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

Let us define the events: - Event 1: A white ball is drawn from Bag I. - Event 2: A black ball is drawn from Bag I. - Event 3: A white ball is drawn from Bag II after transferring the ball from Bag I. We want to find the probability of drawing a white ball from Bag II after transferring a ball from Bag I. This can be done using the law of total probability.
Step 1: Probability of drawing a white ball from Bag I The probability of drawing a white ball from Bag I is: \[ P(\text{White from Bag I}) = \frac{4}{9} \quad \text{(since there are 4 white balls out of 9 total balls in Bag I)} \] The probability of drawing a black ball from Bag I is: \[ P(\text{Black from Bag I}) = \frac{5}{9} \quad \text{(since there are 5 black balls out of 9 total balls in Bag I)} \]
Step 2: Conditional probability of drawing a white ball from Bag II If a white ball is transferred to Bag II, Bag II will contain 7 white balls and 7 black balls, for a total of 14 balls. The probability of drawing a white ball 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 contain 6 white balls and 8 black balls, for a total of 14 balls. The probability of drawing a white ball from Bag II is: \[ P(\text{White from Bag II} | \text{Black transferred}) = \frac{6}{14} = \frac{3}{7} \] 
Step 3: Total probability Now, using the law of total probability, we can calculate the total probability of drawing a white ball from Bag II: \[ P(\text{White from Bag II}) = P(\text{White from Bag I}) \cdot P(\text{White from Bag II} | \text{White transferred}) + P(\text{Black from Bag I}) \cdot P(\text{White from Bag II} | \text{Black transferred}) \] Substitute the values: \[ P(\text{White from Bag II}) = \frac{4}{9} \cdot \frac{1}{2} + \frac{5}{9} \cdot \frac{3}{7} \] \[ = \frac{4}{18} + \frac{15}{63} = \frac{14}{63} + \frac{15}{63} = \frac{29}{63} \] Thus, the probability that the ball drawn is white is: \[ \boxed{\frac{29}{63}} \]