Question

A telephone survey uses a random digit dialing machine to call subjects. The random digit dialing machine is expected to reach a live person 25% of the time. What is the probability (rounded to two decimal places) that no calls are successful in ten attempts?

Answer 1

5.63% probability (rounded to two decimal places) that no calls are successful in ten attempts

Explanation:

For each call, there are only two possible outcomes. Either they are succesful, or they are not. The probability of a call being sucessful 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)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

The random digit dialing machine is expected to reach a live person 25% of the time.

This means that
`p = 0.25`

What is the probability (rounded to two decimal places) that no calls are successful in ten attempts?

This is P(X = 0) when n = 10. So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(10,0).(0.25)^(0).(0.75)^(10) = 0.0563`

5.63% probability (rounded to two decimal places) that no calls are successful in ten attempts