Question

A business manager who needs to make many phone calls has estimated that when she calls a client, the probability that she will reach the client right away is 60%. If she does not reach the client on the first call, the probability that she will reach the client with a subsequent call in the next hour is 20%. a. Find the probability that the manager reaches her client in two or fewer calls. b. Find the probability that the manager reaches her client on the second call but not on the first call. c. Find the probability that the manager is unsuccessful on two consecutive calls.

Answer 1

Answer:

a.  0.68 or 68%

b. 0.08 or 8%

c. 0.32 or 32%

Step-by-step explanation:

Probability of reaching client on the first call = 60%

Probability of reaching client on the second call = 20%

a. The probability that the manager reaches her client in two or fewer calls is the probability of reaching a client with one or two calls:

$P(X\leq2) = P(X=1) +P(X=2)\\P(X\leq2) = 0.60+(1-0.60)*0.2\\P(X\leq2) = 0.68 = 68\%$

b. The probability that the manager reaches her client on the second call but not on the first call is:

$P(X=2) = (1-0.6)*0.2 =0.08 = 8\%$

c. The probability that manager is unsuccessful on two consecutive calls (needing more than two calls), P(X>2), is:

$P(X>2) = 1-P(X\leq2) = 1-0.68\\P(X>2) = 0.32 = 32\%$