Question ID: #569
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:
- (1) 3
- (2) 4
- (3) 5
- (4) 6
Solution:
Let $E_1$ be the event that a white ball is transferred from Bag 1 to Bag 2.
$$ P(E_1) = \frac{4}{9} $$
Let $E_2$ be the event that a black ball is transferred from Bag 1 to Bag 2.
$$ P(E_2) = \frac{5}{9} $$
Let $W$ be the event that a white ball is drawn from Bag 2.
If $E_1$ occurs, Bag 2 contains $(n+1)$ white and 3 black balls (Total $n+4$).
$$ P(W|E_1) = \frac{n+1}{n+4} $$
If $E_2$ occurs, Bag 2 contains $n$ white and 4 black balls (Total $n+4$).
$$ P(W|E_2) = \frac{n}{n+4} $$
Using the Theorem of Total Probability:
$$ P(W) = P(E_1)P(W|E_1) + P(E_2)P(W|E_2) $$
$$ \frac{29}{45} = \frac{4}{9} \left( \frac{n+1}{n+4} \right) + \frac{5}{9} \left( \frac{n}{n+4} \right) $$
Multiply by 45 to simplify:
$$ 29 = 5 \left[ 4 \left( \frac{n+1}{n+4} \right) + 5 \left( \frac{n}{n+4} \right) \right] $$
Wait, cancelling the denominator 45 against 9 leaves a factor of 5 on the LHS? No, $45 \times \frac{1}{9} = 5$. So:
$$ 29 = 5 \left[ \frac{4n + 4 + 5n}{n+4} \right] $$
$$ 29 = 5 \left[ \frac{9n + 4}{n+4} \right] $$
$$ 29(n+4) = 5(9n+4) $$
$$ 29n + 116 = 45n + 20 $$
$$ 16n = 96 $$
$$ n = 6 $$
Ans. (4)
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