Question

1. When a call is received at an airline’s call center, the probability that it comes from abroad is 0.32, and the probability that it is to make a change to an existing reservation is 0.38.
a. Suppose that you are told that the probability that a call is both from abroad and is to make a change to an
existing reservation is 0.15. Calculate the probability that a randomly selected call is either from abroad or is to make a change to an existing reservation.
b. Suppose now that you are not given the information in part (a), but you are told that the events "the call is from abroad" and "the call is to make a change to an
existing reservation" are independent. What is the probability that a randomly selected call is either from abroad or is to make a change to an existing reservation?

Answer 1

Answer:

a)0.55

b)0.58

Step-by-step explanation:

a) Lets consider that 100 call received from call center. We expect that 32 of them is from abroad and 38 of them is to make a change to an existing reservation. It normally makes that 70 of 100 call is aboard or to make a change. But there are 15 call is in common. Then we need to subtract this number from total calls:

70-15=55 calls we expecting out of 100 calls that both abroad or to make a change.

and the probability is 0.55

b) If we know that they are independent We can use below formula to find in common calls.

$P(X=x,Y=y)=P(X=x)P(Y=y)\\P(X=x,Y=y)=0.32*0.38\\P(X=x,Y=y)=0.12$

There must be 12 calls in common from out of 100 calls. Now we need to subtract 12 from total calls we expected.

70-12=58 calls we expecting out of 100 calls that both abroad or to make a change.

and the probability is 0.58