A bag contains 4 red balls, 6 green balls and 5 blue balls. If three balls are picked at random, what is the probability that two of them are green and one of them is blue in colour ?

A bag contains 4 red balls, 6 green balls and 5 blue balls. If three balls are picked at random, what is the probability that two of them are green and one of them is blue in colour ?
1). $$\frac{20}{91}$$
2). $$\frac{10}{91}$$
3). $$\frac{15}{91}$$
4). $$\frac{5}{91}$$

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Guest · Answer 1 · 2019-04-08 19:53:59

Solution

Probability of drawing blue ball in first attempt = 5/15

Probability of drawing two green balls in the next two attempts = (6/14)(5/13)
Probability of drawing 2 green and 1 blue ball = (5/15)(6/14)(5/13) = 150/2730

Probability of drawing green ball in first attempt = 6/15
Probability of drawing blue ball in the next attempt = (5/14)
Probability of drawing green ball in the next attempt = (5/13)
Probability of drawing 2 red and 1 green ball = (6/15)(5/14)(5/13) = 150/2730

Probability of drawing two green balls in first two attempts = (6/15)(5/14)
Probability of drawing blue ball in the next attempt =(5/13)
Probability of drawing 2 red and 1 green ball = (6/15)(5/14)(5/13) = 150/2730

Probability of drawing 2 red balls and 1 green ball= 150/2730 + 150/2730 + 150/2730 = 3(150/2730) = 150/910 = 15/91
Option C is the correct answer

Guest · Answer 2 · 2019-05-01 11:18:16

Solution

Probability of drawing blue ball in first attempt = 5/15

Probability of drawing two green balls in the next two attempts = (6/14)(5/13)
Probability of drawing 2 green and 1 blue ball = (5/15)(6/14)(5/13) = 150/2730

Probability of drawing green ball in first attempt = 6/15
Probability of drawing blue ball in the next attempt = (5/14)
Probability of drawing green ball in the next attempt = (5/13)
Probability of drawing 2 red and 1 green ball = (6/15)(5/14)(5/13) = 150/2730

Probability of drawing two green balls in first two attempts = (6/15)(5/14)
Probability of drawing blue ball in the next attempt =(5/13)
Probability of drawing 2 red and 1 green ball = (6/15)(5/14)(5/13) = 150/2730

Probability of drawing 2 red balls and 1 green ball= 150/2730 + 150/2730 + 150/2730 = 3(150/2730) = 150/910 = 15/91
Option C is the correct answer

A bag contains 4 red balls, 6 green balls and 5 blue balls. If three balls are picked at random, what is the probability that two of them are green and one of them is blue in colour ?

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