The phone lines to an airline reservation system are occupied 61% of the time. Assume that the events that the lines are occupied on successive calls are independent. If 11 calls are placed, what is the probability that the lines are occupied for exactly 5 calls

Answer:

0.1373 = 13.73% probability that the lines are occupied for exactly 5 calls

Step-by-step explanation:

For each call, there are only two possible outcomes. Either the lines are occupied, or they are not. The events that the lines are occupied on successive calls are independent. This means that 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 phone lines to an airline reservation system are occupied 61% of the time.

This means that [tex]p = 0.61[/tex]

If 11 calls are placed, what is the probability that the lines are occupied for exactly 5 calls

This is P(X = 5) when n = 11. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 5) = C_{11,5}.(0.61)^{5}.(0.39)^{6} = 0.1373[/tex]

0.1373 = 13.73% probability that the lines are occupied for exactly 5 calls

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