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(D) Total number of experts $= 2 + 3 + 4 = 9$.We need to select $3$ experts out of $9$ to resign. The total number of ways to choose $3$ experts is gi

Vedclass · Accepted Answer

$\frac{2}{7}$

Vedclass · Answer

(D) Total number of experts $= 2 + 3 + 4 = 9$.We need to select $3$ experts out of $9$ to resign. The total number of ways to choose $3$ experts is given by ${}^9C_3 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.We want the probability that the $3$ experts belong to different institutions. This means we must select $1$ expert from institution $A$,$1$ from $B$,and $1$ from $C$.The number of ways to select $1$ expert from each is ${}^2C_1 	imes {}^3C_1 	imes {}^4C_1 = 2 	imes 3 	imes 4 = 24$.The required probability is $\frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{24}{84}$.Simplifying the fraction,$\frac{24}{84} = \frac{2}{7}$.

$A$ committee consists of $9$ experts taken from three institutions $A, B$ and $C$,of which $2$ are from $A, 3$ from $B$ and $4$ from $C$. If three experts resign,then the probability that they belong to different institutions is

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