A business manager who needs to make many phone calls has estimated that when she calls a client, the probability that she will

Question

2 months ago

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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.

Mathematics

1 answer:

Svet_ta [12.7K] · Answer 1 · 2026-03-25T11:54:01+03:00

Answer:

a. 0.68 or 68%

b. 0.08 or 8%

c. 0.32 or 32%

Step-by-step explanation:

The probability of contacting the client on the first call is 60%

The likelihood of reaching the client on the second call is 20%

a. The chance of the manager successfully connecting with her client within two calls is the sum of the chances for 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 connects during the second call but not the first is:

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

c. The probability that the manager fails to connect in two consecutive calls (requiring more calls) is P(X>2):

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