Question

A class of twelve children has two more boys, than girls. A group of three children are randomly picked from this class to accompany the teacher on a field trip. What is the probability that the group accompanying the teacher contains more girls than boy?

• A. 0
• B. $\frac{325}{864}$
• C. $\frac{525}{864}$
• D. $\frac{5}{12}$

Answer 1

The correct option is B $\frac{325}{864}$

$\because B=G+2\text{and}B+G=12\Rightarrow B=7,G=5$
Total ways of picking 3 children $={}^{12}{C}_3$

Favourable ways (when No. of girls are > No. of Boys)

= Either (2G and 1B) or (3G)

$=({}^5{C}_2\times {}^7{C}_1)+{}^5{C}_3=10\times 7+10=80$

$\text{Req. Prob}=\frac{\text{fav}}{\text{Total}}=\frac{80}{{}^{12}{C}_3}=\frac{80}{220}$

$=0.36=\frac{325}{864}$