Question

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 :

• A. 21/46
• B. 25/117
• C. 3/25
• D. 1/50

Answer 1

The Correct Option is A

To solve the problem, we need to determine the probability of selecting 2 red balls and 1 green ball when three balls are drawn at random from a bag containing 15 red balls and 10 green balls.

The total number of balls in the bag is: 

\(15 + 10 = 25\)

The number of ways to select 3 balls out of 25 is given by the combination formula:

\(\binom{25}{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300\)

The number of ways to select 2 red balls out of 15 is:

\(\binom{15}{2} = \frac{15 \times 14}{2 \times 1} = 105\)

The number of ways to select 1 green ball out of 10 is:

\(\binom{10}{1} = 10\)

The number of ways to select 2 red balls and 1 green ball is:

\(105 \times 10 = 1050\)

Thus, the probability of selecting 2 red balls and 1 green ball is:

\(\frac{1050}{2300} = \frac{21}{46}\)

Therefore, the correct answer is \(\frac{21}{46}\).