1. A mail-order computer business has six telephone lines. Let X ...

7. Customers at a gas station select either regular (A), premium (B), or
diesel fuel (C). Assume that successive customers make independent
choices, with P (A) = 0.3, P (B) = 0.2, and P (C) = 0.5.
(a) Among the next 100 customers, what are the mean and variance
of the number who select regular fuel? Explain your reasoning.
Ans The no. who select regular do so with P (A) = 0.3, therefore
who do not select regular do so with 1 − P (A) = P (B) + P (C) =
0.7. Define Xi = 1 if i-th customer chooses regular, 0 otherwise.
This is therefore a Bernoulli r.v. with parameter, p = 0.3.
Therefore the total no. of customers among the 100 customers
who choose regular, is a Binomial r.v. = Xi. Therefore the
mean and variance are 100 × 0.3 = 30 and 100 × 0.3 × 0.7 = 21.
(b) Answer part (a) for the number among the 100 who select a
nondiesel fuel.
8. An airport limousine can accommodate up to four passengers on any
one trip. The company will accept a maximum of six reservations
for a trip, and a passenger must have a reservation. From previous
records, 20% of all those making reservations do not appear for the
trip. Answer the following questions, assuming independence wherever
appropriate.
(a) If six reservations are made, what is the probability that at least
one individual with a reservation cannot be accommodated on
the trip?
Ans At least one individual with a reservation cannot be accommodated
on the trip is equivalent to 5 or 6 individuals turning
up. This is a Binomial r.v., as can be seen by defining Xi = 1
if an individual turns up, and taking the total no. turning up as
Y =
i Xi. We now have to find, P [Y = 5] + P [Y = 6].
6
5
0.8 5 (1 − 0.8) 6−5 +
6
6
0.8 6 (1 − 0.8) 6−6
= 0.393216 + 0.262144 = 0.65536
(b) If six reservations are made, what is the expected number of
available places when the limousine departs?
Ans The available no. of places Z = max[4 − Y, 0]. Therefore,
P [Z = k] = P [Y = 4 − k], k = 1, . . . , 4, and P [Z = 0] = P [Y =
4] + P [Y = 5] + P [Y = 6]. The expected value is therefore,
1.P [Y = 3] + 2.P [Y = 2] + 3.P [Y = 1] + 4.P [Y = 0]
= 1.P [Y = 3] + 2.P [Y = 2] + 3.P [Y = 1] + 4.P [Y = 0]
6