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.
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:
\[
n = 6
\]
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Approach Solution -2
Step 1: Understand the problem.
Bag 1 contains 4 white and 5 black balls.
Bag 2 contains n white and 3 black balls.
A ball is transferred from Bag 1 to Bag 2, and then a ball is drawn from Bag 2.
The probability that the drawn ball is white is given as \( \frac{29}{45} \).
Step 2: Define events.
Let:
E₁ = Event that a white ball is transferred from Bag 1.
E₂ = Event that a black ball is transferred from Bag 1.
Probability of transferring a white ball from Bag 1:
\[
P(E₁) = \frac{4}{9}, \quad P(E₂) = \frac{5}{9}.
\]
Step 3: Calculate probability of drawing a white ball from Bag 2.
Case 1: If a white ball is transferred, Bag 2 will have \((n + 1)\) white and 3 black balls.
\[
P(\text{White from Bag 2} | E₁) = \frac{n + 1}{(n + 1) + 3} = \frac{n + 1}{n + 4}.
\]
Case 2: If a black ball is transferred, Bag 2 will have \(n\) white and \((3 + 1) = 4\) black balls.
\[
P(\text{White from Bag 2} | E₂) = \frac{n}{n + 4}.
\]
