Question

Urn I contains 2 red and 8 black balls, while urn II contains 6 red and 4 black balls. One ball is drawn at random from one of the urns, and it is found to be black. Find the probability that it was drawn from urn II.

(a) 2/3
(b) 3/5
(c) 1/2
(d) 2/5

Answer 1

The correct probability that a black ball was drawn from urn II, given it was found to be black, is 1/3 which is not listed in the provided options.

Step-by-step explanation:

The question asks for the probability that a black ball drawn at random is from urn II, given urn I contains 2 red and 8 black balls, and urn II contains 6 red and 4 black balls. This is a classic conditional probability problem which can be solved using Bayes' theorem.

First, let's find the overall probability of drawing a black ball. The probability of choosing urn I is 1/2 (since there are only two urns) and the probability of drawing a black ball from urn I is 8/10. Similarly, the probability of choosing urn II is also 1/2 and the probability of drawing a black ball from urn II is 4/10. Multiplying these and adding gives us the total probability of drawing a black ball:

P(Black) = (1/2)(8/10) + (1/2)(4/10) = 4/10 + 2/10 = 6/10

Next, we need to find the probability that the black ball came from urn II. That's the probability of drawing from urn II multiplied by the probability of drawing a black ball from urn II, divided by the overall probability of drawing a black ball:

P(Urn II | Black) = (Probability of urn II and drawing black) / P(Black)
= (1/2)(4/10) / (6/10) = 2/6 = 1/3

So, option (c) 1/2 is incorrect and the correct answer is none of the provided options.