Question

Suppose that there are 5 boxes, each containing 3 blue pens, 1 red pen and 2 black pens. One pen is drawn at random from each of these 5 boxes. If the random variable \( X_1 \) denotes the total number of blue pens drawn and the random variable \( X_2 \) denotes the total number of red pens drawn, then \( P(X_1 = 2, X_2 = 1) \) equals:

• A. \( \frac{5}{36} \)
• B. \( \frac{5}{18} \)
• C. \( \frac{5}{12} \)
• D. \( \frac{5}{9} \)

Hint:

- To solve such problems, calculate the probability for each condition (blue, red, black pens) and multiply the results by the number of ways each condition can occur.
- Use the combination formula \( \binom{n}{k} \) to calculate the number of ways to choose items from a group.

Answer 1

The Correct Option is A

1) Understanding the Problem:
Each of the 5 boxes contains 3 blue pens, 1 red pen, and 2 black pens. A pen is drawn randomly from each box. We need to calculate the probability that exactly 2 blue pens and exactly 1 red pen are drawn from these 5 boxes. 
2) Probability of Drawing Blue and Red Pens:
The probability of drawing a blue pen from a single box is \( P(\text{Blue}) = \frac{3}{6} = \frac{1}{2} \). 
The probability of drawing a red pen from a single box is \( P(\text{Red}) = \frac{1}{6} \). 
The probability of drawing a black pen from a single box is \( P(\text{Black}) = \frac{2}{6} = \frac{1}{3} \).
3) Applying the Given Conditions:
We want to find the probability that exactly 2 blue pens and exactly 1 red pen are drawn. Out of the 5 boxes, we need to select 2 boxes to draw blue pens, 1 box to draw a red pen, and the remaining 2 boxes will draw black pens.
The number of ways to choose 2 boxes for blue pens from 5 is \( \binom{5}{2} = 10 \). Similarly, the number of ways to choose 1 box for the red pen from the remaining 3 boxes is \( \binom{3}{1} = 3 \). The remaining 2 boxes will automatically have black pens.
4) Calculating the Probability:
The total probability is given by: \[ P(X_1 = 2, X_2 = 1) = \binom{5}{2} \times \binom{3}{1} \times \left( \frac{1}{2} \right)^2 \times \left( \frac{1}{6} \right) \times \left( \frac{1}{3} \right)^2 \] Substituting the values: \[ P(X_1 = 2, X_2 = 1) = 10 \times 3 \times \left( \frac{1}{2} \right)^2 \times \left( \frac{1}{6} \right) \times \left( \frac{1}{3} \right)^2 \] \[ P(X_1 = 2, X_2 = 1) = 30 \times \frac{1}{4} \times \frac{1}{6} \times \frac{1}{9} \] \[ P(X_1 = 2, X_2 = 1) = \frac{30}{216} = \frac{5}{36} \] Thus, the correct answer is (A) \( \frac{5}{36} \).