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1. In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that the selected students are 2 boys and 1 girls, 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 let E be the event of selecting 2 boys and 1 girl.
Then, n(S) = number of ways of selecting 3 students out of 25
=

^{25} C_{3} =

\frac{25 \times 24 \times 23}{3 \times 2 \times 1}

= 2300
And, n(E) =

(^{15} C_{2} \times^{10} C_{1})

= (\frac{15 \times 14}{2 \times 1} \times 10)

= 1050

∴ P (E) = \frac{n (E)}{n (S)} = \frac{1050}{2300} = \frac{21}{46}

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