In a class,there are 15 boys and 10 girls. Th | vedclass

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

Question

(A) Total number of students $= 15 + 10 = 25$. Number of ways to select $3$ students out of $25$ is given by ${}^{25}C_{3} = \frac{25 	imes 24 	imes

Vedclass · Accepted Answer

$\frac{21}{46}$

Vedclass · Answer

(A) Total number of students $= 15 + 10 = 25$. Number of ways to select $3$ students out of $25$ is given by ${}^{25}C_{3} = \frac{25 	imes 24 	imes 23}{3 	imes 2 	imes 1} = 2300$. Number of ways to select $1$ girl out of $10$ is ${}^{10}C_{1} = 10$. Number of ways to select $2$ boys out of $15$ is ${}^{15}C_{2} = \frac{15 	imes 14}{2 	imes 1} = 105$. Required probability $= \frac{{}^{10}C_{1} 	imes {}^{15}C_{2}}{{}^{25}C_{3}} = \frac{10 	imes 105}{2300} = \frac{1050}{2300} = \frac{21}{46}$.

In a class,there are $15$ boys and $10$ girls. Three students are selected at random. The probability that $1$ girl and $2$ boys are selected,is

Similar Questions

Two cards are drawn without replacement from a well-shuffled pack. Find the probability that one of them is an ace of heart.

$A$ and $B$ are two events such that $P(A) = 0.8$,$P(B) = 0.6$,and $P(A \cap B) = 0.5$. Then the value of $P(A/B)$ is:

Two numbers are selected randomly from the set $S = \{ 1, 2, 3, 4, 5, 6 \}$ without replacement one by one. The probability that the minimum of the two numbers is less than $4$ is

$A$ card is drawn from a pack of $52$ cards. Find the probability of getting a spade,an ace,or a red card.

If any four numbers are selected and they are multiplied,then the probability that the last digit of the product will be $1, 3, 5$ or $7$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module