Question

A bag contains $5$ white, $4$ black and $2$ red balls. Balls are drawn one by one without replacement. The probability that the $5^{th}$ ball is a red ball, is

• A. $\frac{2}{11}$
• B. $\frac{4}{11}$
• C. $\frac{3}{7}$
• D. $\frac{6}{11}$

Answer 1

Answer: (A)

If the $5^{th}$ ball is a red ball $\Rightarrow$ either the first four ball are of other colour and $5^{th}$ ball is red or the first four ball consist of $1$ red ball, $3$ other ball and $5^{th}$ ball is red ball
Hence, the required probability $=\frac{\_{}^9{C}_4^{}}{\_{}^{11}{C}_4^{}}\times \frac{2}{7}+\frac{\_{}^9{C}_3^{}\times \_{}^2{C}_1^{}}{\_{}^{11}{C}_4^{}}\times \frac{1}{7}$
$=\frac{9\times 8\times 7\times 6}{11\times 10\times 9\times 8}\times \frac{2}{7}+\frac{9\times 8\times 7\times 2\times 4}{11\times 10\times 9\times 8}\times \frac{1}{7}$
$=\frac{6}{55}+\frac{4}{55}=\frac{10}{55}=\frac{2}{11}$