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1. 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-

A. $\frac{1}{9}$

B. $\frac{1}{5}$

C. $\frac{1}{12}$

D. $\frac{10}{21}$

Solution:

n(S) = number of ways of choosing 4 persons out of 9

=^{9} C_{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)

= (^{4} C_{2} \times^{5} C_{2})

= (\frac{4 \times 3}{2 \times 1} \times \frac{5 \times 4}{2 \times 1})

= 60

∴ P (E) = \frac{n (E)}{n (S)} = \frac{60}{126} = \frac{10}{21}

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