Question
Four persons are chosen at random from a group of 3 men, 2 women and 4 children. The chance that exactly 2 of them are children, is-
Answer: Option D
n(S) = number of ways of choosing 4 persons out of 9
$$ = {}^9\mathop C\nolimits_4 $$ $$ = \frac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}}$$ = 126
n(E) = number of ways of choosing 2 children out of 4 and 2 persons out of (3 + 2) personal
n(E) $$ = \left( {{}^4\mathop C\nolimits_2 \times {}^5\mathop C\nolimits_2 } \right)$$ $$ = \left( {\frac{{4 \times 3}}{{2 \times 1}} \times \frac{{5 \times 4}}{{2 \times 1}}} \right)$$ = 60
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{60}}{{126}} = \frac{{10}}{{21}}$$
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n(S) = number of ways of choosing 4 persons out of 9
$$ = {}^9\mathop C\nolimits_4 $$ $$ = \frac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}}$$ = 126
n(E) = number of ways of choosing 2 children out of 4 and 2 persons out of (3 + 2) personal
n(E) $$ = \left( {{}^4\mathop C\nolimits_2 \times {}^5\mathop C\nolimits_2 } \right)$$ $$ = \left( {\frac{{4 \times 3}}{{2 \times 1}} \times \frac{{5 \times 4}}{{2 \times 1}}} \right)$$ = 60
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{60}}{{126}} = \frac{{10}}{{21}}$$
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