If four persons are chosen at random from a group of 3 men, 2 women and 4 children. Then probability

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14.Probability

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If four persons are chosen at random from a group of $3$ men, $2$ women and $4 $ children. Then the probability that exactly two of them are children, is

A

$\frac{{10}}{{21}}$

B

$\frac{8}{{63}}$

C

$\frac{5}{{21}}$

D

$\frac{9}{{21}}$

Solution

(a) Total number of ways $ = {}^9{C_4},$

$2 $ children are chosen in ${}^4{C_2}$ ways and other $2$ persons are chosen in ${}^5{C_2}$ ways.

Hence required probability $= \frac{{{}^4{C_2} \times {}^5{C_2}}}{{{}^9{C_4}}} = \frac{{10}}{{21}}.$

Standard 11

Mathematics

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