Question

an urn contains 4 red balls and 6 white balls. a ball is drawn from the urn and a ball of the other color is then put into the urn. a second ball is then drawn from the urn. (a) find the probability that the second ball is red. (b) if both balls were of the same color, what is the probability that they were both red.

Answer 1

Explanation:

Suppose a white ball was first drawn:

Prob(white) = 3/5.

Then a red ball will be placed in the urn, so we would have 5 red and 5 white in the urn.

Prob(red drawn) = 1/2.

So prob(white then red ) = 3/5 * 1/2 = 3/10.

OR

Suppose a red ball was first drawn

Prob(red) = 2/5

Then there will be 7 white and 3 red after a white was added so

Prob(red drawn) = 3/10.

So prob(red then red ) = 2/5 * 3/10 = 6/50 = 3/25

(a) Required probability = 3/10 + 3/25 = 21/50.