Question

In a class, there are $10$boys and$8$ girls. When $3$students are selected at random, the probability that$2$ girls and $1$ boy are selected is

• A. $\frac{35}{102}$
• B. $\frac{15}{102}$
• C. $\frac{55}{102}$
• D. $\frac{25}{102}$

Answer 1

The correct option is A

$\frac{35}{102}$

Explanation for the correct option:

Finding the probability:

Total number of students in class

$$\begin{gathered} =10+8 \\ =18 \end{gathered}$$

Number of possible combinations of three students${=}^{18}{C}_3$

Now, for $2$ boys and $1$ girl, possible combinations $={}^{8}C_2\times {}^{10}C_1$

Probability $=\frac{\mathrm{Possible}\mathrm{combination}}{\mathrm{Total}\mathrm{combination}}$

$$\begin{gathered} =\frac{{}^{8}C_2\times {}^{10}C_1}{{}^{18}C_3} \\ =\frac{8\times 5\times 3\times 2}{18\times 17\times 16}\left[{\because }^n{C}_r=\frac{n!}{r!\left(n-r\right)!}\right] \\ =\frac{35}{102} \end{gathered}$$

Hence, the correct option is A.