Question

A bag contains $3$ green and $7$ white balls. Two balls are drawn one by one at random without replacement. If the second ball drawn is green, what is the probability that the first ball drawn is also green?

• A. $\frac{5}{9}$
• B. $\frac{4}{9}$
• C. $\frac{2}{9}$
• D. $\frac{8}{9}$

Answer 1

Answer: (C)

Let $E_1$, $E_2$ and $A$ be the events defined as follows :
$E_1 =$ first ball drawn is green,
$E_2 =$ first ball drawn is white and
$A =$ second ball drawn is green
As the bag contains $3$ green and $7$ white balls,
$P(E_1) =\frac{3}{10}$ and $P(E_2) =\frac{7}{10}$
$P(A|E_1) = $ probability of drawing second green ball when one green ball has already been drawn
$= \frac{2}{9}$
$P(A|E_2) = $ probability of drawing second green ball when one white ball has already been drawn
$= \frac{3}{9} = \frac{1}{3}$
We want to find $P(E_1|A)$.
By Bayes' theorem, we have

$P\left(E_{1}|A\right) = \frac{P\left(E_{1}\right)P\left(A |E_{1}\right)}{P \left(E_{1}\right)P\left(A|E_{1}\right) + P\left(E_{2}\right)P\left(A|E_{2}\right)}$
$=\frac{ \frac{3}{10}\cdot\frac{2}{9}}{\frac{3}{10}\cdot\frac{2}{9}+\frac{7}{10}\cdot\frac{1}{3}}= \frac{\frac{2}{3}}{\frac{2}{3}+\frac{7}{3}}$
$= \frac{2}{3}\times\frac{3}{9}= \frac{2}{9}$