Question

A class consists of 10 boys and 8 girls. Three students are selected at random. What is the probability that the selected group has

(i) all boys?

(ii) all girls?

(iii) 1 boy and 2 girls?

(iv) at least one girls?

(v) at most one girls?

Answer 1

Total number of students = (10 + 8) = 18
Let S be the sample space.

Then n(S) = number of ways of selecting 3 students out of $18{=}^{18}{C}_3$ ways

(i) Out of 10 boys, all buys can be selected in ${}^{10}{C}_3$ ways.

$\therefore$ Favourable number of events, $n\left(E\right){=}^{10}{C}_3$

Hence, required probability = $\frac{{}^{10}{C}_3}{{}^{18}{C}_3}$

$=\frac{10\times 9\times 8}{18\times 17\times 16}=\frac{5}{34}$

(ii) Out of eight girls, three girls can be selected in ${}^8{C}_3$ ways

$\therefore$ Favourable number probability $=\frac{{}^8{C}_3}{{}^{18}{C}_3}$

$=\frac{8\times 7\times 6}{18\times 17\times 16}=\frac{7}{102}$

(iii) One boy and two girls can be selected in ${}^{10}{C}_1{\times }^8{C}_2.$

$\therefore$ Favourable number of events

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

Hence, required probability $=\frac{{}^{10}{C}_1{\times }^8{C}_2}{{}^{18}{C}_3}$

$=\frac{10\times 28}{816}=\frac{35}{102}$

(iv) Probability of at least one girl = 1-P (no girl)
=1-P (all 3 are boys)

$=1-\frac{{}^{10}{C}_3}{{}^{18}{C}_3}=1-\frac{5}{34}=\frac{29}{34}$

(v) Let E be the event with at most one girl in the group.
Then E={0 girl, 1 girl}

$\therefore$ Favourable number of events. $n\left(E\right){=}^8{C}_0{\times }^{10}{C}_3{\times }^8{C}_1{\times }^{10}{C}_2$

Hence, the required probability is given by

$=\frac{{}^8{C}_0{\times }^{10}{C}_3{\times }^8{C}_1{\times }^{10}{C}_2}{{}^{18}{C}_3}$

$=\frac{1{\times }^{10}{C}_3{+}^8{C}_1{\times }^{10}{C}_2}{{18}_3^C}$

$=\frac{1{\times }^{10}{C}_3{+}^8{C}_3{+}^8{C}_1{\times }^{10}{C}_2}{{}^{18}{C}_3}$

$=\frac{1\times 20+8\times 45}{816}=\frac{480}{816}=\frac{10}{17}$