A committee has to be made of 5 members from | vedclass

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Question

(b) Total number of ways

$ = {}^4{C_1} 	imes {}^6{C_4} + {}^4{C_2} 	imes {}^6{C_3} + {}^4{C_3} 	imes {}^6{C_2} + {}^4{C_4} 	imes {}^6{C_1} + {}

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) Total number of ways

$ = {}^4{C_1} 	imes {}^6{C_4} + {}^4{C_2} 	imes {}^6{C_3} + {}^4{C_3} 	imes {}^6{C_2} + {}^4{C_4} 	imes {}^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} 	imes {}^6{C_4} + {}^4{C_2} 	imes {}^6{C_3} + {}^4{C_3} 	imes {}^6{C_2} + {}^4{C_4} 	imes {}^6{C_1} = 246$

Hence required probability $ = \frac{{246}}{{252}} = \frac{{41}}{{42}}$.

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

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