Question:
Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability that the ball drawn is white is \(\frac{29}{45}\), then n is equal to:
Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability that the ball drawn is white is \(\frac{29}{45}\), then n is equal to:
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In probability problems involving multiple stages, apply conditional probability rules carefully to avoid errors.
Updated On: Mar 24, 2025
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The Correct Option is D
Solution and Explanation
Step 1: Probability Calculation.
Probability of choosing a white ball from Bag 1 and adding it to Bag 2: \[ P(W \text{ from Bag 1}) = \frac{4}{9} \] Probability of choosing a black ball from Bag 1 and adding it to Bag 2: \[ P(B \text{ from Bag 1}) = \frac{5}{9} \] Now, probability of choosing a white ball from Bag 2: \[ P(W \text{ from Bag 2}) = \frac{n + 1}{n + 4} \times \frac{4}{9} + \frac{n}{n + 4} \times \frac{5}{9} = \frac{29}{45} \] Cross multiplying and simplifying, we find: \[ n = 6 \]
Probability of choosing a white ball from Bag 1 and adding it to Bag 2: \[ P(W \text{ from Bag 1}) = \frac{4}{9} \] Probability of choosing a black ball from Bag 1 and adding it to Bag 2: \[ P(B \text{ from Bag 1}) = \frac{5}{9} \] Now, probability of choosing a white ball from Bag 2: \[ P(W \text{ from Bag 2}) = \frac{n + 1}{n + 4} \times \frac{4}{9} + \frac{n}{n + 4} \times \frac{5}{9} = \frac{29}{45} \] Cross multiplying and simplifying, we find: \[ n = 6 \]
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