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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
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B
1113
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C
1325
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D
2132
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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)
$$ \Rightarrow P =\dfrac{10}{21}$$ 

1117093_462511_ans_d1ab282db5ca4eb8bc0d740d1e7722dd.JPG

Maths

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