Download Engr. 323 H.W. #5 Prob. 62 Fergeson 3/5

Engr. 323
H.W. #5
Prob. 62
Fergeson
3/5
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?
There are C3,10 (10 choose 3) different ways, that for exactly three calls, the lines are
occupied, that is
C3,10 = 10!/3!7! = 120 different ways that for exactly 3 calls the
lines are occupied.
Next, we know the lines are occupied 40% of the time.
Therefore, we have (120)(0.4)3(0.6)7 ≈0.21499
Alternatively, (referring to Ch. 3 Sec. 6 Pg. 125)
You could have recognized that we have a random experiment consisting of
n = 10 successive calls in which:
1.) The calls are independent
2.) Each call results in only two possible outcomes, either busy (b) or open (0), &
3.) The probability of a busy signal during each call, denoted as p = 0.4, remains
constant.
Such a situation is called a binomial experiment.
The random variable X is the number of calls that result in a busy signal, and has a
binomial distribution with parameters p = 0.4 and n = 10.
Furthermore, the probability mass function of a binomial distribution is given by:
f X ( x; p, n) = (C x ,n )( p) x (1 − p) n− x
x = 0,1,2,3,...., n = 10
Then, p = 0.4(the probability that a busy signal is received)
n = 10 (the number of successive/repeated calls)
The probability that exactly three calls receive a busy signal is P(X = 3)
= f X (3) = (C 3,10 )(0.4) 3 (0.6) 7 ≈0.215 = P ( X = 3)
Engr. 323
H.W. #5
Prob. 62
Fergeson
4/5
b.) What is the probability that for at least one call the lines are not occupied?
Here the question is: What is the probability that at least one call does not receive a busy
signal(the minimum number ≥ 1)? That is, there could be more than one open line; which
gives:
10
P ( X ≥ 1) = ∑ C x ,10 (0.4) x (0.6)10− x = 1 − P ( X < 1) = 1 − P ( X = 0)
x =1
= 1 − [(C 0,10 )(0.4) 0 (0.6)10 ] = 1 − 0.006 ≈0.9939 = P ( X ≥ 1)
c.) What is the expected number of calls in which the lines are all occupied?
The expected number (mean) of calls which result in a busy signal is 40% of the calls
made… That is:
µ X = E ( X ) = np = 10(0.4) = 4 calls (from Ch. 3 Sec. 6 Pg. 128)
E(X) = 4 calls