Question

Bag I contains 4 white and 5 black balls. Bag II contains 6 white and 7 black balls. A ball drawn randomly from Bag I is transferred to Bag II and then a ball is drawn randomly from Bag II. Find the probability that the ball drawn is white.

Hint:

Use the law of total probability when an event depends on the outcome of previous events, such as transferring balls between bags.

Answer 1

Define events: - Event 1: Draw a white ball from Bag I. - Event 2: Draw a black ball from Bag I. - Event 3: Draw a white ball from Bag II after transferring a ball from Bag I. We aim to compute the probability of drawing a white ball from Bag II post-transfer using the law of total probability.
Step 1: Initial Probabilities in Bag I. The probability of drawing a white ball from Bag I is: \[ P(\text{White from Bag I}) = \frac{4}{9} \quad \text{(4 white balls out of 9 total)} \] The probability of drawing a black ball from Bag I is: \[ P(\text{Black from Bag I}) = \frac{5}{9} \quad \text{(5 black balls out of 9 total)} \]
Step 2: Conditional Probabilities for Bag II. If a white ball is transferred to Bag II, Bag II will have 7 white and 7 black balls (total 14). The probability of drawing white from Bag II is: \[ P(\text{White from Bag II} | \text{White transferred}) = \frac{7}{14} = \frac{1}{2} \] If a black ball is transferred to Bag II, Bag II will have 6 white and 8 black balls (total 14). The probability of drawing white from Bag II is: \[ P(\text{White from Bag II} | \text{Black transferred}) = \frac{6}{14} = \frac{3}{7} \]
Step 3: Total Probability Calculation. Applying the law of total probability to find the overall probability of drawing a white ball from Bag II: \[ P(\text{White from Bag II}) = P(\text{White from Bag I}) \cdot P(\text{White from Bag II} | \text{White transferred}) + P(\text{Black from Bag I}) \cdot P(\text{White from Bag II} | \text{Black transferred}) \] Substituting values: \[ P(\text{White from Bag II}) = \frac{4}{9} \cdot \frac{1}{2} + \frac{5}{9} \cdot \frac{3}{7} \] \[ = \frac{4}{18} + \frac{15}{63} = \frac{14}{63} + \frac{15}{63} = \frac{29}{63} \] The probability of drawing a white ball is: \[ \boxed{\frac{29}{63}} \]