Question

A box contains 4 white balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is

• A. 1/3
• B. 3/7
• C. 1/2
• D. 4/7

Answer 1

The correct option is C 1/2

P ($1^{st}$ ball is white and $2^{nd}$ is red),

$p_1=\frac{4}{7}\times \frac{3}{6}=\frac{2}{7}$

$\begin{array}{c}4W,\\ 3R\end{array}$
$P\left(1^{st}balliswhite\right),p_2=\frac{4}{7}$

P(2nd ball is red/1st ball is white)

$=\frac{p_1}{p_2}=\frac{\frac{2}{7}}{\frac{4}{7}}=\frac{1}{2}$

Method II:
After removing 1st white ball, remaining balls in bag are 3W & 3R. So

P(red ball in 2nd draw)$=\frac{3}{6}=\frac{1}{2}$