# Solved (Free): The phone lines to an airline reservation system are occupied 50% of the time. Assume that the events that the lines are occupied

#### ByDr. Raju Chaudhari

Apr 1, 2021

The phone lines to an airline reservation system are occupied 50% of the time. Assume that the events that the lines are occupied on successive calls are independent.

Assume that 10 calls are placed to the airline.

(a) What is the probability that for exactly three calls the lines are occupied?
(b) What is the probability that for at least one call the lines are not occupied?
(c) What is the expected number of calls in which the lines are all occupied?

#### Solution

Let $X$ denote the number of times the line is occupied. Assume that the events that the lines are occupied on successive calls are independent. Given that the probability that the liines are occupied is $p= 0.5$.

Then, $X$ has a binomial distribution with $n = 10$ and $p = 0.5$.

That is $X\sim B(n=10, p=0.5)$.

The probability mass function of $X$ is

 \begin{aligned} P(X=x) = \binom{10}{x} (0.5)^x (1-0.5)^{10-x}, \\ & \quad x=0,1,\cdots, 10. \end{aligned}

a. The probability that for exactly three calls the lines are occupied is

 \begin{aligned} P(X= 3) & =\binom{10}{3} (0.5)^{3} (1-0.5)^{10-3}\\ & = 0.1172\\ \end{aligned}

b. The probability that the call is occupied is $p=0.5$ and the probability that the call is not occupiped is $q = 1-p = 0.5$. The binomial distribution is a symmetric distribution.

The probability that for at least one call ($\geq 1$) the lines are not occupied is

 \begin{aligned} P(X\geq 1) & =1- P(X=0)\\ &= 1-\binom{10}{0} (0.5)^{0} (1-0.5)^{10-0}\\ & = 1-0.001\\ & =0.999 \end{aligned}

c. The expected number of calls in which the lines are all occupied is $E(X) = n*p = 10 \times 0.5 = 5$.