Question

(a) A urn contains one red marble, one green marble and one blue marble. An experiment consists of taking one marble from the urn, then returning it and drawing a second marble. What is the probability that exactly one red marble is selected?

(b) What is the probability that exactly one red marble is selected if the experiment is without replacement (marble is not returned)?

Answer 1

(a) WITH REPLACEMENT

Case 1:

P(First Trial: 1 Red) = 1/3 = 0.3333

P(Second Trial: 0 Red) = 2/3 = 0.6667

So,

P(First Trial: 1 Red & Second Trial: 0 Red) = 0.3333 X 0.6667 = 0.2222

Case 2:

P(First Trial: 0 Red) = 2/3 = 0.6667

P(Second Trial: 1 Red) = 1/3 = 0.3333

So,

P(First Trial: 0 Red & Second Trial: 1 Red) = 0.6667 X 0.3333 = 0.2222

So,

P(exactly one red marble is selected) = 0.2222 X 2 = 0.4444

(b) WITHOUT REPLACEMENT

Case 1:
P(First Trial: 1 Red) = 1/3 = 0.3333

P(Second Trial: 0 Red) = 1

So,

P(First Trial: 1 Red & Second Trial: 0 Red) = 0.3333

Case 2:

P(First Trial: 0 Red) = 2/3 = 0.6667

P(Second Trial: 1 Red) = 1/2 = 0.5

So,

P(First Trial: 0 Red & Second Trial: 1 Red) = 0.6667 X 0.5 = 0.3333

So,

P(exactly one Red) = 0.3333 X 2 = 0.6666