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

Question

I am unable to understand the following statement.

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.

$\textbf{What is the probability that for exactly three calls the lines are occupied?}$

I am not exactly confused with how I will find the answer. I am simply having difficulties understanding the question.

Here is my current understanding of the statement:

Each of the phone lines of an airline reservation system are occupied 40% of the time on a certain time interval. The probability that one phone line is occupied is independent of the vacancy of other phone lines, thus, the probability that any phone lines will be occupied at a given time interval will always be 40%. $\textbf{The airline has 10 phone lines.}$

The last sentence stating "Assume that 10 calls are placed to the airline" confuses me. Does this mean that the airline has 10 phone lines? Or are there 10 phone calls to the airline. The latter interpretation seems to be pointless however. I just want to make sure that I am not missing something.

Answer 1

I guess that there is one thing called "The phone lines to an airline reservation system" and they as a whole are occupied for 40% of time.

For each call this thing can be either occupied (40%) or not and for each call this event is independent.

We take 10 calls.

What is the probability that for exactly three calls the lines (thing) are occupied?

Should be simple application of binomial law.

Answer 2

I believe the question is actually pointing to a timeframe - 3 consecutive (or concurrent) calls and the probability that each of the phones is busy for either 3 calls in a row or 3 incoming callers

Answer 3

I think that you're interpretation of the situation is correct.

Let $X$ be the discrete random variable 'no. of lines occupied'

Then $X \sim \mathrm{Bin}(x;0.4,10)$

$$P(X=3)=\cfrac{10!}{(10-3)!\cdot3!}\times 0.4^3\times0.6^7$$