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 indepent.Assume that 10 calls are placed to the airline.
(a)What is the probability for exeactly 3 calls the lines are occupied
(b)What is the probability that for at least 1 call the lines are
occupied
(c)What is the expected number of calls which the lines are occupoed

(a) the number of different ways exactly 3 calls find occupied lines
is

10!/(3!*(10-3)!), or 120.  

The probability of each of these occuring is .4^3 * .6^7, or .00179

So the answer is 120*.00179 = .215, or 21.5% chance.

(b) The probability that the lines are occupied for at least one call
is 1 minus the probability that the lines are occupied for none of the
calls.  That probability is .6^10, so the answer is 1-.6^10 = .994, or
99.4% chance.

(c) The expected number is just 40% of 10, or 4.

racecar may have the correct answer, but in the real world, the calls
do not come at random intervals, but are modified by plane arrival and
departure scheduals, news, ads etc, so at times there may be a dozen
people trying unsuccessfully to reach the reservation desk, and only
one will be talking typically, most of the time.   Neil