Question

$10$ boys and $2$ girls are divided into $3$ groups of $4$ each. The probability that the girls will be in different groups is

• A. $\frac{3}{5}$
• B. $\frac{4}{5}$
• C. $\frac{3}{11}$
• D. $\frac{8}{11}$

Answer 1

The correct option is D $\frac{8}{11}$

Total number of ways $=\frac{{}^{12}{C}_4{\times }^8{C}_4{\times }^4{C}_4}{3!}$
Number of ways when girls are in same group $=\frac{1{\times }^{10}{C}_2{\times }^8{C}_4{\times }^4{C}_4}{2!}$
Probability girls are in same group $=\frac{{}^{10}{C}_2\times 3!}{{}^{12}{C}_4\times 2!}=\frac{4\times 3\times 3}{12\times 11}=\frac{3}{11}$
Probability girls are in different groups $=1-\frac{3}{11}=\frac{8}{11}$