Question

help please! show all work!

Cars pass an interaction according to a Poisson process with rate X = 2 per minute. There are only 2 types of cars, and each passing car is, independently, with probability 0.4 and 0.6, of type A and type B, respectively. During a 2-minutes time period, there are 2 type A cars have passed the interaction. Find the probability that the first type A car passed during the first minute and the second type A car passed during the second minute.

Answer 1

rate of type A = rate of car * P(type A)

= 2*0.4 = 0.8

P(k events in time period) = e-events time period. (time * time period)k k!

here,

event/time = 0.8 type A car per min

P(2 type A cars in 2 min) = e^(-0.8*2) * (0.8*2)^2 / (2!) = 0.2584

P(1 type A car in 1st min) = e^(-0.8*1) * (0.8*1)^1 / (1!) = 0.3595

P(1 type A car in 2nd min) = e^(-0.8*1) * (0.8*1)^1 / (1!) = 0.3595

P(1 type A car in 1st min AND 1 type A car in 2nd min) = P(1 type A car in 1st min)*P(1 type A car in 2nd min)

= 0.3595*0.3595

= 0.1292

P(1 type A car in 1st min AND 1 type A car in 2nd min | 2 type A cars in 2 min)

= P(1 type A car in 1st min AND 1 type A car in 2nd min) / P(2 type A cars in 2 min)

= 0.1292 / 0.2584

= 0.5

(please UPVOTE)