Question

. Ace Taxi divides a city into two zones:A and B. If a driver picks up a client in A, he will drop them off in A with probability 0.25 and drop them off in B with probability 0.75. If a driver picks up a client in zone B, he will drop them off in zone A with chance 0.4 and drop them off in zone B with chance 0.6. The company rules dictate that a driver must remain in a given zone for the next client.

(a) Given that the driver starts in zone A compute the probabilities that the driver will be in zones B after two clients have been served.

(b) Given that the driver starts in zone B compute the probabilities that the driver will be in zones A after two clients have been served.

Answer 1

We know that  if a driver picks up a client in zone A, he drops them in A with probability 0.25 and drops them in B with probability 0.75.

We know that  if a driver picks up a client in zone B, he drops them in A with probability 0.4 and drops them in B with probability 0.6.

a.

We know that the driver starts in zone A.

Thus, we must find the P(B | starts in A)

There are two ways this can go:

1. He gets the first ride in A.

= 0.25*0.75

= 0.1875

2. He gets the first ride in B.

= 0.75*0.60

= 0.45

Thus, the required probability is 0.1875+0.45

= 0.6375

b.

We know that the driver starts in zone B.

Thus, we must find the P(A | starts in B)

There are two ways this can go:

1. He gets the first ride in A.

= 0.4*0.25

= 0.10

2. He gets the first ride in B.

= 0.60*0.40

= 0.24

Thus, the required probability is 0.10+0.24

= 0.34