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

A
1021
B
1113
C
1325
D
2132

Solution

The correct option is A $$\dfrac {10}{21}$$4 person chosen - 2 of them are children $$\Rightarrow ^{4}C_{2}.$$also, choosing rest 2 from $$\left [ (3+2+4-4) = 5 \right ]$$ person will be $$= ^{5}C_{2}$$(only from men and woman)$$\therefore$$ Prob $$\dfrac{^{4}C_{2}.^{5}C_{2}}{^{8}C_{4}}$$  ($$^{9}C_{4}\rightarrow$$ selection of 4 from 9 person)$$\Rightarrow P =\dfrac{10}{21}$$ Maths

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