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].

1. 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]