An urn contains 3 red and 5 blue balls Suppose that 5 balls

An urn contains 3 red and 5 blue balls. Suppose that 5 balls are selected at random and with replacement. What is the probability that the first 2 are red and the rest are blue?

Solution

Note that we are selecting with replacement. Then, as there are 2 red balls and 7 blue balls, P(red) = 2/(2+7) = 2/9 and P(blue) = 7/(2+7) = 7/9

As we are selecting 7 balls total, and the first 2 are red and the rest are blue 7-2=5 are blue.

Then, P(2 red, then 5 blue) = (2/9)^2*(7/9)^5 = 67228/4782969 = 0.0140557047306809