n^2-1}{12} \). Always double-check your calculations with these shortcuts.

 

Question 9. A pair of dice is rolled twice. Let Z denote the number of times, a total of 9 is obtained. Find the mean and variance of the random variable X.
Answer: When a pair of dice is rolled, the total number of possible outcomes is \( 6 \times 6 = 36 \).
A "success" is defined as getting a total of 9. The favorable cases for a total of 9 are: (3,6), (4,5), (5,4), (6,3). There are 4 such cases.
The probability of success (\( p \)) in a single roll of two dice is \( \frac{4}{36} = \frac{1}{9} \).
The probability of failure (\( q \)) is \( 1 - p = 1 - \frac{1}{9} = \frac{8}{9} \).
The pair of dice is rolled twice. Let \( X \) be the random variable representing the number of times a total of 9 is obtained. \( X \) can take values 0, 1, or 2.
Now we calculate the probability for each value of \( X \):
For \( X=0 \) (no total of 9): This means two failures (qq).
\[ P(X=0) = q \times q = \frac{8}{9} \times \frac{8}{9} = \frac{64}{81} \]
For \( X=1 \) (one total of 9): This means (pq or qp).
\[ P(X=1) = (p \times q) + (q \times p) = \left(\frac{1}{9} \times \frac{8}{9}\right) + \left(\frac{8}{9} \times \frac{1}{9}\right) \] \[ P(X=1) = \frac{8}{81} + \frac{8}{81} = \frac{16}{81} \]
For \( X=2 \) (two totals of 9): This means two successes (pp).
\[ P(X=2) = p \times p = \frac{1}{9} \times \frac{1}{9} = \frac{1}{81} \]
Next, we create a table to calculate the mean (\( \mu \)) and variance (\( \sigma^2 \)):

Sum of \( X_i P(X_i) \):
\[ \sum X_i P(X_i) = 0 + \frac{16}{81} + \frac{2}{81} = \frac{18}{81} \]
Sum of \( X_i^2 P(X_i) \):
\[ \sum X_i^2 P(X_i) = 0 + \frac{16}{81} + \frac{4}{81} = \frac{20}{81} \]
Calculate the Mean (\( \mu \)):
\[ \mu = \sum X_i P(X_i) = \frac{18}{81} = \frac{2}{9} \]
Calculate the Variance (\( \sigma^2 \)) using \( \sigma^2 = \sum X_i^2 P(X_i) - \mu^2 \):
\[ \sigma^2 = \frac{20}{81} - \left(\frac{2}{9}\right)^2 \] \[ \sigma^2 = \frac{20}{81} - \frac{4}{81} \] \[ \sigma^2 = \frac{16}{81} \]
So, the mean \( \mu = \frac{2}{9} \) and the variance \( \sigma^2 = \frac{16}{81} \). This calculation is consistent with a binomial distribution with parameters \( n=2 \) and \( p = \frac{1}{9} \).
In simple words: When you roll two dice twice, a 'success' is when the numbers add up to 9. We find the chance of getting 0, 1, or 2 successes. Then, we use these probabilities to calculate the average number of successes (mean) and how much these results typically vary (variance).

🎯 Exam Tip: Be careful with combinations and permutations when calculating probabilities for repeated events. Clearly define success and failure probabilities, \( p \) and \( q \).

 

Question 10. A die is tossed twice. A success is getting an odd number on a random toss. Find the variance of the number of successes.
Answer: When a die is tossed, the odd numbers are {1, 3, 5}. There are 3 odd numbers out of 6 possible outcomes.
The probability of success (\( p \)), which is getting an odd number, is \( \frac{3}{6} = \frac{1}{2} \).
The probability of failure (\( q \)), which is not getting an odd number (getting an even number), is \( 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \).
The die is tossed twice. Let \( X \) be the random variable representing the number of successes (getting an odd number). \( X \) can take values 0, 1, or 2.
Now we find the probability for each value of \( X \):
For \( X=0 \) (no successes): This means two failures (qq).
\[ P(X=0) = q \times q = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]
For \( X=1 \) (one success): This means (pq or qp).
\[ P(X=1) = (p \times q) + (q \times p) = \left(\frac{1}{2} \times \frac{1}{2}\right) + \left(\frac{1}{2} \times \frac{1}{2}\right) \] \[ P(X=1) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \]
For \( X=2 \) (two successes): This means two successes (pp).
\[ P(X=2) = p \times p = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]
Now we calculate the mean (\( \mu \)) and variance (\( \sigma^2 \)):
Calculate the Mean (\( \mu \)) using \( \mu = \sum X P(X) \):
\[ \mu = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2)) \] \[ \mu = \left(0 \times \frac{1}{4}\right) + \left(1 \times \frac{1}{2}\right) + \left(2 \times \frac{1}{4}\right) \] \[ \mu = 0 + \frac{1}{2} + \frac{2}{4} \] \[ \mu = 0 + \frac{1}{2} + \frac{1}{2} \] \[ \mu = 1 \]
Calculate the Variance (\( \sigma^2 \)) using \( \sigma^2 = \sum X^2 P(X) - \mu^2 \).
First, find \( \sum X^2 P(X) \):
\[ \sum X^2 P(X) = (0^2 \times P(X=0)) + (1^2 \times P(X=1)) + (2^2 \times P(X=2)) \] \[ \sum X^2 P(X) = \left(0 \times \frac{1}{4}\right) + \left(1 \times \frac{1}{2}\right) + \left(4 \times \frac{1}{4}\right) \] \[ \sum X^2 P(X) = 0 + \frac{1}{2} + 1 \] \[ \sum X^2 P(X) = \frac{3}{2} \]
Now, substitute into the variance formula:
\[ \sigma^2 = \frac{3}{2} - (1)^2 \] \[ \sigma^2 = \frac{3}{2} - 1 \] \[ \sigma^2 = \frac{1}{2} \]
So, the variance of the number of successes is \( \frac{1}{2} \). This is a binomial distribution with \( n=2 \), \( p=\frac{1}{2} \), and \( q=\frac{1}{2} \).
In simple words: When you roll a die twice, getting an odd number is a success. Since half the numbers on a die are odd, the chance of success is 1/2. We find the probabilities of getting 0, 1, or 2 odd numbers in two rolls. Then, we use these to calculate the mean (average odd numbers) and the variance, which shows how spread out the number of odd results typically are.

🎯 Exam Tip: For simple binomial distributions like this, the mean \( np \) and variance \( npq \) formulas are very useful for quick verification of your detailed calculations. Ensure you correctly identify \( n \), \( p \), and \( q \).