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 boys and 2 girls?
(iv) at least one girl?
(v) at most one girl?

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 = ¹⁸C₃ ways

(i)
Out of 10 boys, three boys can be selected in ¹⁰C₃ ways.
∴ Favourable number of events, n(E) = ¹⁰C₃
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 ⁸C₃ ways.
∴ Favourable number of events, n(E) = ⁸C₃
Hence, required 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 ¹⁰C₁ × ⁸C₂.
∴ Favourable number of events = ¹⁰C₁ × ⁸C₂
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}
∴ Favourable number of events, n(E) = ⁸C₀ × ¹⁰C₃ × ⁸C₁ × ¹⁰C₂
Hence, the required probability is given by
$$\begin{gathered} \frac{{}^{8}C_0\times {}^{10}C_3+{}^{8}C_1\times {}^{10}C_2}{{}^{18}C_3} \\ =\frac{1\times {}^{10}C_3+{}^{8}C_1\times {}^{10}C_2}{{}^{18}C_3} \\ =\frac{1\times 120+8\times 45}{816} \\ =\frac{480}{816} \\ =\frac{10}{17} \end{gathered}$$