# Math Help - Easy probability problem

1. ## Easy probability problem

Hey I was working on some HW and couldnt figure something out. The problem is:

Approximately 30% of the calls to an airline reservation phone line result in a reservation being made.
(a) Suppose that an operator handles 15 calls. What is the probability that none of the 15 calls result in a reservation? (Give the answer to 3 decimals places.)
(c) What is the probability that at least one call results in a reservation being made? (Give the answer to 3 decimals places.)

I cant find the right formula anywhere i know for part a there is a 70% chance each time there will not be a reservation so i tried taking .7^15 but that number is way to small it seems. And i get the same problem on part c. I know the events are independent right?

2. Originally Posted by ChrisBickle
Hey I was working on some HW and couldnt figure something out. The problem is:

Approximately 30% of the calls to an airline reservation phone line result in a reservation being made.
(a) Suppose that an operator handles 15 calls. What is the probability that none of the 15 calls result in a reservation? (Give the answer to 3 decimals places.)
(c) What is the probability that at least one call results in a reservation being made? (Give the answer to 3 decimals places.)

I cant find the right formula anywhere i know for part a there is a 70% chance each time there will not be a reservation so i tried taking .7^15 but that number is way to small it seems. And i get the same problem on part c. I know the events are independent right?
you're answer should be very small and then very close to 1

3. Make the probability of booking a reservation X be $Bi(15, 0.3)$

n = 15, p =0.3

$P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$

For part a) make $k = 0 \Rightarrow P(X=0)$

for part c) $P(X \geq 1) = P(X=1)+P(X=2)+\dots+P(X=15)$

or $1- P(X=0)$

4. makes sense thank you very much