Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is $\frac{m}{n}$ where $\gcd(m,n)=1$, then $m+n$ is equal to:
- (1) 14
- (2) 4
- (3) 11
- (4) 13
Solution:
Let $B_1$ be the event that the first ball is black, and $B_2$ be the event that the second ball is black.
We need to find $P(B_1 | B_2) = \frac{P(B_1 \cap B_2)}{P(B_2)}$.
Total balls = 10 (4 White, 6 Black).
$P(B_1 \cap B_2)$ (Both Black):
$$ = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} $$
$P(B_2)$ (Second is Black):
This can happen in two ways: (White, Black) or (Black, Black).
$$ P(B_2) = P(W_1 B_2) + P(B_1 B_2) $$
$$ = \left(\frac{4}{10} \times \frac{6}{9}\right) + \left(\frac{6}{10} \times \frac{5}{9}\right) $$
$$ = \frac{24}{90} + \frac{30}{90} = \frac{54}{90} $$
Required Probability:
$$ P(B_1 | B_2) = \frac{30/90}{54/90} = \frac{30}{54} = \frac{5}{9} $$
Here $m=5, n=9$. Since $\gcd(5,9)=1$, these are the required values.
$m+n = 5 + 9 = 14$.
Ans. (1)