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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:
Options:
A .  $$\frac{21}{46}$$
B .  $$\frac{25}{117}$$
C .  $$\frac{1}{50}$$
D .  $$\frac{3}{25}$$
E .  None of these
Answer: Option A
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
= $${}^{25}\mathop C\nolimits_3 = $$   $$\frac{{25 \times 24 \times 23}}{{3 \times 2 \times 1}}$$   = 2300
And, n(E) = $$\left( {{}^{15}\mathop C\nolimits_2 \times {}^{10}\mathop C\nolimits_1 } \right)$$   $$ = \left( {\frac{{15 \times 14}}{{2 \times 1}} \times 10} \right)$$    = 1050
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{1050}}{{2300}} = \frac{{21}}{{46}}$$

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