An urn contains 4 red balls and 6 green balls Two balls are

An urn contains 4 red balls and 6 green balls. Two balls are drawn one after another without replacement.

Let A = event both balls are the same color and let B = event both balls are green. Calculate Pr(B|A)

Solution

There are 4C2 = 6 ways to get both red, and 6C2 = 15 ways to get both green.

Thus, there are 6+15 = 21 ways to get same colored balls.

Thus,

P(B|A) = 15/21 = 5/7 or 0.714285714 [ANSWER]