# A bag contains 3 black, 3 white and 2 red balls. One by one three balls are drawn without replacement. The probability that the third ball is red is equal to

$\begin {array} {1 1} (A)\;\large\frac{3}{56} & \quad (B)\;\large\frac{1}{56} \\ (C)\;\large\frac{12}{56} & \quad (D)\\large\frac{14}{56} \end {array}$

## 1 Answer

Three possible cases are
$Ist \: draw\: \: \: \: \: IInd \: draw\: \: \: \: \: IIIrd\: draw$
$Red\: \: \: \: \: \: \: \: \: \: \: \: \: \: Non\: Red\: \: \: \: \: \: \: \: Red$
$Non\: Red\: \: \: \: \: \: \: \: \: \: \: Red\:\: \: \: \: \: \: \: \: \: \: Red$
$Non\: Red\: \: \: \: \:\: \: \: \: Non\: Red\: \: \: \: \: Red$
$\therefore$ Required probability = $\large\frac{2}{8} \times \large\frac{6}{7} \times \large\frac{1}{6}+ \large\frac{6}{8} \times \large\frac{2}{7} \times \large\frac{1}{6}+ \large\frac{6}{8} \times \large\frac{5}{7} \times \large\frac{2}{6}$
$= \large\frac{14}{56}$
Ans : (D)

answered Jan 16, 2014

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