Answer:
0.064 = 6.4% probability that none of the 10 calls result in a reservation.
Step-by-step explanation:
For each call, there are only two possible outcomes. Either it results in a reservation, or it does not. The probability of a call resulting in a reservation is independent of other calls. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
24% of the calls to an airline reservation phone line result in a reservation being made.
This means that ![p = 0.24](https://tex.z-dn.net/?f=p%20%3D%200.24)
Suppose that an operator handles 10 calls. What is the probability that none of the 10 calls result in a reservation?
This is P(X = 0) when n = 10. So
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{10,0}.(0.24)^{0}.(0.76)^{10} = 0.064](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B10%2C0%7D.%280.24%29%5E%7B0%7D.%280.76%29%5E%7B10%7D%20%3D%200.064)
0.064 = 6.4% probability that none of the 10 calls result in a reservation.