The probability of forming a 12-person commit | vedclass

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Question

(A) Total number of ways to select $12$ persons from $16$ $(4+2+10)$ is $^{16}C_{12} = ^{16}C_4 = \frac{16 	imes 15 	imes 14 	imes 13}{4 	imes 3 \

Vedclass · Accepted Answer

$\frac{129}{182}$

Vedclass · Answer

(A) Total number of ways to select $12$ persons from $16$ $(4+2+10)$ is $^{16}C_{12} = ^{16}C_4 = \frac{16 	imes 15 	imes 14 	imes 13}{4 	imes 3 	imes 2 	imes 1} = 1820$. We need at least $3$ engineers and at least $1$ doctor. The possible cases are: $1)$ $3$ Engineers,$1$ Doctor,$8$ Professors: $^4C_3 	imes ^2C_1 	imes ^{10}C_8 = 4 	imes 2 	imes 45 = 360$. $2)$ $3$ Engineers,$2$ Doctors,$7$ Professors: $^4C_3 	imes ^2C_2 	imes ^{10}C_7 = 4 	imes 1 	imes 120 = 480$. $3)$ $4$ Engineers,$1$ Doctor,$7$ Professors: $^4C_4 	imes ^2C_1 	imes ^{10}C_7 = 1 	imes 2 	imes 120 = 240$. $4)$ $4$ Engineers,$2$ Doctors,$6$ Professors: $^4C_4 	imes ^2C_2 	imes ^{10}C_6 = 1 	imes 1 	imes 210 = 210$. Total favorable ways = $360 + 480 + 240 + 210 = 1290$. Required probability = $\frac{1290}{1820} = \frac{129}{182}$.

The probability of forming a $12$-person committee from $4$ engineers,$2$ doctors,and $10$ professors containing at least $3$ engineers and at least $1$ doctor is:

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