Urn 1 contains 5 red, 4 white, and 6 blue marbles; Urn 2 contains 12 red, 2 white, and 3 blue marbles; and Urn 3 contains 20 red, 1 white, and 1 blue marble. One of the three urns is chosen at random, and then one marble is drawn at random from that urn

.a. Determine the probability that a red marble is drawn.

b. Determine the probability that a red marble is drawn, given that Urn 2 was chosen.

c. What is the probability that Urn 2 was chosen, given that a red marble was drawn?

d. What is the probability that Urn 1 was chosen, given that a red marble was NOT drawn?

e. What is the probability that Urn 3 was NOT chosen, given that a red marble was NOT drawn?

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Ryan S. | Mathematics and StatisticsMathematics and Statistics

Let R be the event that a red marble is chosen.

Let R' be the event that a red marble is not chosen.

Let I be a random variable that takes on the value 1 when urn 1 is chosen, 2 when urn 2 is chosen, 3 when urn 3 is chosen.

a) P(R) = P(R|I=1)*P(I=1)+P(R|I=2)*P(I=2)+P(R|I=3)*P(I=3)

P(R)=(5/15)*(1/3)+(12/17)*(1/3)+(20/22)*(1/3)=.649

b) P(R|I=2)=12/17=.706

c) Bayes Theorem says the following: P(I=2|R)=[P(R|I=2)*P(I=2)]/P(R) = (12/17)*(1/3)/.649

d) P(I=1|R')=[P(R'|I=1)*P(I=1)]/P(R') = (10/15)*(1/3)/(1-.649)=.634

e) P(I=1or2|R') = 1 - P(I=3|R') = 1 - [P(R'|I=3)*P(I=3)]/P(R') = (2/22)*(1/3)/(1-.649)= 1 - .086= .914

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