In order to get a ticket you would have to pass through location , which has probability of .2, when the radar "trap" was operating, which has probability .4, so that both happening (passing through while it is operating) has probability (.2)(.4)= .08 OR pass through location , which has probability .1, when the radar "trap" is operating, which has probability .3, so that both happening has probability (.1)(.3)= .03, or ..., etc. The total probability of all those "ors" is the sum of each probability.

I like to handle these kids of problems "en masse". That is, suppose the person drives to work 100 times. In those 100 times, he will pass .4(100)= 40 times and will get a ticket .2(40)= 8 times. He will pass .3(100)= 30 times and will get a ticket .1(30)= 3 times. He will pass .2(100)= 20 times and will get a ticket .5(20)= 10 times. He will pass .3(100)= 30 times and will get a ticket .2(30)= 6 times.2. if a person received a speeding ticket on his way to work, what is the probability that he passed the trap located at L_{2}

In total, then, he got 8+ 3+ 10+ 6= 27 tickets (which also gives an answer to (1)), 3 of them at . Given that he got a ticket, the probability it was from is .

(Let's hope that with that many speeding tickets he loses his license and stays off the road!)