Question

Approximately 15% of the calls to an airline reservation phone line result in a reservation being made.
(a) Suppose that an operator handles 10 calls. What is the probability that none of the 10 calls result in a reservation? (Give the answer to 3 decimals places.)

Answer 1

probability of a call resulting in a reservation = .15
probability of a call not resulting in a reservation = .85

15 calls are made.

what is the probability that at least 1 of them will be a reservation?

that equals 1 minus the probability that all of them do not result in a reservation.


since the probability of any one call not resulting in a reservation equals .85, then the probability that all 15 calls will not result in a reservation equals .85^15.

that number is .087354219.

the probability that at least 1 call will result in a reservation is 1 minus the probability that no calls will result in a reservation.

that probability = .912645781.

that's the probability that at least 1 call will result in a reservation.

Answer 2

but it's 10 calls?

Answer 3

(a) Suppose that an operator handles ***10 calls***. What is the probability that none of the 10 calls result in a reservation? (Give the answer to 3 decimals places.)

Answer 4

then take my equation and do it for 10 calls?

Answer 5

okay

Answer 6

What's part C, then
'
(c) What is the probability that at least one call results in a reservation being made? (Give the answer to 3 decimals places.)'

Answer 7

\[ \large
P(\text{ k successes in n trials }) =
\left(\begin{matrix}n \\ k\end{matrix}\right)p^{k}q^{n-k} \\ \large
n = 10;~~ k = 0; ~~p = 0.15; ~~q = 0.85
\]

Answer 8

what does that mean?

Answer 9

what do I have to do with the n and k? Multiply them? Divide?

Answer 10

Within the parenthesis ^

Answer 11

Have you done binomial probability distribution yet?
\[
\left(\begin{matrix}n \\ k\end{matrix}\right) = \frac{n!}{(n-k)! * k!}
\]

Answer 12

thank you! And no...