Problem I Consider two urns. The first contains two white and seven black marble

Question

Problem I Consider two urns. The first contains two white and seven black marbles, and the second contains four white and two black marbles. We flip a coin and then draw a marble from one of the two ums: We draw fom the first um if the result of the coin flip is heads and fiom the second urn if it is tails (a) What is the probability that we select a white marble? (b) What is the probability that the outcome of the toss was heads given that a white marble was selected? What is the probability that the outcome of the toss was heads given that a black marble is selected?

Explanation / Answer

Let

A = first urn was selected
A' = second urn was selected
W, B = white or black marble is selected, respectively

a)

By Bayes' Rule,

P(W) = P(A) P(W|A) + P(A') P(W|A') = (1/2)*(2/9) + (1/2)*(4/6) = 0.444444444 [ANSWER]

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b)

That means the first urn is selected.

Hence,

P(A|W) = P(A) P(W|A) / P(W)

= (1/2)*(2/9)/0.4444444444

= 0.25 [ANSWER]

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c)

That means the first urn is selected.

Hence,

P(A|B) = P(A) P(B|A) / P(B)

As P(B) = 1 - P(W), then

P(A|B) = P(A) P(B|A) / P(B)

= (1/2)*(7/9)/(1-0.4444444444)

= 0.7 [ANSWER]

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