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

Answer: (A)

$\frac{35}{102}$

Total Students $=10+8=18$

Total boys $=10$

Total circles $=8$

$\therefore$ Total no. of combination of three students ${=}^{18}{C}_3$

2 girls and 1 boy(possible combination) ${=}^8{C}_2{\times }^{10}{C}_1$

$\Rightarrow$ Probability $=\frac{Possiblecombination}{Totalcombination}$

$=\frac{{}^8{C}_2{\times }^{10}{C}_1}{{}^{18}{C}_3}$

$=\frac{8\times 7\times 5\times 3\times 2}{18\times 17\times 16}$

$=\frac{35}{102}$