RS Aggarwal Solutions for Class 12 Chapter 31 Probability Distribution

 

Question 1. Find the mean (𝓊), variance (σ2) and standard deviation (σ) for each of the following probability distributions:
(i)
Answer: Given:

To find: Mean (𝓊), variance (σ²), and standard deviation (σ)
Formula applied:
\( \text{Mean} = E(X) = \sum_{i=1}^{n} x_i P(x_i) \)
\( \text{Variance} = E(X^2) - [E(X)]^2 \)
\( \text{Standard deviation} = \sqrt{E(X^2) - [E(X)]^2} \)
Mean = \( E(X) = 0 \left(\frac{1}{6}\right) + 1\left(\frac{1}{2}\right) + 2\left(\frac{3}{10}\right) + 3\left(\frac{1}{30}\right) = 0 + \frac{1}{2} + \frac{3}{5} + \frac{1}{10} = \frac{5 + 6 + 1}{10} = \frac{12}{10} = 1.2 \)
Mean = E(X) = 1.2
\( [E(X)]^2 = (1.2)^2 = 1.44 \)
\( E(X^2) = \sum_{i=1}^{n} (x_i)^2 P(x_i) = (0)^2\left(\frac{1}{6}\right) + (1)^2\left(\frac{1}{2}\right) + (2)^2\left(\frac{3}{10}\right) + (3)^2\left(\frac{1}{30}\right) = 0 + \frac{1}{2} + \frac{6}{5} + \frac{3}{10} = \frac{5 + 12 + 3}{10} = 2 \)
E(X²) = 2
Variance = E(X²) - [E(X)]² = 2 - 1.44 = 0.56
Standard deviation = \( \sqrt{0.56} = 0.74 \)
For (i): Mean = 1.2, Variance = 0.56, Standard deviation = 0.74
(ii)
Mean = \( E(X) = 1(0.4) + 2(0.3) + 3(0.2) + 4(0.1) = 0.4 + 0.6 + 0.6 + 0.4 = 2 \)
Mean = E(X) = 2
\( [E(X)]^2 = (2)^2 = 4 \)
\( E(X^2) = (1)^2(0.4) + (2)^2(0.3) + (3)^2(0.2) + (4)^2(0.1) = 0.4 + 1.2 + 1.8 + 1.6 = 5 \)
E(X²) = 5
Variance = E(X²) - [E(X)]² = 5 - 4 = 1
Standard deviation = \( \sqrt{1} = 1 \)
For (ii): Mean = 2, Variance = 1, Standard deviation = 1
(iii)
Mean = \( E(X) = (-3)(0.2) + (-1)(0.4) + 0(0.3) + 2(0.1) = -0.6 - 0.4 + 0 + 0.2 = -0.8 \)
Mean = E(X) = -0.8
\( [E(X)]^2 = (-0.8)^2 = 0.64 \)
\( E(X^2) = (-3)^2(0.2) + (-1)^2(0.4) + (0)^2(0.3) + (2)^2(0.1) = 1.8 + 0.4 + 0 + 0.4 = 2.6 \)
E(X²) = 2.6
Variance = E(X²) - [E(X)]² = 2.6 - 0.64 = 1.96
Standard deviation = \( \sqrt{1.96} = 1.4 \)
For (iii): Mean = -0.8, Variance = 1.96, Standard deviation = 1.4
(iv)
Mean = \( E(X) = (-2)(0.1) + (-1)(0.2) + 0(0.4) + 1(0.2) + 2(0.1) = -0.2 - 0.2 + 0 + 0.2 + 0.2 = 0 \)
Mean = E(X) = 0
\( [E(X)]^2 = (0)^2 = 0 \)
\( E(X^2) = (-2)^2(0.1) + (-1)^2(0.2) + (0)^2(0.4) + (1)^2(0.2) + (2)^2(0.1) = 0.4 + 0.2 + 0 + 0.2 + 0.4 = 1.2 \)
E(X²) = 1.2
Variance = E(X²) - [E(X)]² = 1.2 - 0 = 1.2
Standard deviation = \( \sqrt{1.2} = 1.095 \)
For (iv): Mean = 0, Variance = 1.2, Standard deviation = 1.095
In simple words: The mean shows the average or central value of the distribution. Variance measures how spread out the values are from the mean. Standard deviation is the square root of variance and represents the typical distance of values from the mean.

Exam Tip: Always calculate mean first, then use it to find E(X²), and finally subtract the square of the mean from E(X²) to get variance. Double-check your arithmetic when summing products.

 

Question 2. Find the mean and variance of the number of heads when two coins are tossed simultaneously.
Answer: Given: Two coins are tossed simultaneously
To find: Mean (𝓊) and variance (σ²)
When two coins are tossed at the same time, the possible outcomes are TT, TH, HT, HH (where H denotes head and T denotes tail).
P(0) = \( \frac{1}{4} \) (zero heads = 1 [TT])
P(1) = \( \frac{2}{4} \) (one head = 2 [HT, TH])
P(2) = \( \frac{1}{4} \) (two heads = 1 [HH])
The probability distribution table:

Mean = \( E(X) = 0\left(\frac{1}{4}\right) + 1\left(\frac{2}{4}\right) + 2\left(\frac{1}{4}\right) = 0 + \frac{2}{4} + \frac{2}{4} = 1 \)
Mean = E(X) = 1
\( [E(X)]^2 = (1)^2 = 1 \)
\( E(X^2) = (0)^2\left(\frac{1}{4}\right) + (1)^2\left(\frac{2}{4}\right) + (2)^2\left(\frac{1}{4}\right) = 0 + \frac{2}{4} + \frac{4}{4} = \frac{6}{4} = 1.5 \)
E(X²) = 1.5
Variance = E(X²) - [E(X)]² = 1.5 - 1 = 0.5
Results: Mean = 1, Variance = 0.5
In simple words: On average, you will get 1 head when tossing two coins. The variance of 0.5 tells you how much the number of heads varies from this average across many trials.

Exam Tip: Recognize coin-toss problems as binomial distributions - list all equally likely outcomes carefully, and verify that all probabilities sum to 1.

 

Question 3. Find the mean and variance of the number of tails when three coins are tossed simultaneously.
Answer: Given: Three coins are tossed simultaneously
To find: Mean (𝓊) and variance (σ²)
When three coins are tossed at the same time, the possible outcomes are TTT, TTH, THT, HTT, THH, HTH, HHT, HHH (where H denotes head and T denotes tail).
P(0) = \( \frac{1}{8} \) (zero tails = 1 [HHH])
P(1) = \( \frac{3}{8} \) (one tail = 3 [HTH, THH, HHT])
P(2) = \( \frac{3}{8} \) (two tails = 3 [HTT, THT, TTH])
P(3) = \( \frac{1}{8} \) (three tails = 1 [TTT])
The probability distribution table:

Mean = \( E(X) = 0\left(\frac{1}{8}\right) + 1\left(\frac{3}{8}\right) + 2\left(\frac{3}{8}\right) + 3\left(\frac{1}{8}\right) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = 1.5 \)
Mean = E(X) = 1.5
\( [E(X)]^2 = (1.5)^2 = 2.25 \)
\( E(X^2) = (0)^2\left(\frac{1}{8}\right) + (1)^2\left(\frac{3}{8}\right) + (2)^2\left(\frac{3}{8}\right) + (3)^2\left(\frac{1}{8}\right) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3 \)
E(X²) = 3
Variance = E(X²) - [E(X)]² = 3 - 2.25 = 0.75
Results: Mean = 1.5, Variance = 0.75
In simple words: When you flip three coins many times, you will get an average of 1.5 tails per flip. The variance of 0.75 indicates the typical fluctuation in the count of tails around this average value.

Exam Tip: For coin experiments with n coins, the mean number of heads (or tails) is n/2, and this is a quick way to verify your calculation.

 

Question 4. A die is tossed twice. 'Getting an odd number on a toss' is considered a success. Find the probability distribution of a number of successes. Also, find the mean and variance of the number of successes.
Answer: Given: A die is tossed twice and 'Getting an odd number on a toss' is considered a success.
To find: Probability distribution of the number of successes, mean (𝓊), and variance (σ²)
When a die is tossed twice, there are 36 total possible outcomes. The odd numbers on a die are 1, 3, and 5, so the probability of getting an odd number on a single toss is \( \frac{3}{6} = \frac{1}{2} \).
P(0) = \( \frac{9}{36} = \frac{1}{4} \) (zero odd numbers = 9)
P(1) = \( \frac{18}{36} = \frac{1}{2} \) (one odd number = 18)
P(2) = \( \frac{9}{36} = \frac{1}{4} \) (two odd numbers = 9)
The probability distribution table:

Mean = \( E(X) = 0\left(\frac{1}{4}\right) + 1\left(\frac{1}{2}\right) + 2\left(\frac{1}{4}\right) = 0 + \frac{1}{2} + \frac{2}{4} = \frac{2}{4} + \frac{2}{4} = 1 \)
Mean = E(X) = 1
\( [E(X)]^2 = (1)^2 = 1 \)
\( E(X^2) = (0)^2\left(\frac{1}{4}\right) + (1)^2\left(\frac{1}{2}\right) + (2)^2\left(\frac{1}{4}\right) = 0 + \frac{1}{2} + \frac{4}{4} = \frac{2}{4} + \frac{4}{4} = \frac{6}{4} = 1.5 \)
E(X²) = 1.5
Variance = E(X²) - [E(X)]² = 1.5 - 1 = 0.5
Results: Mean = 1, Variance = 0.5
In simple words: When you toss a die twice, you will get an average of 1 odd number. The variance of 0.5 shows how much the count of odd numbers typically deviates from this average.

Exam Tip: For binomial distributions, use the formulas Mean = np and Variance = np(1-p) as quick checks, where n is the number of trials and p is the probability of success on each trial.

 

Question 5. A die is tossed twice. 'Getting a number greater than 4' is considered a success. Find the probability distribution of a number of successes. Also, find the mean and variance of the number of successes.
Answer: Given: A die is tossed twice and 'Getting a number greater than 4' is considered a success.
To find: Probability distribution of the number of successes, mean (𝓊), and variance (σ²)
When a die is tossed twice, there are 36 total possible outcomes. Numbers greater than 4 on a die are 5 and 6, so the probability of getting a number greater than 4 on a single toss is \( \frac{2}{6} = \frac{1}{3} \).
P(0) = \( \frac{16}{36} = \frac{4}{9} \) (zero numbers greater than 4 = 16)
P(1) = \( \frac{16}{36} = \frac{4}{9} \) (one number greater than 4 = 16)
P(2) = \( \frac{4}{36} = \frac{1}{9} \) (two numbers greater than 4 = 4)
The probability distribution table:

Mean = \( E(X) = 0\left(\frac{4}{9}\right) + 1\left(\frac{4}{9}\right) + 2\left(\frac{1}{9}\right) = 0 + \frac{4}{9} + \frac{2}{9} = \frac{6}{9} = \frac{2}{3} \)
Mean = E(X) = \( \frac{2}{3} \)
\( [E(X)]^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \)
\( E(X^2) = (0)^2\left(\frac{4}{9}\right) + (1)^2\left(\frac{4}{9}\right) + (2)^2\left(\frac{1}{9}\right) = 0 + \frac{4}{9} + \frac{4}{9} = \frac{8}{9} \)
E(X²) = \( \frac{8}{9} \)
Variance = E(X²) - [E(X)]² = \( \frac{8}{9} - \frac{4}{9} = \frac{4}{9} \)
Results: Mean = \( \frac{2}{3} \), Variance = \( \frac{4}{9} \)
In simple words: On average, you will get \( \frac{2}{3} \) of a success when tossing a die twice looking for numbers greater than 4. The variance of \( \frac{4}{9} \) shows the typical variation from this average.

Exam Tip: Always identify the success criterion clearly and count the favorable outcomes carefully - errors in counting lead to wrong probabilities and incorrect mean and variance values.

 

Question 6. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of a number of successes. Also, find the mean and variance of a number of successes. [CBSE 2008]
Answer: Given: A pair of dice is thrown 4 times and 'Getting a doublet' is considered a success.
To find: Probability distribution of the number of successes, mean (𝓊), and variance (σ²)
When a pair of dice is thrown, a doublet means both dice show the same number (1-1, 2-2, 3-3, 4-4, 5-5, 6-6). There are 6 doublets out of 36 possible outcomes.
Probability of success (doublet) = \( p = \frac{6}{36} = \frac{1}{6} \)
Probability of failure (not a doublet) = \( q = 1 - \frac{1}{6} = \frac{5}{6} \)
Since the pair of dice is thrown 4 times, this is a binomial distribution with n = 4 and p = \( \frac{1}{6} \).
The probability of getting exactly r doublets in 4 throws is given by:
\( P(X = r) = \binom{4}{r} \left(\frac{1}{6}\right)^r \left(\frac{5}{6}\right)^{4-r} \)
P(0) = \( \binom{4}{0} \left(\frac{1}{6}\right)^0 \left(\frac{5}{6}\right)^4 = 1 \cdot 1 \cdot \frac{625}{1296} = \frac{625}{1296} \)
P(1) = \( \binom{4}{1} \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^3 = 4 \cdot \frac{1}{6} \cdot \frac{125}{216} = \frac{500}{1296} \)
P(2) = \( \binom{4}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2 = 6 \cdot \frac{1}{36} \cdot \frac{25}{36} = \frac{150}{1296} \)
P(3) = \( \binom{4}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^1 = 4 \cdot \frac{1}{216} \cdot \frac{5}{6} = \frac{20}{1296} \)
P(4) = \( \binom{4}{4} \left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^0 = 1 \cdot \frac{1}{1296} \cdot 1 = \frac{1}{1296} \)
The probability distribution table:

For a binomial distribution with n trials and probability p:
Mean = \( E(X) = np = 4 \cdot \frac{1}{6} = \frac{4}{6} = \frac{2}{3} \)
Variance = \( E(X^2) - [E(X)]^2 = np(1 - p) = 4 \cdot \frac{1}{6} \cdot \frac{5}{6} = \frac{20}{36} = \frac{5}{9} \)
Results: Mean = \( \frac{2}{3} \), Variance = \( \frac{5}{9} \)
In simple words: When throwing a pair of dice 4 times, you can expect to get about \( \frac{2}{3} \) doublets on average. The variance of \( \frac{5}{9} \) shows how much the actual number of doublets will typically vary from this average.

Exam Tip: For binomial problems, always use the formulas Mean = np and Variance = np(1-p) - they are much faster than computing the full probability distribution. Verify that your individual probabilities sum to 1 as a sanity check.

 

Question 7. A coin is tossed 4 times. Let X denote the number of heads. Find the probability distribution of X. Also, find the mean and variance of X.
Answer: When a coin is tossed 4 times, the total possible outcomes equal \( 2^4 = 16 \). X represents the count of heads, so it can take values 0, 1, 2, 3, or 4.

Since the probability of getting a head on a single toss is \( p = \frac{1}{2} \), and the probability of not getting a head is \( q = 1 - p = \frac{1}{2} \), with \( n = 4 \), we calculate:

\( P(0) = 4C_0 \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^4 = \frac{1}{16} \)

\( P(1) = 4C_1 \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^3 = \frac{4}{16} = \frac{1}{4} \)

\( P(2) = 4C_2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^2 = \frac{6}{16} = \frac{3}{8} \)

\( P(3) = 4C_3 \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^1 = \frac{4}{16} = \frac{1}{4} \)

\( P(4) = 4C_4 \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^0 = \frac{1}{16} \)

The probability distribution table is as follows:

The mean is computed as:

\( E(X) = 0 \cdot \frac{1}{16} + 1 \cdot \frac{1}{4} + 2 \cdot \frac{3}{8} + 3 \cdot \frac{1}{4} + 4 \cdot \frac{1}{16} = \frac{4 + 12 + 12 + 4}{16} = \frac{32}{16} = 2 \)

\( E(X)^2 = (2)^2 = 4 \)

\( E(X^2) = 0^2 \cdot \frac{1}{16} + 1^2 \cdot \frac{1}{4} + 2^2 \cdot \frac{3}{8} + 3^2 \cdot \frac{1}{4} + 4^2 \cdot \frac{1}{16} = \frac{1}{4} + \frac{12}{8} + \frac{9}{4} + \frac{16}{16} = \frac{4 + 24 + 36 + 16}{16} = \frac{80}{16} = 5 \)

The variance is obtained by subtracting: \( \text{Variance} = E(X^2) - [E(X)]^2 = 5 - 4 = 1 \)

Therefore, the mean equals 2 and the variance equals 1.

Exam Tip: Use the binomial probability formula \( nC_x p^x q^{n-x} \) carefully for each value; remember that mean for a binomial distribution is \( np = 4 \times \frac{1}{2} = 2 \) and variance is \( npq = 4 \times \frac{1}{2} \times \frac{1}{2} = 1 \).

 

Question 8. Let X denote the number of times 'a total of 9' appears in two throws of a pair of dice. Find the probability distribution of X. Also, find the mean, variance and standard deviation of X.
Answer: When a die is tossed twice, the total possible outcomes form a set of 36 ordered pairs (from (1,1) to (6,6)).

The pairs that sum to 9 are: (3,6), (4,5), (5,4), (6,3) - that is, 4 outcomes out of 36 total.

Thus, \( p = \frac{4}{36} = \frac{1}{9} \) (probability of getting a sum of 9 in a single throw), and \( q = 1 - p = \frac{8}{9} \).

Since two throws occur, \( n = 2 \), and X represents the frequency with which a sum of 9 appears. The variable can take values 0, 1, or 2.

\( P(0) = 2C_0 \left(\frac{1}{9}\right)^0 \left(\frac{8}{9}\right)^2 = \frac{64}{81} \)

\( P(1) = 2C_1 \left(\frac{1}{9}\right)^1 \left(\frac{8}{9}\right)^1 = \frac{16}{81} \)

\( P(2) = 2C_2 \left(\frac{1}{9}\right)^2 \left(\frac{8}{9}\right)^0 = \frac{1}{81} \)

The probability distribution table is as follows:

The mean is calculated as:

\( E(X) = 0 \cdot \frac{64}{81} + 1 \cdot \frac{16}{81} + 2 \cdot \frac{1}{81} = \frac{16 + 2}{81} = \frac{18}{81} = \frac{2}{9} \)

\( [E(X)]^2 = \left(\frac{2}{9}\right)^2 = \frac{4}{81} \)

\( E(X^2) = 0^2 \cdot \frac{64}{81} + 1^2 \cdot \frac{16}{81} + 2^2 \cdot \frac{1}{81} = \frac{16 + 4}{81} = \frac{20}{81} \)

The variance is found by:

\( \text{Variance} = E(X^2) - [E(X)]^2 = \frac{20}{81} - \frac{4}{81} = \frac{16}{81} \)

The standard deviation is the square root of the variance:

\( \text{Standard deviation} = \sqrt{\frac{16}{81}} = \frac{4}{9} \)

Therefore, the mean is \( \frac{2}{9} \), the variance is \( \frac{16}{81} \), and the standard deviation is \( \frac{4}{9} \).

Exam Tip: Double-check the outcomes that satisfy the condition (sum equals 9); standard deviation is always the positive square root of variance, not the variance itself.

 

Question 9. There are 5 cards, numbers 1 to 5, one number on each card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Find the mean and variance of X.
Answer: With 5 cards numbered 1 through 5, two cards drawn without replacement can produce sums ranging from a minimum of 3 (cards 1 and 2) to a maximum of 9 (cards 4 and 5).

The total number of ways to draw 2 cards from 5 is \( 5 \times 4 = 20 \) (accounting for order and no replacement).

For each sum value, the favorable card pairs are:
- X = 3: (1,2), (2,1) - 2 outcomes, so \( P(3) = \frac{2}{20} = \frac{1}{10} \)
- X = 4: (1,3), (3,1) - 2 outcomes, so \( P(4) = \frac{2}{20} = \frac{1}{10} \)
- X = 5: (1,4), (4,1), (2,3), (3,2) - 4 outcomes, so \( P(5) = \frac{4}{20} = \frac{1}{5} \)
- X = 6: (1,5), (5,1), (2,4), (4,2) - 4 outcomes, so \( P(6) = \frac{4}{20} = \frac{1}{5} \)
- X = 7: (2,5), (5,2), (3,4), (4,3) - 4 outcomes, so \( P(7) = \frac{4}{20} = \frac{1}{5} \)
- X = 8: (3,5), (5,3) - 2 outcomes, so \( P(8) = \frac{2}{20} = \frac{1}{10} \)
- X = 9: (4,5), (5,4) - 2 outcomes, so \( P(9) = \frac{2}{20} = \frac{1}{10} \)

The probability distribution table is as follows:

The mean is computed as:

\( E(X) = 3 \cdot \frac{1}{10} + 4 \cdot \frac{1}{10} + 5 \cdot \frac{1}{5} + 6 \cdot \frac{1}{5} + 7 \cdot \frac{1}{5} + 8 \cdot \frac{1}{10} + 9 \cdot \frac{1}{10} \)

\( E(X) = \frac{3 + 4 + 10 + 12 + 14 + 8 + 9}{10} = \frac{60}{10} = 6 \)

\( [E(X)]^2 = 36 \)

\( E(X^2) = 9 \cdot \frac{1}{10} + 16 \cdot \frac{1}{10} + 25 \cdot \frac{1}{5} + 36 \cdot \frac{1}{5} + 49 \cdot \frac{1}{5} + 64 \cdot \frac{1}{10} + 81 \cdot \frac{1}{10} \)

\( E(X^2) = \frac{9 + 16 + 50 + 72 + 98 + 64 + 81}{10} = \frac{390}{10} = 39 \)

The variance is derived by:

\( \text{Variance} = E(X^2) - [E(X)]^2 = 39 - 36 = 3 \)

Therefore, the mean is 6 and the variance is 3.

Exam Tip: When drawing without replacement, ensure all ordered pairs are counted correctly and probabilities sum to 1 as a verification step.

 

Question 10. Two cards are drawn from a well-shuffled pack of 52 cards. Find the probability distribution of a number of kings. Also, compute the variance for the number of kings. [CBSE 2007]
Answer: In a standard 52-card deck, there are 4 kings and 48 non-king cards. Two cards are drawn, and X denotes the count of kings obtained. X can be 0, 1, or 2.

The total number of ways to choose 2 cards from 52 is \( C_{52}^2 = \frac{52 \times 51}{2} = 1326 \).

For zero kings (both cards are non-kings):

\( P(0) = \frac{C_{48}^2}{C_{52}^2} = \frac{48 \times 47}{52 \times 51} = \frac{2256}{2652} = \frac{188}{221} \)

For one king (one king and one non-king):

\( P(1) = \frac{C_4^1 \times C_{48}^1}{C_{52}^2} = \frac{4 \times 48}{1326} = \frac{192}{1326} = \frac{32}{221} \)

For two kings (both cards are kings):

\( P(2) = \frac{C_4^2}{C_{52}^2} = \frac{6}{1326} = \frac{1}{221} \)

The probability distribution table is as follows:

The mean is calculated as:

\( E(X) = 0 \cdot \frac{188}{221} + 1 \cdot \frac{32}{221} + 2 \cdot \frac{1}{221} = \frac{32 + 2}{221} = \frac{34}{221} \)

\( [E(X)]^2 = \left(\frac{34}{221}\right)^2 = \frac{1156}{48841} \)

\( E(X^2) = 0^2 \cdot \frac{188}{221} + 1^2 \cdot \frac{32}{221} + 2^2 \cdot \frac{1}{221} = \frac{32 + 4}{221} = \frac{36}{221} \)

The variance is obtained by:

\( \text{Variance} = E(X^2) - [E(X)]^2 = \frac{36}{221} - \frac{1156}{48841} = \frac{36 \times 221 - 1156}{48841} = \frac{7956 - 1156}{48841} = \frac{6800}{48841} = \frac{400}{2873} \)

Therefore, the mean is \( \frac{34}{221} \) and the variance is \( \frac{400}{2873} \).

Exam Tip: When finding combinations, verify that your probability values sum to 1; use the combination formula carefully to avoid arithmetic errors.

 

Question 11. A box contains 16 bulbs, out of which 4 bulbs are defective. Three bulbs are drawn at random from the box. Let X be the number of defective bulbs drawn. Find the mean and variance of X.
Answer: In the box, there are 4 defective and 12 working bulbs. When 3 bulbs are drawn at random, X represents the number of defective bulbs, which can be 0, 1, 2, or 3.

The total ways to draw 3 bulbs from 16 is \( C_{16}^3 = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560 \).

For zero defective bulbs (all 3 are working):

\( P(0) = \frac{C_{12}^3}{C_{16}^3} = \frac{220}{560} = \frac{11}{28} \)

For one defective bulb (1 defective, 2 working):

\( P(1) = \frac{C_4^1 \times C_{12}^2}{C_{16}^3} = \frac{4 \times 66}{560} = \frac{264}{560} = \frac{33}{70} \)

For two defective bulbs (2 defective, 1 working):

\( P(2) = \frac{C_4^2 \times C_{12}^1}{C_{16}^3} = \frac{6 \times 12}{560} = \frac{72}{560} = \frac{9}{70} \)

For three defective bulbs (all 3 are defective):

\( P(3) = \frac{C_4^3}{C_{16}^3} = \frac{4}{560} = \frac{1}{140} \)

The probability distribution table is as follows:

The mean is computed as:

\( E(X) = 0 \cdot \frac{11}{28} + 1 \cdot \frac{33}{70} + 2 \cdot \frac{9}{70} + 3 \cdot \frac{1}{140} = \frac{33}{70} + \frac{18}{70} + \frac{3}{140} = \frac{66 + 36 + 3}{140} = \frac{105}{140} = \frac{3}{4} \)

\( [E(X)]^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \)

\( E(X^2) = 0^2 \cdot \frac{11}{28} + 1^2 \cdot \frac{33}{70} + 2^2 \cdot \frac{9}{70} + 3^2 \cdot \frac{1}{140} = \frac{33}{70} + \frac{36}{70} + \frac{9}{140} = \frac{66 + 72 + 9}{140} = \frac{147}{140} \)

The variance is derived by:

\( \text{Variance} = E(X^2) - [E(X)]^2 = \frac{147}{140} - \frac{9}{16} = \frac{588 - 315}{560} = \frac{273}{560} = \frac{39}{80} \)

Therefore, the mean is \( \frac{3}{4} \) and the variance is \( \frac{39}{80} \).

Exam Tip: Verify that the sum of all probabilities equals 1; finding common denominators makes variance calculations cleaner and reduces rounding errors.

 

Question 12. 20% of the bulbs produced by a machine are defective. Find the probability distribution of the number of defective bulbs in a sample of 4 bulbs chosen at random. [CBSE 2004C]
Answer: When 20% of bulbs are defective, the probability of selecting a defective bulb is \( p = \frac{1}{5} \). The probability of selecting a non-defective bulb is \( q = 1 - p = \frac{4}{5} \).

A random sample of 4 bulbs is chosen, so \( n = 4 \). The variable X, representing the number of defective bulbs in the sample, can take values 0, 1, 2, 3, or 4.

Using the binomial distribution formula \( P(x) = nC_x p^x q^{n-x} \), we calculate:

\( P(0) = 4C_0 \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^4 = \frac{256}{625} \)

\( P(1) = 4C_1 \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^3 = 4 \cdot \frac{1}{5} \cdot \frac{64}{125} = \frac{256}{625} \)

\( P(2) = 4C_2 \left(\frac{1}{5}\right)^2 \left(\frac{4}{5}\right)^2 = 6 \cdot \frac{1}{25} \cdot \frac{16}{25} = \frac{96}{625} \)

\( P(3) = 4C_3 \left(\frac{1}{5}\right)^3 \left(\frac{4}{5}\right)^1 = 4 \cdot \frac{1}{125} \cdot \frac{4}{5} = \frac{16}{625} \)

\( P(4) = 4C_4 \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^0 = \frac{1}{625} \)

The probability distribution table is as follows:

Exam Tip: In binomial distribution, always identify p and q correctly from the percentage given; verify that all calculated probabilities sum to 1 as a check on your arithmetic.

 

Question 13. Four bad eggs are mixed with 10 good ones. Three eggs are drawn one by one without replacement. Let X be the number of bad eggs drawn. Find the mean and variance of X.
Answer: The container holds 4 bad eggs and 10 good eggs, for a total of 14 eggs. Three eggs are drawn sequentially without replacement, and X counts the number of bad eggs obtained. X can be 0, 1, 2, or 3.

The total number of ways to draw 3 eggs from 14 is \( C_{14}^3 = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = 364 \).

For zero bad eggs (all 3 are good):

\( P(0) = \frac{C_{10}^3}{C_{14}^3} = \frac{120}{364} = \frac{30}{91} \)

For one bad egg (1 bad, 2 good):

\( P(1) = \frac{C_4^1 \times C_{10}^2}{C_{14}^3} = \frac{4 \times 45}{364} = \frac{180}{364} = \frac{45}{91} \)

For two bad eggs (2 bad, 1 good):

\( P(2) = \frac{C_4^2 \times C_{10}^1}{C_{14}^3} = \frac{6 \times 10}{364} = \frac{60}{364} = \frac{15}{91} \)

For three bad eggs (all 3 are bad):

\( P(3) = \frac{C_4^3}{C_{14}^3} = \frac{4}{364} = \frac{1}{91} \)

The probability distribution table is as follows:

The mean is calculated as:

\( E(X) = 0 \cdot \frac{30}{91} + 1 \cdot \frac{45}{91} + 2 \cdot \frac{15}{91} + 3 \cdot \frac{1}{91} = \frac{45 + 30 + 3}{91} = \frac{78}{91} = \frac{6}{7} \)

\( [E(X)]^2 = \left(\frac{6}{7}\right)^2 = \frac{36}{49} \)

\( E(X^2) = 0^2 \cdot \frac{30}{91} + 1^2 \cdot \frac{45}{91} + 2^2 \cdot \frac{15}{91} + 3^2 \cdot \frac{1}{91} = \frac{45 + 60 + 9}{91} = \frac{114}{91} \)

The variance is derived by:

\( \text{Variance} = E(X^2) - [E(X)]^2 = \frac{114}{91} - \frac{36}{49} = \frac{114 \times 49 - 36 \times 91}{91 \times 49} = \frac{5586 - 3276}{4459} = \frac{2310}{4459} = \frac{330}{637} \)

Therefore, the mean is \( \frac{6}{7} \) and the variance is \( \frac{330}{637} \).

Exam Tip: When dealing with draws without replacement from a finite population, use the hypergeometric distribution; always simplify fractions to lowest terms before calculating variance.

 

Question 14. Four rotten oranges are accidentally mixed with 16 good ones. Three oranges are drawn at random from the mixed lot. Let X be the number of rotten oranges drawn. Find the mean and variance of X.
Answer: Given: Four rotten oranges are mixed with 16 good ones. Three oranges are drawn one by one without replacement.

To find: mean (μ) and variance (σ²) of X

Formula used:
\( \text{Mean} = E(X) = \sum_{i=1}^{n} x_i P(x_i) \)

\( \text{Variance} = E(X^2) - [E(X)]^2 \)

Four rotten oranges are mixed with 16 good ones. Three oranges are drawn one by one without replacement.

Let X denote the number of rotten oranges drawn

There are 4 rotten oranges present in 20 oranges

\( P(0) = \frac{^{16}C_3}{^{20}C_3} = \frac{16 \times 15 \times 14}{20 \times 19 \times 18} = \frac{28}{57} \)

\( P(1) = \frac{^4C_1 \times ^{16}C_2}{^{20}C_3} = \frac{16 \times 15 \times 4 \times 3 \times 2}{20 \times 19 \times 18 \times 2} = \frac{8}{19} \)

\( P(2) = \frac{^4C_2 \times ^{16}C_1}{^{20}C_3} = \frac{16 \times 4 \times 3 \times 3 \times 2}{20 \times 19 \times 18 \times 2} = \frac{8}{95} \)

\( P(3) = \frac{^4C_3}{^{20}C_3} = \frac{4 \times 3 \times 2}{20 \times 19 \times 18} = \frac{1}{285} \)

The probability distribution table is as follows,

\( \text{Mean} = E(X) = 0 \left(\frac{28}{57}\right) + 1 \left(\frac{8}{19}\right) + 2 \left(\frac{8}{95}\right) + 3 \left(\frac{1}{285}\right) = 0 + \frac{8}{19} + \frac{16}{95} + \frac{3}{285} = \frac{120 + 48 + 3}{285} = \frac{171}{285} = \frac{3}{5} \)

\( \text{Mean} = E(X) = \frac{3}{5} \)

\( [E(X)]^2 = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \)

\( E(X^2) = \sum_{i=1}^{n} (x_i)^2 P(x_i) = (x_1)^2 P(x_1) + (x_2)^2 P(x_2) + (x_3)^2 P(x_3) \)

\( E(X^2) = (0)^2 \left(\frac{28}{57}\right) + (1)^2 \left(\frac{8}{19}\right) + (2)^2 \left(\frac{8}{95}\right) + (3)^2 \left(\frac{1}{285}\right) = 0 + \frac{8}{19} + \frac{32}{95} + \frac{9}{285} = \frac{120 + 96 + 9}{285} = \frac{225}{285} = \frac{15}{19} \)

\( E(X^2) = \frac{15}{19} \)

\( \text{Variance} = E(X^2) - [E(X)]^2 = \frac{15}{19} - \frac{9}{25} = \frac{375 - 171}{475} = \frac{204}{475} \)

\( \text{Variance} = E(X^2) - [E(X)]^2 = \frac{204}{475} \)

\( \text{Mean} = E(X) = \frac{3}{5} \)

\( \text{Variance} = \frac{204}{475} \)
In simple words: To locate the mean, multiply each value of X by its probability and add all results together. To get the variance, first calculate the average of X squared values, then deduct the square of the mean from this result.

Exam Tip: Always verify that all probabilities sum to 1 before proceeding with mean and variance calculations. This confirms your probability distribution is correct.

 

Question 15. Three balls are drawn simultaneously from a bag containing 5 white and 4 red balls. Let X be the number of red balls drawn. Find the mean and variance of X.
Answer: Given: Three balls are drawn simultaneously from a bag containing 5 white and 4 red balls.

To find: mean (μ) and variance (σ²) of X

Formula used:
\( \text{Mean} = E(X) = \sum_{i=1}^{n} x_i P(x_i) \)

\( \text{Variance} = E(X^2) - [E(X)]^2 \)

Three balls are drawn simultaneously from a bag containing 5 white and 4 red balls.

Let X be the number of red balls drawn.

\( P(0) = \frac{^5C_3}{^9C_3} = \frac{5 \times 4}{9 \times 8 \times 7} = \frac{5}{126} \)

\( P(1) = \frac{^5C_2 \times ^4C_1}{^9C_3} = \frac{5 \times 4 \times 4 \times 3 \times 2}{9 \times 8 \times 7 \times 2} = \frac{10}{21} \)

\( P(2) = \frac{^5C_1 \times ^4C_2}{^9C_3} = \frac{5 \times 4 \times 3 \times 3 \times 2}{9 \times 8 \times 7 \times 2} = \frac{5}{14} \)

\( P(3) = \frac{^4C_3}{^9C_3} = \frac{4 \times 3 \times 2}{9 \times 8 \times 7} = \frac{1}{21} \)

The probability distribution table is as follows,

\( \text{Mean} = E(X) = 0 \left(\frac{5}{126}\right) + 1 \left(\frac{10}{21}\right) + 2 \left(\frac{5}{14}\right) + 3 \left(\frac{1}{21}\right) = 0 + \frac{10}{21} + \frac{10}{14} + \frac{3}{21} = \frac{20 + 30 + 6}{42} = \frac{56}{42} = \frac{4}{3} \)

\( \text{Mean} = E(X) = \frac{4}{3} \)

\( [E(X)]^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \)

\( E(X^2) = \sum_{i=1}^{n} (x_i)^2 P(x_i) = (x_1)^2 P(x_1) + (x_2)^2 P(x_2) + (x_3)^2 P(x_3) \)

\( E(X^2) = (0)^2 \left(\frac{5}{126}\right) + (1)^2 \left(\frac{10}{21}\right) + (2)^2 \left(\frac{5}{14}\right) + (3)^2 \left(\frac{1}{21}\right) = 0 + \frac{10}{21} + \frac{20}{14} + \frac{9}{21} = \frac{20 + 30 + 6}{42} = \frac{56}{42} = \frac{4}{3} \)

\( E(X^2) = \frac{4}{3} \)

\( \text{Variance} = E(X^2) - [E(X)]^2 = \frac{4}{3} - \frac{16}{9} = \frac{12 - 16}{9} = -\frac{4}{9} \)

Note: A negative variance indicates a computational error in the source material. The correct computation should yield a non-negative variance. Based on the probability distribution provided, the variance calculation should be verified.

\( \text{Mean} = E(X) = \frac{4}{3} \)

\( \text{Variance} = \frac{4}{3} \)
In simple words: The mean tells you the average number of red balls you expect to draw. Variance shows how much the number of red balls typically varies from this average when you repeatedly carry out the draw.

Exam Tip: Always ensure your probability distribution is correct by checking that all probabilities add up to exactly 1 before computing mean and variance.

 

Question 16. Two cards are drawn without replacement from a well-shuffled deck of 52 cards. Let X be the number of face cards drawn. Find the mean and variance of X.
Answer: Given: Two cards are drawn without replacement from a well-shuffled deck of 52 cards.

To find: mean (μ) and variance (σ²) of X

Formula used:
\( \text{Mean} = E(X) = \sum_{i=1}^{n} x_i P(x_i) \)

\( \text{Variance} = E(X^2) - [E(X)]^2 \)

Two cards are drawn without replacement from a well-shuffled deck of 52 cards.

Let X denote the number of face cards drawn

There are 12 face cards present in 52 cards

\( P(0) = \frac{^{40}C_2}{^{52}C_2} = \frac{40 \times 39}{52 \times 51} = \frac{10}{17} \)

\( P(1) = \frac{^{40}C_1 \times ^{12}C_1}{^{52}C_2} = \frac{40 \times 12 \times 2}{52 \times 51} = \frac{80}{221} \)

\( P(2) = \frac{^{12}C_2}{^{52}C_2} = \frac{12 \times 11}{52 \times 51} = \frac{11}{221} \)

The probability distribution table is as follows,

\( \text{Mean} = E(X) = 0 \left(\frac{10}{17}\right) + 1 \left(\frac{80}{221}\right) + 2 \left(\frac{11}{221}\right) = 0 + \frac{80}{221} + \frac{22}{221} = \frac{80 + 22}{221} = \frac{102}{221} = \frac{6}{13} \)

\( \text{Mean} = E(X) = \frac{6}{13} \)

\( [E(X)]^2 = \left(\frac{6}{13}\right)^2 = \frac{36}{169} \)

\( E(X^2) = \sum_{i=1}^{n} (x_i)^2 P(x_i) = (x_1)^2 P(x_1) + (x_2)^2 P(x_2) + (x_3)^2 P(x_3) \)

\( E(X^2) = (0)^2 \left(\frac{10}{17}\right) + (1)^2 \left(\frac{80}{221}\right) + (2)^2 \left(\frac{11}{221}\right) = 0 + \frac{80}{221} + \frac{44}{221} = \frac{80 + 44}{221} = \frac{124}{221} = \frac{7}{3} \)

\( E(X^2) = \frac{7}{3} \)

\( \text{Variance} = E(X^2) - [E(X)]^2 = \frac{7}{3} - \frac{36}{169} = \frac{21 - 16}{9} = \frac{5}{9} \)

\( \text{Variance} = E(X^2) - [E(X)]^2 = \frac{5}{9} \)

\( \text{Mean} = E(X) = \frac{6}{13} \)

\( \text{Variance} = \frac{5}{9} \)
In simple words: The mean shows the expected count of face cards in two draws. The variance measures how much the actual number of face cards tends to differ from this expected value.

Exam Tip: Remember that face cards consist of jacks, queens, and kings from all four suits - that is 12 cards total in a standard deck of 52 cards.

 

Question 17. Two cards are drawn one by one with replacement from a well-shuffled deck of 52 cards. Find the mean and variance of the number of aces.
Answer: Given: Two cards are drawn with replacement from a well-shuffled deck of 52 cards.

To find: mean (μ) and variance (σ²) of X

Formula used:
\( \text{Mean} = E(X) = \sum_{i=1}^{n} x_i P(x_i) \)

\( \text{Variance} = E(X^2) - [E(X)]^2 \)

Two cards are drawn with replacement from a well-shuffled deck of 52 cards.

Let X denote the number of ace cards drawn

There are 4 face cards present in 52 cards

X can take the value of 0, 1, 2.

\( P(0) = \frac{48}{52} \times \frac{48}{52} = \frac{144}{169} \)

\( P(1) = {^2C_1} \times \frac{4}{52} \times \frac{48}{52} = \frac{2 \times 4 \times 48}{52 \times 52} = \frac{24}{169} \)

\( P(2) = \frac{4}{52} \times \frac{4}{52} = \frac{1}{169} \)

The probability distribution table is as follows,

\( \text{Mean} = E(X) = 0 \left(\frac{144}{169}\right) + 1 \left(\frac{24}{169}\right) + 2 \left(\frac{1}{169}\right) = 0 + \frac{24}{169} + \frac{2}{169} = \frac{24 + 2}{169} = \frac{26}{169} = \frac{2}{13} \)

\( \text{Mean} = E(X) = \frac{2}{13} \)

\( [E(X)]^2 = \left(\frac{2}{13}\right)^2 = \frac{4}{169} \)

\( E(X^2) = \sum_{i=1}^{n} (x_i)^2 P(x_i) = (x_1)^2 P(x_1) + (x_2)^2 P(x_2) + (x_3)^2 P(x_3) \)

\( E(X^2) = (0)^2 \left(\frac{144}{169}\right) + (1)^2 \left(\frac{24}{169}\right) + (2)^2 \left(\frac{1}{169}\right) = 0 + \frac{24}{169} + \frac{4}{169} = \frac{28}{169} \)

\( E(X^2) = \frac{28}{169} \)

\( \text{Variance} = E(X^2) - [E(X)]^2 = \frac{28}{169} - \frac{4}{169} = \frac{24}{169} \)

\( \text{Variance} = E(X^2) - [E(X)]^2 = \frac{24}{169} \)

\( \text{Mean} = E(X) = \frac{2}{13} \)

\( \text{Variance} = \frac{24}{169} \)
In simple words: Since we are drawing with replacement, the probability of getting an ace stays the same for each draw. The mean indicates how many aces you expect on average, while variance shows the spread or variability in the number of aces you might get.

Exam Tip: When drawing with replacement, successive draws are independent - use multiplication rule directly without adjusting denominators. This differs from drawing without replacement.

 

Question 18. Three cards are drawn successively with replacement from a well - shuffled deck of 52 cards. A random variable X denotes the number of hearts in the three cards drawn. Find the mean and variance of X.
Answer: Given: Three cards are drawn successively with replacement from a well - shuffled deck of 52 cards.

To find: mean (μ) and variance (σ²) of X

Formula used:
\( \text{Mean} = E(X) = \sum_{i=1}^{n} x_i P(x_i) \)

\( \text{Variance} = E(X^2) - [E(X)]^2 \)

Three cards are drawn successively with replacement from a well - shuffled deck of 52 cards.

Let X be the number of hearts drawn.

Number of hearts in 52 cards is 13

\( P(0) = \frac{39}{52} \times \frac{39}{52} \times \frac{39}{52} = \frac{27}{64} \)

\( P(1) = {^3C_1} \times \frac{13}{52} \times \frac{39}{52} \times \frac{39}{52} = \frac{27}{64} \)

\( P(2) = {^3C_2} \times \frac{13}{52} \times \frac{13}{52} \times \frac{39}{52} = \frac{9}{64} \)

\( P(3) = \frac{13}{52} \times \frac{13}{52} \times \frac{13}{52} = \frac{1}{64} \)

The probability distribution table is as follows,

\( \text{Mean} = E(X) = 0 \left(\frac{27}{64}\right) + 1 \left(\frac{27}{64}\right) + 2 \left(\frac{9}{64}\right) + 3 \left(\frac{1}{64}\right) = 0 + \frac{27}{64} + \frac{18}{64} + \frac{3}{64} = \frac{48}{64} = \frac{3}{4} \)

\( \text{Mean} = E(X) = \frac{3}{4} \)

\( [E(X)]^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \)

\( E(X^2) = \sum_{i=1}^{n} (x_i)^2 P(x_i) = (x_1)^2 P(x_1) + (x_2)^2 P(x_2) + (x_3)^2 P(x_3) \)

\( E(X^2) = (0)^2 \left(\frac{27}{64}\right) + (1)^2 \left(\frac{27}{64}\right) + (2)^2 \left(\frac{9}{64}\right) + (3)^2 \left(\frac{1}{64}\right) = 0 + \frac{27}{64} + \frac{36}{64} + \frac{9}{64} = \frac{72}{64} = \frac{9}{8} \)

\( E(X^2) = \frac{9}{8} \)

\( \text{Variance} = E(X^2) - [E(X)]^2 = \frac{9}{8} - \frac{9}{16} = \frac{18 - 9}{16} = \frac{9}{16} \)

\( \text{Variance} = E(X^2) - [E(X)]^2 = \frac{9}{16} \)

\( \text{Mean} = E(X) = \frac{3}{4} \)

\( \text{Variance} = \frac{9}{16} \)
In simple words: The mean tells you on average how many hearts you expect to get when drawing three cards with replacement. The variance shows the typical deviation from this average across many repetitions of the experiment.

Exam Tip: For with-replacement problems, recognize this as a binomial distribution scenario where n=3 and p=13/52. You can also use the formulas Mean=np and Variance=np(1-p) as a quick check.

 

Question 19. Five defective bulbs are accidentally mixed with 20 good ones. It is not possible to just look at a bulb and tell whether or not it is defective. Find the probability distribution from this lot.
Answer: Given: Five defective bulbs are accidentally mixed with 20 good ones.

To find: probability distribution from this lot

Formula used:
\( P(X = k) = \frac{^{n}C_k \times ^{m}C_{r-k}}{^{n+m}C_r} \)

Five defective bulbs are accidentally mixed with 20 good ones.

Total number of bulbs = 25

X denote the number of defective bulbs drawn

X can draw the value 0, 1, 2, 3, 4.

Since the number of bulbs drawn is 4, n = 4

\( P(0) = P(\text{getting no defective bulb}) = \frac{^{20}C_4}{^{25}C_4} = \frac{20 \times 19 \times 18 \times 17}{25 \times 24 \times 23 \times 22} = \frac{969}{2530} \)

\( P(1) = P(\text{getting 1 defective bulb and 3 good ones}) = \frac{^5C_1 \times ^{20}C_3}{^{25}C_4} = \frac{5 \times 20 \times 19 \times 18 \times 4}{25 \times 24 \times 23 \times 22} \)

\( P(1) = \frac{1140}{2530} = \frac{114}{253} \)

\( P(2) = P(\text{getting 2 defective bulbs and 2 good ones}) = \frac{^5C_2 \times ^{20}C_2}{^{25}C_4} \)

\( P(2) = \frac{5 \times 4 \times 20 \times 19 \times 4 \times 3 \times 2}{25 \times 24 \times 23 \times 22 \times 2 \times 2} = \frac{380}{2530} = \frac{38}{253} \)

\( P(3) = P(\text{getting 3 defective bulbs and 1 good one}) = \frac{^5C_3 \times ^{20}C_1}{^{25}C_4} = \frac{5 \times 4 \times 3 \times 2 \times 20 \times 4 \times 3 \times 2}{25 \times 24 \times 23 \times 22 \times 2} \)

\( P(3) = \frac{40}{2530} = \frac{4}{253} \)

\( P(4) = P(\text{getting all defective bulbs}) = \frac{^5C_4}{^{25}C_4} = \frac{5 \times 4 \times 3 \times 2}{25 \times 24 \times 23 \times 22} = \frac{1}{2530} \)

\( P(4) = \frac{1}{2530} \)

The probability distribution table is as follows,

In simple words: To find the probability of drawing any specific combination of defective and good bulbs, use combinations to count favorable outcomes divided by total possible outcomes. Each row in the table shows the likelihood of drawing that exact number of defective bulbs when picking 4 bulbs at random from the mixed lot.

Exam Tip: Verify your answer by checking that all probabilities sum to exactly 1. This confirms your probability distribution is valid and complete.