A bag contains 10 black and 5 white balls. Two balls are drawn randomly one by one without replacement from the bag. If it is known that at least one ball is black then find the probability that both are black?
For the above question , I thought that since atleast one ball we already know is black then of remaining 14 balls(9 of which will be black) that ball drawn must be black so Probability = 9/14
But the correct way
A : Atleast one ball is black ,P(A)=19/21
B : Both balls are black, P(B) = 3/7
then P(B/A) = 9/19 is the correct probability. I not able to figure out what is wrong with the first method?
The probability that at least one ball is black = $\Pr(B)$
$$\Pr(B) = \frac{5}{15} \cdot \frac{10}{14} + \frac{10}{15} \cdot \frac{5}{14} + \frac{10}{15} \cdot \frac{9}{14} = \frac{19}{21}$$
The probability that both balls are black and at least one ball is black = The probability that both balls are black = $\Pr(E \cap B) = \frac{10}{15} \cdot \frac{9}{14}$
The probability that both balls are black given that at least one ball is black = $P(E|B)$
$$\Pr(E|B) = \frac{P(E \cap B)}{P(B)} = \frac{9}{19}$$
Phew ...