Question

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, 30% of all those making reservations do not appear for the trip. Answer the following questions, assuming independence wherever appropriate. (Round your answers to three decimal places.)
(a) If six reservations are made, what is the probability that at least one individual with a reservation cannot be accommodated on the trip?
(b) If six reservations are made, what is the expected number of available places when the limousine departs?

Answer 1

Answer:

(a) The probability that at least one individual with a reservation cannot be accommodated on the trip is 0.4202.

(b) The expected number of available places when the limousine departs is 0.338.

Step-by-step explanation:

Let the random variable Y represent the number of passenger reserving the trip shows up.

The probability of the random variable Y is, p = 0.70.

The success in this case an be defined as the number of passengers who show up for the trip.

The random variable Y follows a Binomial distribution with probability of success as 0.70.

(a)

It is provided that n = 6 reservations are made.

Compute the probability that at least one individual with a reservation cannot be accommodated on the trip as follows:

P (At least one individual cannot be accommodated) = P (X = 5) + P (X = 6)

$={6 \choose 5}\ (0.70)^{5}\ (1-0.70)^{6-5}+{6 \choose 6}\ (0.70)^{6}\ (1-0.70)^{6-6}\\\\=0.302526+0.117649\\\\=0.420175\\\\\approx 0.4202$

Thus, the probability that at least one individual with a reservation cannot be accommodated on the trip is 0.4202.

(b)

The formula to compute the expected value is:

$E(Y) = \sum X\cdot P(X)$

$P (X=0)={6 \choose 0}\ (0.70)^{0}\ (1-0.70)^{6-0}=0.000729\\\\P (X=1)={6 \choose 1}\ (0.70)^{1}\ (1-0.70)^{6-1}=0.010206\\\\P (X=2)={6 \choose 2}\ (0.70)^{2}\ (1-0.70)^{6-2}=0.059535\\\\P (X=3)={6 \choose 3}\ (0.70)^{3}\ (1-0.70)^{6-3}=0.18522\\\\P (X=4)={6 \choose 4}\ (0.70)^{4}\ (1-0.70)^{6-4}=0.324135$

Compute the expected number of available places when the limousine departs as follows:

$E(Y) = \sum X\cdot P(X)$

         $=(4\cdot 0.000729)+(3\cdot 0.010206)+(2\cdot 0.059535)+(1\cdot 0.18522)\\+(0\cdot 0.324135)\\\\=0.002916+0.030618+0.11907+0.18522+0\\\\=0.337824\\\\\approx 0.338$

Thus, the expected number of available places when the limousine departs is 0.338.