In a bag, there are 8 ted balls and 7 green balls. Three balls are picked at random. What is the probability that two balls are red and one ball is green in colour ?

The balls are drawn without replacement.

Probability of drawing green ball in first attempt = 7/15
Probability of drawing two red balls in the next two attempt = (8/14)(7/13)
Probability of drawing 2 red and 1 green ball = (7/15)(8/14)(7/13) = 392/2730

Probability of drawing red ball in first attempt = 8/15
Probability of drawing green ball in the next attempt = (7/14)
Probability of drawing red ball in the next attempt = (7/13)
Probability of drawing 2 red and 1 green ball = (8/15)(7/14)(7/13) = 392/2730

Probability of drawing two red balls in first two attempts = (8/15)(7/14)
Probability of drawing red ball in the next attempt =(7/13)
Probability of drawing 2 red and 1 green ball = (8/15)(7/14)(7/13) = 392/2730

Probability of drawing 2 red balls and 1 green ball= 392/2730 + 392/2730 + 392/2730 = 3(392/2730) = 392/910 = 28/65
Option A is the correct answer.