A committee of 5 is to be chosen from a group | vedclass

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(C) Total number of ways to choose $5$ people from $9$ is ${}^9C_5 = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} = 126$.Case $1$

Vedclass · Accepted Answer

$\frac{4}{9}$

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(C) Total number of ways to choose $5$ people from $9$ is ${}^9C_5 = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} = 126$.Case $1$: The couple serves together.We need to choose $3$ more people from the remaining $7$ people. Number of ways $= {}^7C_3 = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$.Case $2$: The couple does not serve at all.We need to choose $5$ people from the remaining $7$ people. Number of ways $= {}^7C_5 = {}^7C_2 = \frac{7 	imes 6}{2 	imes 1} = 21$.Total favorable ways $= 35 + 21 = 56$.Required probability $= \frac{56}{126} = \frac{4}{9}$.

$A$ committee of $5$ is to be chosen from a group of $9$ people. The probability that a certain married couple will either serve together or not at all,is

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