Question

A bag contains $3$ white, $3$ black, and $2$ red balls. One by one three balls are drawn without replacing them, then find the probability that the third ball is red.

• A. $\frac{1}{4}$
• B. $\frac{3}{4}$
• C. $\frac{1}{3}$
• D. None of these

Answer 1

Answer: (A)

$\frac{1}{4}$

Let $R$ stand for drawing red ball, $B$ for black ball, and $W$ for drawing white ball. Then, the required probability $=P\left(WWR\right)+P\left(BBR\right)+P\left(WBR\right)+P\left(BWR\right)+P\left(WRR\right)+P\left(BRR\right)+P\left(RWR\right)+P\left(RBR\right)$
$=\frac{3\times 2\times 2}{8\times 7\times 6}+\frac{3\times 2\times 2}{8\times 7\times 6}+\frac{3\times 3\times 2}{8\times 7\times 6}+\frac{3\times 3\times 2}{8\times 7\times 6}+\frac{3\times 2\times 1}{8\times 7\times 6}+\frac{3\times 2\times 1}{8\times 7\times 6}+\frac{2\times 3\times 1}{8\times 7\times 6}+\frac{2\times 3\times 1}{8\times 7\times 6}$

$=\frac{2}{56}+\frac{2}{56}+\frac{3}{56}+\frac{3}{56}+\frac{1}{56}+\frac{1}{56}+\frac{1}{56}+\frac{1}{56}$ $=\frac{1}{4}$