Question

The phone lines to an airline reservation system are occupied 40% 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

Answer 1

Answer:

a) $P(X=3)$

And we can use the probability mass function and we got:

$P(X=3) = (10C3) (0.4)^3 (1-0.4)^{10-3} = 0.215$

b) For this case we want this probability:

$P(X \geq 1)$

and the probability that the one call would be not occupied is 1-0.4=0.6

And if we use the complement rule we got:

$P(X \geq 1)= 1-P(X$

$P(X=0) = (10C0) (0.6)^0 (1-0.6)^{10-0} = 0.000105$

$P(X \geq 1)= 1-P(X$

c) $E(X) = np = 10*0.4 = 4$

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

a. What is the probability that for exactly three calls, the lines are occupied?

Let X the random variable of interest, on this case we now that:  

$X \sim Binom(n=10, p=0.4)$  

The probability mass function for the Binomial distribution is given as:  

$P(X)=(nCx)(p)^x (1-p)^{n-x}$  

Where (nCx) means combinatory and it's given by this formula:  

$nCx=\frac{n!}{(n-x)! x!}$  

And for this case we want this probability:

$P(X=3)$

And we can use the probability mass function and we got:

$P(X=3) = (10C3) (0.4)^3 (1-0.4)^{10-3} = 0.215$

b. What is the probability that for at least one call, the lines are not occupied?

For this case we want this probability:

$P(X \geq 1)$

and the probability that the one call would be not occupied is 1-0.4=0.6

And if we use the complement rule we got:

$P(X \geq 1)= 1-P(X$

$P(X=0) = (10C0) (0.6)^0 (1-0.6)^{10-0} = 0.000105$

$P(X \geq 1)= 1-P(X$

c. What is the expected number of calls in which the lines are all occupied?

For this case the expected value is given by:

$E(X) = np = 10*0.4 = 4$