Question

Alice and Bob people arrive at the train station sometime between 4:00 and 4:20 with uniform probability and independent of each other. The trains arrive at 4:02, 4:12, 4:22. (i) What is the probability that Alice needs to wait more than 2 minutes? (ii) What is the probability both of these two people take the 4:02 train? (iii) What is the probability that these two people take different trains? (iv) What is the probability that they both take the 4:12 train given that they take the same train?

Answer 1

i)probability that Alice needs to wait more than 2 minutes=P(arrive between 4:02 to 4:10)+P(arrive between 4:12 to 4:20)

=(8+8)/20=16/20=0.8

ii)probability both of these two people take the 4:02 train =P(both arrive before 4:02) =(2/20)*(2/20)=0.01

iii)

probability that these two people take different trains=1-P(take same trians)=1-P(both arrive before 4:02)-P(arrive between 4:02 and 4:12)-P(arrive between 4:12 and 4:20) =1-(2/20)*(2/20)-(10/20)*(10/20)-(8/20)*(8/20)=0.58

iv)P(both take 4:12 train given take same train)=(10/20)*(10/20)/((2/20)*(2/20)+(10/20)*(10/20)+(8/20)*(8/20))=100/168=0.5952