(D) Total number of experts $= 2 + 3 + 4 = 9$. Number of ways to select $3$ experts out of $9$ is ${}^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \ti

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(D) Total number of experts $= 2 + 3 + 4 = 9$. Number of ways to select $3$ experts out of $9$ is ${}^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$. We want to select $3$ experts such that they belong to different institutions,meaning one from $A$,one from $B$,and one from $C$. Number of ways to select one expert from each institution $= {}^2C_1 \times {}^3C_1 \times {}^4C_1 = 2 \times 3 \times 4 = 24$. Required probability $= \frac{24}{84} = \frac{2}{7}$.

Class 11 Mathematics Probability Advanced Use of permutations and combinations in probability

Medium

$A$ 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

A
$\frac{1}{729}$
B
$\frac{1}{24}$
C
$\frac{1}{21}$
D
$\frac{2}{7}$

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Probability

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Advanced Use of permutations and combinations in probability

Let $m, n$ be two distinct integers chosen randomly from the set $\{0, 1, 2, \ldots, 99\}$. Then,the probability that $4^m + 4^n + 3$ is divisible by $5$ lies in the interval

KVPY 2017

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MediumTS EAMCET 2025

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Easy

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The numbers $2, 3, 5, 7, 11, 13$ are written on six distinct paper chits. If $3$ of them are chosen at random,then the probability that the sum of the numbers on the obtained chits is divisible by $3$ is:

DifficultTS EAMCET 2024

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$A$ 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

Similar Questions

Let $m, n$ be two distinct integers chosen randomly from the set $\{0, 1, 2, \ldots, 99\}$. Then,the probability that $4^m + 4^n + 3$ is divisible by $5$ lies in the interval

Let $\omega$ be a complex cube root of unity with $\omega \neq 1$. $A$ fair die is thrown three times. If $r_1, r_2,$ and $r_3$ are the numbers obtained on the die,what is the probability that $\omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0$?

If three smallest squares are chosen at random on a chess board,then the probability of getting them in such a way that they are all together in a row or in a column is

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The numbers $2, 3, 5, 7, 11, 13$ are written on six distinct paper chits. If $3$ of them are chosen at random,then the probability that the sum of the numbers on the obtained chits is divisible by $3$ is:

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