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)

$\frac{2}{7}$

Answer 2

Three persons can be chosen out of 8 in $^8C_3 = 56$ 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 $= \frac{16}{56}= \frac{2}{7}$