Question

From a group of $5$ boys and $3$ girls three persons are chosen at random; Find the probability that there are more girls than boys

• A. $\frac{3}{8}$
• B. $\frac{4}{7}$
• C. $\frac{5}{8}$
• D. $\frac{2}{7}$

Answer 1

Answer: (D)

Three persons can be chosen out of 8 in ${}^8{C}_3=56$ ways. The number of girls is more than that of the boys if either 3 girls are chosen or two girls and one boy is chosen. This can be done in ${}^3{C}_3{+}^3{C}_2{\times }^5{C}_1$
$=1+3\times 5=16$ ways.
$\therefore$ Required probability $=\frac{16}{56}=\frac{2}{7}$