Find the indicated probabilities using the geometric distribution, the Poisson distribution, or the binomial distribution. Then determine if the events are unusual. If convenient, use the appropriate probability table or technology to find the probabilities. Assume the probability that you will make a sale on any given telephone call is 0.14 . Find the probability that you (a) make your first sale on the fifth call, (b) make your sale on the first, second, or third call, and (c) do not make a sale on the first three calls. (a) $\mathrm{P}($ make your first sale on the fifth call $)=0.077$ (Round to three decimal places as needed.) (b) $P$ (make your sale on the first, second, 5 r third call $)=0.364$ (Round to three decimal places as needed.) (c) $P($ do not make a sale on the first three calls $)=$ (Round to three decimal places as needed.)

Step 1 ：The problem is asking for the probability of different scenarios given a success probability of 0.14 for each call.

Step 2 ：For part (a), we are asked to find the probability that the first sale is made on the fifth call. This is a geometric distribution problem. The formula for the geometric distribution is \(P(X=k) = (1-p)^{(k-1)} * p\), where p is the probability of success, k is the number of trials until the first success, and \(P(X=k)\) is the probability that the first success occurs on the kth trial.

Step 3 ：Using the formula, we find that \(P_a = 0.0765811424\).

Step 4 ：For part (b), we are asked to find the probability that a sale is made on the first, second, or third call. This is a binomial distribution problem. The formula for the binomial distribution is \(P(X=k) = C(n, k) * p^k * (1-p)^{(n-k)}\), where p is the probability of success, n is the number of trials, k is the number of successes, and \(P(X=k)\) is the probability of k successes in n trials. However, since we are looking for the probability of a sale on the first, second, or third call, we need to sum the probabilities for k=1, k=2, and k=3.

Step 5 ：Using the formula, we find that \(P_b = 0.36394400000000005\).

Step 6 ：For part (c), we are asked to find the probability that no sale is made on the first three calls. This is also a binomial distribution problem, but in this case, we are looking for the probability of 0 successes in 3 trials, so k=0.

Step 7 ：Using the formula, we find that \(P_c = 0.636056\).

Step 8 ：Final Answer: (a) The probability that you make your first sale on the fifth call is approximately \(\boxed{0.077}\). (b) The probability that you make your sale on the first, second, or third call is approximately \(\boxed{0.364}\). (c) The probability that you do not make a sale on the first three calls is approximately \(\boxed{0.636}\).