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1. In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

A. $\frac{21}{46}$

B. $\frac{25}{117}$

C. $\frac{1}{50}$

D. $\frac{3}{25}$

Solution:

Let S be the sample space and E be the event of selecting 1 girl and 2 boys
Then, n(S) = Number ways of selecting 3 students out of 25

\begin{aligned} =^{25} C_{3} \\ = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} \\ = 2300 \\ n (E) =^{10} C_{1} \times^{15} C_{2} \\ = 10 \times \frac{15 \times 14}{2 \times 1} \\ = 1050 \\ ∴ P (E) = \frac{n (E)}{n (S)} = \frac{1050}{2300} = \frac{21}{46} \end{aligned}

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