Question

A bag A contains 3 red, 2 white and 2 black balls and another bag B contains 1 red, 2 white and 4 black balls. A die is thrown to select a bag. If an odd prime number appears on the die, a ball is drawn from bag A; otherwise from bag B. If the ball drawn is black, then the probability that it is drawn from bag B is

• A. \(\frac{6}{7}\)
• B. \(\frac{3}{7}\)
• C. \(\frac{1}{5}\)
• D. \(\frac{4}{5}\)

Hint:

When probability asks “given that”, immediately think Bayes theorem and reverse the conditional probability.

Answer 1

The Correct Option is A

Concept: Use Bayes theorem \[ P(A|B)=\frac{P(A)\cdot P(B|A)}{P(B)} \] where conditional probability reverses known information.

Step 1: Probability of selecting bags.
Odd prime outcomes on die: \[ 2,3,5 \] Condition says odd prime selects bag A. Thus favorable die outcomes: \[ 3,5 \] Hence \[ P(A)=\frac26=\frac13 \] Remaining outcomes choose bag B \[ P(B)=\frac46=\frac23 \]

Step 2: Probability of black ball from each bag.
Bag A contains total \[ 7 \] Black balls \[ 2 \] Thus \[ P(Black|A)=\frac27 \] Bag B total \[ 7 \] Black balls \[ 4 \] Thus \[ P(Black|B)=\frac47 \]

Step 3: Total probability of black ball.
\[ P(Black) = P(A)P(Black|A)+P(B)P(Black|B) \] \[ = \frac13\times\frac27+\frac23\times\frac47 \] \[ = \frac{2}{21}+\frac{8}{21} = \frac{10}{21} \]

Step 4: Apply Bayes theorem.
Required probability \[ P(B|Black) = \frac{P(B)P(Black|B)}{P(Black)} \] \[ = \frac{\frac23\times\frac47}{\frac{10}{21}} \] \[ = \frac{8}{10} = \frac45 \] Hence \[ \boxed{\frac45} \]