Question

A bag contains 4 red and 5 black balls. Another bag contains 3 red and 6 black balls. If one ball is drawn from the first bag and two balls from the second bag at random, the probability that out of the three, two are black and one is red, is:

• A. \( \frac{20}{27} \)
• B. \( \frac{17}{18} \)
• C. \( \frac{25}{54} \)
• D. \( \frac{25}{108} \)

Hint:

For probability problems involving selection from multiple sources, break the solution into cases and use combinatorial counting. The total probability is the sum of the probabilities of all successful cases.

Answer 1

The Correct Option is C

Step 1: Define Probabilities of Drawing from Each Bag Let:
- \( B_1 \) be the first bag containing 4 red and 5 black balls.
- \( B_2 \) be the second bag containing 3 red and 6 black balls.
We need to find the probability of drawing exactly two black balls and one red ball.
Step 2: Possible Cases to Get (2 Black, 1 Red)
There are two ways to achieve this:
1. Drawing a black ball from \( B_1 \) and one black, one red from \( B_2 \).
2. Drawing a red ball from \( B_1 \) and two black balls from \( B_2 \).
Case 1: One Black from \( B_1 \) and (One Black, One Red) from \( B_2 \)
- Probability of drawing a black ball from \( B_1 \): \[ P(B_1 = B) = \frac{5}{9}. \] - Probability of drawing one black and one red from \( B_2 \): \[ P(BR \text{ from } B_2) = \frac{\text{Ways to choose 1 black from 6 and 1 red from 3}}{\text{Ways to choose any 2 from 9}}. \] \[ = \frac{\binom{6}{1} \binom{3}{1}}{\binom{9}{2}} = \frac{6 \times 3}{36} = \frac{18}{36} = \frac{1}{2}. \] Thus, the probability for this case: \[ P(\text{Case 1}) = \frac{5}{9} \times \frac{1}{2} = \frac{5}{18}. \] Case 2: One Red from \( B_1 \) and Two Black from \( B_2 \) - Probability of drawing a red ball from \( B_1 \): \[ P(B_1 = R) = \frac{4}{9}. \] - Probability of drawing two black balls from \( B_2 \): \[ P(BB \text{ from } B_2) = \frac{\binom{6}{2}}{\binom{9}{2}} = \frac{15}{36} = \frac{5}{12}. \] Thus, the probability for this case: \[ P(\text{Case 2}) = \frac{4}{9} \times \frac{5}{12} = \frac{20}{108} = \frac{10}{54}. \] Step 3: Compute the Final Probability Adding both cases: \[ P(\text{Final}) = P(\text{Case 1}) + P(\text{Case 2}). \] \[ = \frac{5}{18} + \frac{10}{54}. \] Converting to a common denominator: \[ = \frac{15}{54} + \frac{10}{54} = \frac{25}{54}. \] Thus, the correct answer is \( \mathbf{\frac{25}{54}} \).