Samacheer Kalvi Class 12 Maths Solutions Chapter 11 Probability Distributions Exercise 11.5

Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.5

Question 1. Compute P(X = k) for the binomial distribution, B(n, p) where
(i) n = 6, p = \( \frac { 1 }{ 3 } \), k = 3
(ii) n = 10, p = \( \frac { 1 }{ 5 } \), k = 4
(iii) n = 9, p = \( \frac { 1 }{ 2 } \), k = 7
Answer:
(i) Given: \( n = 6, p = \frac { 1 }{ 3 }, k = 3 \).
The general formula for binomial distribution is \( P(X=x) = ^nC_x p^x q^{n-x} \), where \( x = 0, 1, 2, ..., n \).
First, we find \( q \):
\( q = 1 - p \)
\( q = 1 - \frac { 1 }{ 3 } \)
\( q = \frac { 3-1 }{ 3 } \)
\( q = \frac { 2 }{ 3 } \)
Now, we compute \( P(X = 3) \):
\( P(X = 3) = ^6C_3 \left( \frac { 1 }{ 3 } \right)^3 \left( \frac { 2 }{ 3 } \right)^{6-3} \)
\( P(X = 3) = \frac { 6 \times 5 \times 4 }{ 3 \times 2 \times 1 } \times \left( \frac { 1 }{ 3 } \right)^3 \times \left( \frac { 2 }{ 3 } \right)^3 \)
\( P(X = 3) = 20 \times \frac { 1 }{ 27 } \times \frac { 8 }{ 27 } \)
\( P(X = 3) = \frac { 160 }{ 729 } \)

(ii) Given: \( n = 10, p = \frac { 1 }{ 5 }, k = 4 \).
The formula is \( P(X=x) = ^nC_x p^x q^{n-x} \), where \( x = 0, 1, 2, ..., n \).
First, we find \( q \):
\( q = 1 - p \)
\( q = 1 - \frac { 1 }{ 5 } \)
\( q = \frac { 5-1 }{ 5 } \)
\( q = \frac { 4 }{ 5 } \)
Now, we compute \( P(X = 4) \):
\( P(X = 4) = ^{10}C_4 \left( \frac { 1 }{ 5 } \right)^4 \left( \frac { 4 }{ 5 } \right)^{10-4} \)
\( P(X = 4) = ^{10}C_4 \left( \frac { 1 }{ 5 } \right)^4 \left( \frac { 4 }{ 5 } \right)^6 \)
\( P(X = 4) = \frac { 10 \times 9 \times 8 \times 7 }{ 4 \times 3 \times 2 \times 1 } \times \frac { 1^4 }{ 5^4 } \times \frac { 4^6 }{ 5^6 } \)
\( P(X = 4) = 210 \times \frac { 1 }{ 5^4 } \times \frac { 4^6 }{ 5^6 } \)
\( P(X = 4) = 210 \times \frac { 4^6 }{ 5^{10} } \)

(iii) Given: \( n = 9, p = \frac { 1 }{ 2 }, k = 7 \).
The formula is \( P(X=x) = ^nC_x p^x q^{n-x} \), where \( x = 0, 1, 2, ..., n \).
First, we find \( q \):
\( q = 1 - p \)
\( q = 1 - \frac { 1 }{ 2 } \)
\( q = \frac { 1 }{ 2 } \)
Now, we compute \( P(X = 7) \):
\( P(X = 7) = ^9C_7 \left( \frac { 1 }{ 2 } \right)^7 \left( \frac { 1 }{ 2 } \right)^{9-7} \)
\( P(X = 7) = ^9C_7 \left( \frac { 1 }{ 2 } \right)^7 \left( \frac { 1 }{ 2 } \right)^2 \)
\( P(X = 7) = \frac { 9 \times 8 }{ 2 \times 1 } \times \frac { 1 }{ 2^7 } \times \frac { 1 }{ 2^2 } \)
\( P(X = 7) = 36 \times \frac { 1 }{ 128 } \times \frac { 1 }{ 4 } \)
\( P(X = 7) = 36 \times \frac { 1 }{ 512 } \)
\( P(X = 7) = \frac { 36 }{ 512 } \)
\( P(X = 7) = \frac { 9 }{ 128 } \)
In simple words: For each case, we use the binomial probability formula. First, we find the chance of failure (\( q \)) from the chance of success (\( p \)). Then, we put all the numbers into the formula to find the chance of getting exactly \( k \) successes out of \( n \) tries. This formula helps us understand how likely an event is when we repeat it many times.

🎯 Exam Tip: Remember to simplify the combination \( ^nC_x \) and powers of fractions correctly. Always check that \( p+q=1 \) before calculating.

Question 2. The probability that Mr. Q hits a target any trial is \( \frac { 1 }{ 4 } \). Suppose he tries at the target 10 times. Find the probability that he hits the target (i) exactly 4 times (ii) atleast one time
Answer:
Let \( p \) be the probability of hitting the target.
Given: \( p = \frac { 1 }{ 4 } \).
Number of trials \( n = 10 \).
We can say \( X \) follows a binomial distribution \( B(n, p) \), so \( X \sim B(10, \frac { 1 }{ 4}) \).
The general formula for binomial distribution is \( P(X = x) = ^nC_x p^x q^{n-x} \), where \( x = 0, 1, 2, ..., n \).
First, we find \( q \):
\( q = 1 - p \)
\( q = 1 - \frac { 1 }{ 4 } \)
\( q = \frac { 3 }{ 4 } \)

(i) Probability of hitting the target exactly 4 times:
We need to find \( P(X = 4) \).
\( P(X = 4) = ^{10}C_4 \left( \frac { 1 }{ 4 } \right)^4 \left( \frac { 3 }{ 4 } \right)^{10-4} \)
\( P(X = 4) = ^{10}C_4 \left( \frac { 1 }{ 4 } \right)^4 \left( \frac { 3 }{ 4 } \right)^6 \)
\( P(X = 4) = \frac { 10 \times 9 \times 8 \times 7 }{ 4 \times 3 \times 2 \times 1 } \times \frac { 1^4 }{ 4^4 } \times \frac { 3^6 }{ 4^6 } \)
\( P(X = 4) = 210 \times \frac { 1 }{ 4^4 } \times \frac { 3^6 }{ 4^6 } \)
\( P(X = 4) = 210 \times \frac { 3^6 }{ 4^{10} } \)

(ii) Probability of hitting the target at least one time:
We need to find \( P(X \ge 1) \).
It is easier to calculate \( 1 - P(X < 1) \), which means \( 1 - P(X = 0) \).
\( P(X \ge 1) = 1 - P(X = 0) \)
\( P(X = 0) = ^{10}C_0 \left( \frac { 1 }{ 4 } \right)^0 \left( \frac { 3 }{ 4 } \right)^{10-0} \)
\( P(X = 0) = 1 \times 1 \times \left( \frac { 3 }{ 4 } \right)^{10} \)
\( P(X = 0) = \frac { 3^{10} }{ 4^{10} } \)
So, \( P(X \ge 1) = 1 - \frac { 3^{10} }{ 4^{10} } \)
In simple words: First, we find the chance that Mr. Q hits the target exactly 4 times out of 10 tries using the binomial formula. Then, to find the chance he hits it at least once, we subtract the chance of hitting it zero times from 1. This is often an easier way to solve "at least" problems.

🎯 Exam Tip: For "at least one" probabilities, it's usually faster to calculate \( 1 - P(\text{none}) \). Make sure to simplify fractions and exponents where possible.

Question 3. Using binomial distribution find the mean and variance of X for the following experiments
(i) A fair coin is tossed 100 times and X denotes the number of heads.
(ii) A fair die is tossed 240 times and X denotes the number of times that four appeared.
Answer:
(i) For a fair coin tossed 100 times, \( X \) is the number of heads.
Let \( p \) be the probability of getting heads.
Since it's a fair coin, \( p = \frac { 1 }{ 2 } \).
Number of trials \( n = 100 \).
We have \( X \sim B(n, p) \), so \( X \sim B(100, \frac { 1 }{ 2}) \).
First, we find \( q \):
\( q = 1 - p \)
\( q = 1 - \frac { 1 }{ 2 } \)
\( q = \frac { 1 }{ 2 } \)
Mean of a binomial distribution is given by \( np \):
Mean \( = np = 100 \times \frac { 1 }{ 2 } = 50 \).
Variance of a binomial distribution is given by \( npq \):
Variance \( = npq = 100 \times \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } = 50 \times \frac { 1 }{ 2 } = 25 \).

(ii) For a fair die tossed 240 times, \( X \) is the number of times that four appeared.
Let \( p \) be the probability of getting a 4 when a die is thrown.
For a fair die, there are 6 possible outcomes (1, 2, 3, 4, 5, 6), and only one is a 4.
So, \( p = \frac { 1 }{ 6 } \).
Number of trials \( n = 240 \).
We have \( X \sim B(n, p) \), so \( X \sim B(240, \frac { 1 }{ 6}) \).
First, we find \( q \):
\( q = 1 - p \)
\( q = 1 - \frac { 1 }{ 6 } \)
\( q = \frac { 5 }{ 6 } \)
Mean of a binomial distribution is given by \( np \):
Mean \( = np = 240 \times \frac { 1 }{ 6 } = 40 \).
Variance of a binomial distribution is given by \( npq \):
Variance \( = npq = 240 \times \frac { 1 }{ 6 } \times \frac { 5 }{ 6 } = 40 \times \frac { 5 }{ 6 } = \frac { 200 }{ 6 } = \frac { 100 }{ 3 } \).
In simple words: For binomial distribution, the mean tells us the average number of successes we expect, calculated by multiplying the number of tries by the success probability. The variance tells us how spread out these successes are likely to be, which is calculated by multiplying the number of tries, success probability, and failure probability. These calculations help us understand the typical outcome and variability of repeated experiments.

🎯 Exam Tip: Remember the formulas for mean (\( \mu = np \)) and variance (\( \sigma^2 = npq \)) for binomial distributions. Make sure to correctly identify \( n \) and \( p \) for each problem.

Question 4. The probability that a certain kind of component will survive an electrical test is \( \frac { 3 }{ 4 } \). Find the probability that exactly 3 of the 5 components tested survive.
Answer:
Let \( p \) be the probability that a component survives an electrical test.
Given: \( p = \frac { 3 }{ 4 } \).
Number of components tested \( n = 5 \).
We need to find the probability that exactly 3 components survive, so \( k = 3 \).
We have \( X \sim B(n, p) \), so \( X \sim B(5, \frac { 3 }{ 4}) \).
The general formula for binomial distribution is \( P(X = x) = ^nC_x p^x q^{n-x} \), where \( x = 0, 1, 2, ..., n \).
First, we find \( q \):
\( q = 1 - p \)
\( q = 1 - \frac { 3 }{ 4 } \)
\( q = \frac { 1 }{ 4 } \)
Now, we compute \( P(X = 3) \):
\( P(X = 3) = ^5C_3 \left( \frac { 3 }{ 4 } \right)^3 \left( \frac { 1 }{ 4 } \right)^{5-3} \)
\( P(X = 3) = ^5C_3 \left( \frac { 3 }{ 4 } \right)^3 \left( \frac { 1 }{ 4 } \right)^2 \)
\( P(X = 3) = \frac { 5 \times 4 \times 3 }{ 3 \times 2 \times 1 } \times \frac { 3^3 }{ 4^3 } \times \frac { 1^2 }{ 4^2 } \)
\( P(X = 3) = 10 \times \frac { 27 }{ 64 } \times \frac { 1 }{ 16 } \)
\( P(X = 3) = \frac { 10 \times 27 }{ 64 \times 16 } \)
\( P(X = 3) = \frac { 270 }{ 1024 } \)
\( P(X = 3) = \frac { 135 }{ 512 } \)
In simple words: We are looking for the chance that exactly 3 out of 5 parts pass a test, knowing the chance of one part passing. We use the binomial formula by figuring out the chance of a part failing, and then plugging all the values into the formula. This helps predict how many parts might work properly.

🎯 Exam Tip: Be careful with powers of fractions; remember to apply the power to both the numerator and the denominator. Simplify the combination \( ^nC_x \) carefully.

Question 5. A retailer purchases a certain kind of electronic, device from a manufacturer. The manufacturer indicates that the defective rate of the device is 5%. The inspector of the retailer randomly picks 10 items from a shipment. What is the probability that there will be
(i) atleast one defective item
(ii) exactly two defective items?
Answer:
Let \( p \) be the probability that an electronic device is defective.
Given: Defective rate is 5%, so \( p = 5\% = 0.05 \).
Number of items picked \( n = 10 \).
We have \( X \sim B(n, p) \), so \( X \sim B(10, 0.05) \).
The general formula for binomial distribution is \( P(X = x) = ^nC_x p^x q^{n-x} \), where \( x = 0, 1, 2, ..., n \).
First, we find \( q \):
\( q = 1 - p \)
\( q = 1 - 0.05 \)
\( q = 0.95 \)

(i) Probability of at least one defective item:
We need to find \( P(X \ge 1) \).
It is easier to calculate \( 1 - P(X < 1) \), which means \( 1 - P(X = 0) \).
\( P(X \ge 1) = 1 - P(X = 0) \)
\( P(X = 0) = ^{10}C_0 (0.05)^0 (0.95)^{10-0} \)
\( P(X = 0) = 1 \times 1 \times (0.95)^{10} \)
\( P(X = 0) = (0.95)^{10} \)
So, \( P(X \ge 1) = 1 - (0.95)^{10} \).

(ii) Probability of exactly two defective items:
We need to find \( P(X = 2) \).
\( P(X = 2) = ^{10}C_2 (0.05)^2 (0.95)^{10-2} \)
\( P(X = 2) = ^{10}C_2 (0.05)^2 (0.95)^8 \)
In simple words: When checking for defective items, the binomial distribution helps us. For "at least one defective," it's easier to find the chance of "zero defectives" and subtract that from 1. For "exactly two defectives," we directly use the binomial formula with the given probabilities and number of tries. This method helps businesses predict quality control results.

🎯 Exam Tip: Remember to convert percentages to decimals for probability calculations. For "at least" questions, the complement rule \( P(A) = 1 - P(A') \) is often more efficient.

Question 6. If the probability that fluorescent light has a useful life of atleast 600 hours is 0.9, find the probabilities that among 12 such lights
(i) exactly 10 will have a useful life of atleast 600 hours
(ii) At atleast 11 will have a useful life of atleast 600 hours
(iii) At atleast 2 will not have a useful life of at least 600 hours.
Answer:
Let \( p \) be the probability that a fluorescent light has a useful life of at least 600 hours.
Given: \( p = 0.9 \).
Number of lights \( n = 12 \).
We have \( X \sim B(n, p) \), so \( X \sim B(12, 0.9) \).
The general formula for binomial distribution is \( P(X = x) = ^nC_x p^x q^{n-x} \), where \( x = 0, 1, 2, ..., n \).
First, we find \( q \):
\( q = 1 - p \)
\( q = 1 - 0.9 \)
\( q = 0.1 \)

(i) Probability that exactly 10 lights will have a useful life of at least 600 hours:
We need to find \( P(X = 10) \).
\( P(X = 10) = ^{12}C_{10} (0.9)^{10} (0.1)^{12-10} \)
\( P(X = 10) = ^{12}C_{10} (0.9)^{10} (0.1)^2 \)

(ii) Probability that at least 11 lights will have a useful life of at least 600 hours:
We need to find \( P(X \ge 11) \).
This means \( P(X = 11) + P(X = 12) \).
\( P(X = 11) = ^{12}C_{11} (0.9)^{11} (0.1)^{12-11} = ^{12}C_{11} (0.9)^{11} (0.1)^1 \).
\( P(X = 12) = ^{12}C_{12} (0.9)^{12} (0.1)^{12-12} = ^{12}C_{12} (0.9)^{12} (0.1)^0 \).
So, \( P(X \ge 11) = ^{12}C_{11} (0.9)^{11} (0.1)^1 + ^{12}C_{12} (0.9)^{12} (0.1)^0 \)
\( P(X \ge 11) = 12 \times (0.9)^{11} \times 0.1 + 1 \times (0.9)^{12} \times 1 \)
\( P(X \ge 11) = (0.9)^{11} [12 \times 0.1 + 0.9] \)
\( P(X \ge 11) = (0.9)^{11} [1.2 + 0.9] \)
\( P(X \ge 11) = (0.9)^{11} (2.1) \)
\( P(X \ge 11) = 2.1 (0.9)^{11} \)

(iii) Probability that at least 2 lights will not have a useful life of at least 600 hours:
This means that at least 2 lights will fail to meet the 600-hour useful life mark. This is the same as saying at most 10 lights *will* have a useful life of at least 600 hours.
So, we need to find \( P(X \le 10) \).
It is easier to calculate \( 1 - P(X > 10) \).
This means \( 1 - [P(X = 11) + P(X = 12)] \).
From part (ii), we know \( P(X = 11) + P(X = 12) = 2.1 (0.9)^{11} \).
So, \( P(X \le 10) = 1 - 2.1 (0.9)^{11} \).
In simple words: This problem uses binomial distribution to calculate probabilities about light bulbs lasting a certain time. We find the chance for exactly 10 bulbs, then for at least 11 bulbs (meaning 11 or 12). For "at least 2 not lasting," it's easier to find the chance of "more than 10 lasting" and subtract that from 1. This helps in understanding product reliability.

🎯 Exam Tip: When dealing with "at least" or "at most" probabilities, draw a number line to visualize the range of events. Remember that \( P(X \le k) = 1 - P(X > k) \).

Question 7. The mean and standard deviation of a binomial variate X are respectively 6 and 2. Find
(i) The probability mass function
(ii) P(X = 3)
(iii) P(X ≥ 2).
Answer:
Given: Mean \( np = 6 \).
Standard deviation \( \sigma = 2 \).
We know that the variance \( \sigma^2 = (2)^2 = 4 \).
Also, the variance for a binomial distribution is \( npq \). So, \( npq = 4 \).
Now we can find \( q \) by dividing the variance by the mean:
\( \frac { npq }{ np } = \frac { 4 }{ 6 } \)
\( q = \frac { 2 }{ 3 } \)
Now we can find \( p \) using \( p = 1 - q \):
\( p = 1 - \frac { 2 }{ 3 } \)
\( p = \frac { 1 }{ 3 } \)
Now we find \( n \) using \( np = 6 \):
\( n \times \frac { 1 }{ 3 } = 6 \)
\( n = 6 \times 3 \)
\( n = 18 \)
So, for the binomial variate \( X \), we have \( X \sim B(18, \frac { 1 }{ 3}) \).

(i) The probability mass function (PMF) is given by:
\( P(X = x) = ^{18}C_x \left( \frac { 1 }{ 3 } \right)^x \left( \frac { 2 }{ 3 } \right)^{18-x} \), for \( x = 0, 1, 2, ..., 18 \).

(ii) To find \( P(X = 3) \):
Using the PMF from part (i):
\( P(X = 3) = ^{18}C_3 \left( \frac { 1 }{ 3 } \right)^3 \left( \frac { 2 }{ 3 } \right)^{18-3} \)
\( P(X = 3) = ^{18}C_3 \left( \frac { 1 }{ 3 } \right)^3 \left( \frac { 2 }{ 3 } \right)^{15} \)

(iii) To find \( P(X \ge 2) \):
It is easier to calculate \( 1 - P(X < 2) \).
\( P(X < 2) = P(X = 0) + P(X = 1) \).
\( P(X \ge 2) = 1 - [P(X = 0) + P(X = 1)] \)
First, find \( P(X = 0) \):
\( P(X = 0) = ^{18}C_0 \left( \frac { 1 }{ 3 } \right)^0 \left( \frac { 2 }{ 3 } \right)^{18-0} = 1 \times 1 \times \left( \frac { 2 }{ 3 } \right)^{18} = \left( \frac { 2 }{ 3 } \right)^{18} \).
Next, find \( P(X = 1) \):
\( P(X = 1) = ^{18}C_1 \left( \frac { 1 }{ 3 } \right)^1 \left( \frac { 2 }{ 3 } \right)^{18-1} = 18 \times \frac { 1 }{ 3 } \times \left( \frac { 2 }{ 3 } \right)^{17} = 6 \times \left( \frac { 2 }{ 3 } \right)^{17} \).
So, \( P(X \ge 2) = 1 - \left[ \left( \frac { 2 }{ 3 } \right)^{18} + 6 \left( \frac { 2 }{ 3 } \right)^{17} \right] \)
\( P(X \ge 2) = 1 - \left( \frac { 2 }{ 3 } \right)^{17} \left[ \frac { 2 }{ 3 } + 6 \right] \)
\( P(X \ge 2) = 1 - \left( \frac { 2 }{ 3 } \right)^{17} \left[ \frac { 2 + 18 }{ 3 } \right] \)
\( P(X \ge 2) = 1 - \left( \frac { 2 }{ 3 } \right)^{17} \left( \frac { 20 }{ 3 } \right) \)
In simple words: When we know the average (mean) and how spread out the data is (standard deviation) for a binomial distribution, we can work backward to find its basic parts: the number of trials (\( n \)) and the chance of success (\( p \)). Once we have these, we can write down the main probability rule (PMF) and then calculate the chance of specific outcomes, like getting exactly 3 successes or at least 2 successes. This shows how useful mean and standard deviation are for understanding distributions.

🎯 Exam Tip: Always remember that standard deviation is the square root of variance. To find \( n \) and \( p \) from mean and variance, set up equations \( np = \text{mean} \) and \( npq = \text{variance} \) and solve for \( p \) and then \( n \).

Question 8. If X ~ B(n, p) such that 4P(X = 4) = P(X = 2) and n = 6. Find the distribution, mean and standard deviation of X.
Answer:
Given: \( X \sim B(n, p) \) and \( n = 6 \).
The general probability mass function (PMF) is \( P(X = x) = ^nC_x p^x q^{n-x} \).
Given condition: \( 4P(X = 4) = P(X = 2) \).
Substitute the PMF into the condition with \( n=6 \):
\( 4 \times (^6C_4 p^4 q^{6-4}) = ^6C_2 p^2 q^{6-2} \)
\( 4 \times (^6C_4 p^4 q^2) = ^6C_2 p^2 q^4 \)
We know that \( ^6C_4 = \frac { 6 \times 5 }{ 2 \times 1 } = 15 \) and \( ^6C_2 = \frac { 6 \times 5 }{ 2 \times 1 } = 15 \).
So, \( 4 \times 15 p^4 q^2 = 15 p^2 q^4 \)
Divide both sides by \( 15 p^2 q^2 \) (assuming \( p \ne 0 \) and \( q \ne 0 \)):
\( 4 p^2 = q^2 \)
Since \( p = 1 - q \), substitute this into the equation:
\( 4 (1 - q)^2 = q^2 \)
\( 4 (1 - 2q + q^2) = q^2 \)
\( 4 - 8q + 4q^2 = q^2 \)
Rearrange into a quadratic equation:
\( 3q^2 - 8q + 4 = 0 \)
Factor the quadratic equation:
\( 3q^2 - 6q - 2q + 4 = 0 \)
\( 3q(q - 2) - 2(q - 2) = 0 \)
\( (3q - 2)(q - 2) = 0 \)
This gives two possible values for \( q \):
\( 3q - 2 = 0 \implies q = \frac { 2 }{ 3 } \)
\( q - 2 = 0 \implies q = 2 \)
Since probability \( q \) must be between 0 and 1 (inclusive), \( q=2 \) is not possible.
Therefore, \( q = \frac { 2 }{ 3 } \).
Now find \( p \):
\( p = 1 - q = 1 - \frac { 2 }{ 3 } = \frac { 1 }{ 3 } \).

**Distribution of X:**
The distribution is binomial with \( n = 6, p = \frac { 1 }{ 3 } \).
So, \( X \sim B(6, \frac { 1 }{ 3}) \).
The probability mass function is: \( P(X = x) = ^6C_x \left( \frac { 1 }{ 3 } \right)^x \left( \frac { 2 }{ 3 } \right)^{6-x} \), for \( x = 0, 1, 2, 3, 4, 5, 6 \).

**Mean of X:**
Mean \( = np = 6 \times \frac { 1 }{ 3 } = 2 \).

**Standard Deviation of X:**
Variance \( = npq = 6 \times \frac { 1 }{ 3 } \times \frac { 2 }{ 3 } = 2 \times \frac { 2 }{ 3 } = \frac { 4 }{ 3 } \).
Standard deviation \( = \sqrt{npq} = \sqrt{\frac { 4 }{ 3 }} = \frac { 2 }{ \sqrt{3} } \).
In simple words: We are given a relationship between the probabilities of getting 4 successes and 2 successes, along with the total number of tries. By using the binomial probability formula and solving a quadratic equation, we find the chance of failure (\( q \)) and then the chance of success (\( p \)). With \( n \) and \( p \), we can fully describe the distribution and calculate its average value (mean) and how spread out it is (standard deviation).

🎯 Exam Tip: When given a relationship between probabilities, substitute the binomial PMF and simplify algebraically to solve for \( p \) or \( q \). Always remember that probability values must be between 0 and 1.

Question 9. In a binomial distribution consisting of 5 independent trials, the probability of 1 and 2 successes are 0.4096 and 0.2048 respectively. Find the mean and variance of the random variable.
Answer:
Given: A binomial distribution with \( n = 5 \) independent trials.
\( P(X = 1) = 0.4096 \).
\( P(X = 2) = 0.2048 \).
Let \( p \) be the probability of success and \( q \) be the probability of failure, so \( q = 1 - p \).
The probability mass function is \( P(X = x) = ^nC_x p^x q^{n-x} \).

For \( P(X = 1) \):
\( P(X = 1) = ^5C_1 p^1 q^{5-1} \)
\( P(X = 1) = 5 p q^4 = 0.4096 \) ... (Equation 1)

For \( P(X = 2) \):
\( P(X = 2) = ^5C_2 p^2 q^{5-2} \)
\( P(X = 2) = 10 p^2 q^3 = 0.2048 \) ... (Equation 2)

Now, divide Equation 1 by Equation 2:
\( \frac { 5 p q^4 }{ 10 p^2 q^3 } = \frac { 0.4096 }{ 0.2048 } \)
Simplify the left side:
\( \frac { q }{ 2p } = 2 \)
\( q = 4p \)
Substitute \( q = 1 - p \) into this equation:
\( 1 - p = 4p \)
\( 1 = 5p \)
\( p = \frac { 1 }{ 5 } \)
Now find \( q \):
\( q = 1 - p = 1 - \frac { 1 }{ 5 } = \frac { 4 }{ 5 } \).

**Mean of the random variable:**
Mean \( = np = 5 \times \frac { 1 }{ 5 } = 1 \).

**Variance of the random variable:**
Variance \( = npq = 5 \times \frac { 1 }{ 5 } \times \frac { 4 }{ 5 } = 1 \times \frac { 4 }{ 5 } = \frac { 4 }{ 5 } \).

**The Distribution:**
The distribution is \( X \sim B(5, \frac { 1 }{ 5}) \).
The probability mass function is: \( P(X = x) = ^5C_x \left( \frac { 1 }{ 5 } \right)^x \left( \frac { 4 }{ 5 } \right)^{5-x} \), for \( x = 0, 1, 2, 3, 4, 5 \).
In simple words: When we know the number of trials and the probabilities for two different numbers of successes in a binomial distribution, we can set up two equations. By dividing these equations, we can solve for the chance of success (\( p \)) and failure (\( q \)). Once we have \( n, p, q \), we can easily find the mean (average successes) and variance (spread of successes) of the distribution.

🎯 Exam Tip: When given probabilities for different outcomes in a binomial distribution, setting up a system of equations and dividing them often simplifies the process to find \( p \) and \( q \).