Question

A bag contain 4 red balls and 6 white balls two balls are drawn without replacement
(I)find the probability that the second ball is white given that the first ball is red
(Ii) find the probability that the second ball is red given that the first ball is white

Answer 1

(|)

The first ball was red. So now there are 3 red balls and 6 white balls in bag.

$P(W_{second}|R_{first}) = \frac{6}{3+6} = \frac{2}{3}$

Answer: 2/3

(||)

The first ball was white. So now there are 4 red balls and 5 white balls in bag.

$P(R_{second}|W_{first}) = \frac{4}{4+5} = \frac{4}{9}$

​ Answer: 4/9