Answer :
Answer:
5.63% probability (rounded to two decimal places) that no calls are successful in ten attempts
Step-by-step 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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
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 [tex]p = 0.25[/tex]
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
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}.(0.25)^{0}.(0.75)^{10} = 0.0563[/tex]
5.63% probability (rounded to two decimal places) that no calls are successful in ten attempts