CMAT 2022 Solved Question Paper

Question : 5 of 100

Marks: +1, -0

In a bag there are 15 red balls and 10 green balls. Three balls are selected at random. The probability of selecting 2 red balls and 1 green ball is :

$21/46$
$25/117$
$3/25$
$1/50$

Validate

Solution:

Number of ways of selecting 2 red balls and 1 green ball

=15 \C_2 .10 \C_1

Number of ways of selecting 3 balls

=25 \C_3

Probability of selecting 2 red balls and 1 green ball =

={15 \C_2 .10 \C_1}/{25 \C_3}

21/46

Answer is option A.

Go to Question:

Prev Question

Next Question