Question

Suppose that
Box-I contains 8 red, 3 blue and 5 green balls,
Box-II contains 24 red, 9 blue and 15 green balls,
Box-III contains 1 blue, 12 green and 3 yellow balls,
Box-IV contains 10 green, 16 orange and 6 white balls.
A ball is chosen randomly from Box-I ; call this ball $b$ If $b$ is red then a ball is chosen randomly from Box-II, if $b$ is blue then a ball is chosen randomly from Box-III, and if $b$ is green then a ball is chosen randomly from Box-IV The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen balls is green' has happened, is equal to

• A. $\frac{15}{256}$
• B. $\frac{3}{16}$
• C. $\frac{5}{52}$
• D. $\frac{1}{8}$

Answer 1

The Correct Option is C

The correct option is (C): \(\frac{5}{52}\)

Answer 2

To solve the problem, we need to determine the conditional probability \( P(A | B) \), where:
- \( A \) is the event that one of the chosen balls is white.
- \( B \) is the event that at least one of the chosen balls is green.

Let's calculate the probability step by step:
 Step 1: Determine the probabilities for choosing balls from Box I

Box I contains:
- 8 red balls
- 3 blue balls
- 5 green balls

Total balls in Box I = 8 + 3 + 5 = 16
 Probability of choosing each color from Box I:
- \( P(R) = \frac{8}{16} = \frac{1}{2} \)
- \( P(B) = \frac{3}{16} \)
- \( P(G) = \frac{5}{16} \)

 Step 2: Determine the probabilities of events \( A \) and \( B \)
 Event \( A \) (one of the chosen balls is white):
If a ball \( b \) chosen from Box I is:
- Red: Choose from Box II, which has no white balls.
- Blue: Choose from Box III, which has no white balls.
- Green: Choose from Box IV, which has 6 white balls out of 32.

Thus, \( A \) can only occur if the ball chosen from Box I is green, and then a white ball is chosen from Box IV.

The probability of choosing a white ball from Box IV is:
\[ P(W | G) = \frac{6}{32} = \frac{3}{16} \]

 Step 3: Calculate the probability of event \( B \) (at least one of the chosen balls is green)

Event \( B \) can occur in two ways:
1. \( b \) is green from Box I.
2. \( b \) is not green from Box I, but a green ball is chosen from Box II or Box III.

To calculate \( P(B) \):

Case 1: \( b \) is green from Box I
- Probability = \( P(G) = \frac{5}{16} \)
- The ball chosen from Box IV is either green or white (but not affecting \( B \)).

Case 2: \( b \) is not green from Box I, and another green ball is chosen:
- Probability of \( b \) being red = \( P(R) = \frac{8}{16} = \frac{1}{2} \)
 - From Box II: Probability of choosing a green ball = \( \frac{15}{48} = \frac{5}{16} \)

- Probability of \( b \) being blue = \( P(B) = \frac{3}{16} \)
 - From Box III: Probability of choosing a green ball = \( \frac{12}{16} = \frac{3}{4} \)

So,
\[ P(B) = P(G) + P(R) \times \frac{15}{48} + P(B) \times \frac{12}{16} \]
\[ P(B) = \frac{5}{16} + \left(\frac{1}{2} \times \frac{5}{16}\right) + \left(\frac{3}{16} \times \frac{3}{4}\right) \]
\[ P(B) = \frac{5}{16} + \frac{5}{32} + \frac{9}{64} \]
\[ P(B) = \frac{20}{64} + \frac{10}{64} + \frac{9}{64} = \frac{39}{64} \]

 Step 4: Calculate \( P(A \cap B) \) (one of the chosen balls is white and at least one is green)

This can only happen if:
- \( b \) is green from Box I and a white ball is chosen from Box IV.

\[ P(A \cap B) = P(G) \times P(W | G) = \frac{5}{16} \times \frac{3}{16} = \frac{15}{256} \]

 Step 5: Calculate the conditional probability \( P(A | B) \)

\[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{15}{256}}{\frac{39}{64}} = \frac{15}{256} \times \frac{64}{39} = \frac{15}{156} = \frac{5}{52} \]

Thus, the conditional probability that one of the chosen balls is white given that at least one of the chosen balls is green is \( \boxed{\frac{5}{52}} \).
So The correct Answer is  option (C): \(\frac{5}{52}\)