Let the moment generating function of random variable X be M_X(t) = 37e^2t + 7/7
ID: 3257304 • Letter: L
Question
Let the moment generating function of random variable X be M_X(t) = 37e^2t + 7/78 + 17e^-1/39. (a) Use the mgf M_X(t) to find E(X^2(X - 3)). (b) Find P(X^2 lessthanorequalto 2). In a particular city, accidents at intersections A, B, and C are independent, each with a Poisson distribution. The intersections are equally likely to be chosen by drivers. Intersection A has a rate of one accident per day, intersection B has a rate of one accident per two days, and intersection C has a rate of one accident per three days. Suppose in a particular day, there was one accident. What is the probability that it came from intersection B?Explanation / Answer
The Poisson probability is:
P(x; ) = (e-) (x) / x! where is the rate of the distribution.
For intersection A, = 1 accident per day = 1
For intersection B, = 1 accident per 2 days = 0.5 accident per day
For intersection C, = 1 accident per 3 days = 0.33 accident per day
Let the probability that there is only one accident at intersection A, B and C be Pa, Pb and Pc.
Pa = P(1; 1) = (e-0.5) (0.51) / 1! = 0.3679
Pb = P(1; 0.5) = (e-1) (11) / 1! = 0.3033
Pc = P(1; 0.33) = (e-0.33) (0.331) / 1! = 0.2372
Probability that there is only one accident (at intersection A or B or C)
= Pa(1-Pb)(1-Pc) + (1-Pa)Pb(1-Pc) + (1-Pa)(1-Pb)Pc
= 0.3679*(1-0.3033)*(1-0.2372) + (1-0.3679)*0.3033*(1-0.2372) + (1-0.3679)*(1-0.3033)*0.2372
= 0.4462
Probability that there is only one accident at intersection B = (1-Pa)Pb(1-Pc)
= (1-0.3679)*0.3033*(1-0.2372) = 0.1462
By conditional probability, Given there is only one accident, Probability that there is only one accident at intersection B
= Probability that there is only one accident at intersection B/ Probability that there is only one accident
= 0.1462/0.4462 = 0.3276
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