ku Posted December 14, 2005 Share Posted December 14, 2005 The number of accidents on XYZ highway each day is a Poisson random variable (r.v.) with mean 3. We know that these numbers are independent for different days.(i) What is the probability that no accidents occur today? (ii) What is the probability that there will be exactly 2 accidents on XYZ highway during this weekend? There will be at least 2 accidents? (iii) We have not got complete data for yesterday yet, but it is already known that there occurred at least 2 accidents. What is the probability that there were exactly 3 accidents yesterday given that information? The answer for the (i) is 0.049, which I got by evaluating Poisson pdf [math]\frac{e^{-3}3^{0}}{0!} = e^{-3} \approx 0.049[/math], but I'm not sure what to do for (ii) and (iii). The answers are (ii) 0.0446; 0.9826 and (iii) 0.2798. I just don't know how the answers were obtained. Link to comment Share on other sites More sharing options...

Tartaglia Posted January 20, 2006 Share Posted January 20, 2006 P(X = 2) = 6^2*e^-6/2! = 0.0446 P(X >=2) = 1 - P(X=0) - P(X=1) = 1 - e^-6 - 6e^-6 = 0.98265 P(x = 3 | X >= 2) = P(x = 3)/ P(X >= 2) P(X =3) = 3^3*e-3/3! = 0.22404 P(X >= 2) = 1 - P(X =0) -P(X = 1) = 1 - e^-3 - 3e^-3 = 1 - 4*e^-3 = 0.8008517 P(x = 3 | X >= 2) = 0.22404/0.8008517 = 0.27975 Link to comment Share on other sites More sharing options...

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