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
