7. An urn contains 4 red balls and 8 white balls. 2 balls are drawn at random. Use a tree diagram to construct a probabilistic model for this experiment. Use the model to answer the following questions.

• a) What is the probability of having at least 1 white ball?
• b) What is the probability that the balls are the same color?
• c) What is the probability that the second ball is white?
• d) What is the probability that the second ball is white given that the balls are the same color?
• e) Are the events in b) and c) independent?
• f) What is the expected number of white balls?

We first use the tree diagram to construct the probabilistic model.

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a) To compute the probability of drawing at least 1 white ball, we add up the probabilities of the outcomes in the event.

A = {RW, WR, WW}

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b) To compute the probability that the balls are the same color, we add up the probabilities of the outcomes in the event.

S = {RR, WW}

17/34 = 1/2,   so this is close to   1/2,   as we would suspect, but not quite.

c) To compute the probability that the second ball is white, we add up the probabilities of the outcomes in the event.

B = {RW, WW}

Note that the probability that the second ball is white is the same as the probability that the first ball is white.

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d) The probability that second ball is white given that the balls are the same color is a conditional probability, so we need to use the formula for conditional probability.

The event on top is the event that the second ball is white and the balls are the same color so this is

{WW}

We can look its probability up in the probabilistic model to be 14/33. The bottom comes from part b). This gives us

e) Since the conditional probability   Pr(B|S)   in d) is different from the original probability,   Pr(B),   in c), the events are not independent.   14/17   is approximately   82%   which is bigger than   2/3,   so if the balls are the same color, it makes it more likely that the second ball is white. This is because both red is the least likely event. Both white is much more likely, so if they are both the same color, they are more likely to be both white than both red.

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f) What is the expected number of white balls? We make up a table

 n p np 0 3/33 0 1 16/33 16/33 2 14/33 28/33 totals 44/33 = 4/3

From the model, we see that to get the probability of one white ball, we add up the probabilities of the two outcomes in which there is one white ball, and it adds up to   16/33.   Multiply the numbers of white balls times the probabilities of getting those numbers and add the products to get the expected value.

Note that the answer is   2x(2/3),   the number of drawings times the fraction of white balls as we had noticed in problem 5d.

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