Two balls are selected at random one by one without replacement from a bag containing 4 wh…

Question

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 $\dfrac{m}{n}$, where $\gcd(m,n)=1$, then $m+n$ is equal to:

Accepted Answer

In a bag containing 4 white and 6 black balls, two balls are selected one by one without replacement. We need to find the probability that the first ball is black, given that the second ball is black. Let's define the events:

A: The first ball selected is black.

B: The second ball selected is black.

The probability $ P(A|B) $ is given by the formula:

$ P(A|B) = \frac{P(A \cap B)}{P(B)} $

Let's compute each part separately:

Probability of both balls being black $ P(A \cap B) $:

The chance that the first ball is black is $\frac{6}{10}$. Given that one black ball is removed, the probability that the second is also black is $\frac{5}{9}$.

$    P(A \cap B) = \frac{6}{10} 	imes \frac{5}{9} = \frac{30}{90} = \frac{1}{3}    $

Probability that the second ball is black $ P(B) $:

This can occur if either the first ball was white or black:

First ball white, second black: $\frac{4}{10} 	imes \frac{6}{9}$

Both balls black: $\frac{6}{10} 	imes \frac{5}{9}$

$    P(B) = \frac{4}{10} 	imes \frac{6}{9} + \frac{6}{10} 	imes \frac{5}{9} = \frac{24}{90} + \frac{30}{90} = \frac{54}{90} = \frac{3}{5}    $

Now, using these probabilities:

$ P(A|B) = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3} 	imes \frac{5}{3} = \frac{5}{9} $

Here, $ \frac{5}{9} $ is in its simplest form ($\operatorname{gcd}(5, 9) = 1$), so $ m = 5$ and $ n = 9 $.

Therefore, $ m + n = 5 + 9 = 14 $.

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