Each of urns I and II has 5 red balls, 2 white balls, and 3 green balls. Urn III has 5 red balls, 3 white balls, and 2 green balls. An urn is selected at random and a ball is drawn. If the ball is green G, find the probability that urn III was selected.

Question

Each of urns I and II has 5 red balls, 2 white balls, and 3 green balls. Urn III has 5 red balls, 3 white balls, and 2 green balls. An urn is selected at random and a ball is drawn. If the ball is green G, find the probability that urn III was selected. ​

Quizplus · Accepted Answer

We need to find the probability that urn III was selected given that a green ball was drawn. We can use Bayes' theorem:

P(urn III | G) = P(G | urn III) * P(urn III) / P(G)

P(G | urn III) is the probability of drawing a green ball from urn III, which is 2/10 = 1/5.
P(urn III) is the probability of selecting urn III, which is 1/3 since there are three urns.
P(G) is the overall probability of drawing a green ball, which is the weighted average of the probabilities of drawing a green ball from each urn:

P(G) = P(G | urn I) * P(urn I) + P(G | urn II) * P(urn II) + P(G | urn III) * P(urn III)

= (3/10 * 1/3) + (3/10 * 1/3) + (2/10 * 1/3)

= 1/5

Plugging these values into Bayes' theorem, we get:

P(urn III | G) = (1/5) * (1/3) / (1/5)

= 1/3

So the probability that urn III was selected given that a green ball was drawn is 1/3. The correct answer choice is C.

Each of Urns I and II Has 5 Red Balls