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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$\frac{4}{9}$

Vedclass · Answer

(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

Similar Questions

Four numbers are chosen at random from $\{1, 2, 3, \ldots, 40\}$. The probability that they are not consecutive is:

The coefficients $a, b, c$ in the quadratic equation $ax^2 + bx + c = 0$ are chosen from the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$. The probability of this equation having repeated roots is:

If a person throws $3$ dice,what is the probability that the sum of the numbers obtained is exactly $15$?

From $3n$ consecutive integers, three integers are selected at random. The probability that their sum is divisible by $3$ is

$A$ bag contains $6$ white and $4$ black balls. $A$ die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module