Question

A bag contains 4 green and 6 white balls. Two balls are drawn one by one without replacement.If the second ball drawn is white, what is the probability that the first ball drawn is also white?

Answer 1

Let $E_1$ : first ball drawn is white, $E_2$ : first ball drawn is green, A : second ball drawn is white.

$\therefore P\left(E_1\right)=\frac{6}{10},P\left(E_2\right)=\frac{4}{10},P\left(A|E_1\right)=\frac{5}{9}$ and $P\left(A|E_2\right)=\frac{6}{9}.$

By Bayes' theorem, $P\left(E_1|A\right)=\frac{P\left(E_1\right)P\left(A|E_1\right)}{P\left(E_1\right)P\left(A|E_1\right)+P\left(E_2\right)P\left(A|E_2\right)}$

$\Rightarrow =\frac{\frac{6}{10}\times \frac{5}{9}}{\frac{6}{10}\times \frac{5}{9}+\frac{4}{10}\times \frac{6}{9}}=\frac{5}{9}.$