A committee has to be made of 5 members from 6 men and 4 women. probability that at least one woman

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14.Probability

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A committee has to be made of $5$ members from $6$ men and $4$ women. The probability that at least one woman is present in committee, is

A

$\frac{1}{{42}}$

B

$\frac{{41}}{{42}}$

C

$\frac{2}{{63}}$

D

$\frac{1}{7}$

Solution

(b) Total number of ways

$ = {}^4{C_1} \times {}^6{C_4} + {}^4{C_2} \times {}^6{C_3} + {}^4{C_3} \times {}^6{C_2} + {}^4{C_4} \times {}^6{C_1} + {}^6{C_5}$

$ = 60 + 120 + 60 + 6 + 6 = 252$

No. of ways in which at least one woman exist are

$ = {}^4{C_1} \times {}^6{C_4} + {}^4{C_2} \times {}^6{C_3} + {}^4{C_3} \times {}^6{C_2} + {}^4{C_4} \times {}^6{C_1} = 246$

Hence required probability $ = \frac{{246}}{{252}} = \frac{{41}}{{42}}$.

Standard 11

Mathematics

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