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 boys ?

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

Answer 1

The correct option is B

$\frac{325}{864}$
$B+G=12$
$B=G+2$
$\Rightarrow B=7$
$G=5$

7 Boys and 5 Girls are there is 12 students.
Among 3 students selected boys have to be more than girls.

So only two cases arise.

G B
2 1
3 0

At case (i)

$GGG\to \frac{5}{12}\times \frac{5}{12}\times \frac{5}{12}$

$Case\left(ii\right)\begin{array}{ccc}G& G& B\\ G& B& G\\ B& G& G\end{array}\}\to 3\times \frac{5}{12}\times \frac{5}{12}\times \frac{7}{12}\text{Required prob =}{\left(\frac{5}{12}\right)}^3+3{\left(\frac{5}{12}\right)}^2\times \left(\frac{7}{12}\right)=\frac{650}{1728}=\frac{325}{864}=0.3761574=0.376$