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, 50% 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.)
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) 0.109375 = 0.109 to 3 d.p

b) 1.00 to 3 d.p

Step-by-step explanation:

Probability of someone that made a reservation not showing up = 50% = 0.5

Probability of someone that made a reservation showing up = 1 - 0.5 = 0.5

a) If six reservations are made, what is the probability that at least one individual with a reservation cannot be accommodated on the trip?

For this to happen, 5 or 6 people have to show up since the limousine can accommodate a maximum of 4 people

Let P(X=x) represent x people showing up

probability that at least one individual with a reservation cannot be accommodated on the trip = P(X = 5) + P(X = 6)

P(X = x) can be evaluated using binomial distribution formula

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 6

x = Number of successes required = 5 or 6

p = probability of success = 0.5

q = probability of failure = 0.5

P(X = 5) = ⁶C₅ (0.5)⁵ (0.5)⁶⁻⁵ = 6(0.5)⁶ = 0.09375

P(X = 6) = ⁶C₆ (0.5)⁶ (0.5)⁶⁻⁶ = 1(0.5)⁶ = 0.015625

P(X=5) + P(X=6) = 0.09375 + 0.015625 = 0.109375

b) If six reservations are made, what is the expected number of available places when the limousine departs?

Probability of one person not showing up after reservation of a seat = 0.5

Expected number of people that do not show up = E(X) = Σ xᵢpᵢ

where xᵢ = each independent person,

pᵢ = probability of each independent person not showing up.

E(X) = 6(1×0.5) = 3

If 3 people do not show up, it means 3 people show up and the number of unoccupied seats in a 4-seater limousine = 4 - 3 = 1

So, expected number of unoccupied seats = 1