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Unformatted text preview: 1. Consider the following fictional joint probability distribution of the color of car driven and the number of speeding tickets received per year. Car Color / Tickets per year 0 Tickets 1 Ticket 2 Tickets White 1/10 1/10 Red 1/10 3/10 Blue 1/10 2/10 1/10 (a) What is the probability of receiving 0 tickets? What is the probability of getting 1 ticket? What is the probability of getting 2 tickets? Let T be tickets and C be car color. To find P ( T = 0) we use the formula for the marginal: P ( T = 0) = P ( T = 0 ,C = W ) + P ( T = 0 ,C = R ) + P ( T = 0 ,C = B ) = 1 10 + 0 + 1 10 = 1 5 We can find the probability of 1 ticket and 2 tickets by using the formula for a marginal in a similar way, which yields P ( T = 1) = 4 / 10 and P ( T = 2) = 4 / 10. (b) What is the probability of driving a red car conditional on having received 2 tickets? What is the probability of driving a red car conditional on having received no tickets? For this question, use the formula for conditional probabilities and the answers to part (a) to find that P ( C = R | T = 2) = P ( C = R,T = 2) /P ( T = 2) = (3 / 10) / (4 / 10) = 3 / 4. Similarly, P ( C = R | T = 0) = P ( C = R,T = 0) /P ( T = 0) = 0 / (1 / 5) = 0. (c) Are the number of tickets received and the color of the car independent? Independence holds if for all a and b , P ( C = a,T = b ) = P ( C = a ) P ( T = b ). However, from part (a) P ( T = 0) = 1 / 5, while we can also find using the formula for the marginal that P ( C = R ) = 4 / 10. However, this implies that P ( C = R,T = 0) = 0 6 = 4 10 1 5 = P ( C = R ) P ( T = 0) , and therefore we conclude that they are not independent. (d) What is the mean of the number of tickets? What is the variance of the number of tickets?...
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- Spring '10