Miss Dombo, a Ticketing Manager with an Airline flying international routes, has realised
that not all passengers show up for their reserved seat. It is clear to her that each passenger’s
behaviour is independent of the other. Having fixed the chances of a passenger not showing
up at ten in a hundred, she then instructs the Ticketing Office to sell 123 tickets for a flight
that holds only 118 passengers. On the basis of this information, compute, for a particular
flight, the probability that
(a) every passenger who shows up can take the flight. [7]
(b) the flight departs with empty seat(s). [5]
(c) the flight departs with exactly five empty seats. [3]
To solve this problem, we can use the concept of binomial probability.
Given information:
- The probability of a passenger not showing up is 10 in 100, or 0.1.
- There are 123 tickets sold for a flight that holds only 118 passengers.
Now let's calculate the probabilities step by step:
(a) The probability that every passenger who shows up can take the flight:
Since each passenger's behavior is independent, the probability that a passenger shows up is 1 - 0.1 = 0.9. We need all 118 passengers who show up to be able to take the flight.
The probability that each passenger can take the flight is given by:
P(Each passenger takes the flight) = (0.9)^(118) = 0.2098
Therefore, the probability that every passenger who shows up can take the flight is approximately 0.2098.
(b) The probability that the flight departs with empty seat(s):
Since there are 123 tickets sold for a flight that holds only 118 passengers, there will be either 0, 1, 2, 3, or 5 empty seats.
The probability that the flight departs with empty seat(s) can be calculated as the sum of probabilities of having 0, 1, 2, 3, or 5 empty seats:
P(Flight departs with empty seat(s)) = P(0 empty seats) + P(1 empty seat) + P(2 empty seats) + P(3 empty seats) + P(5 empty seats)
Let's calculate each probability separately:
P(0 empty seats) = (0.9)^118
P(1 empty seat) = 118C1 * (0.9)^117 * (0.1)
P(2 empty seats) = 118C2 * (0.9)^116 * (0.1)^2
P(3 empty seats) = 118C3 * (0.9)^115 * (0.1)^3
P(5 empty seats) = 118C5 * (0.9)^113 * (0.1)^5
Calculating each probability above yields:
P(0 empty seats) = 0.036
P(1 empty seat) = 0.118
P(2 empty seats) = 0.243
P(3 empty seats) = 0.270
P(5 empty seats) = 0.216
Therefore, the probability that the flight departs with empty seat(s) is approximately 0.036 + 0.118 + 0.243 + 0.270 + 0.216 = 0.883.
(c) The probability that the flight departs with exactly five empty seats:
We have already calculated P(5 empty seats) = 0.216 in the previous part.
Therefore, the probability that the flight departs with exactly five empty seats is approximately 0.216.
To solve this problem, we need to use the concept of probability. Let's define some variables for easier understanding:
Let P(A) be the probability of event A happening.
Let n be the number of passengers who show up for the flight.
Let p be the probability that a passenger does not show up.
(a) To find the probability that every passenger who shows up can take the flight, we need to find the probability that the number of passengers who show up (n) is less than or equal to the number of available seats (118).
Since each passenger's behavior is independent of the others, we can use the binomial probability formula:
P(n <= 118) = ∑ [P(A) * P(B)]
Where A = "Exactly n passengers show up" and B = "All n passengers can take the flight."
We can calculate the probability of exactly n passengers showing up as P(A) = (1 - p)^n * p^(118 - n). And the probability that all n passengers can take the flight as P(B) = 1.
Since p = 10/100 = 0.1 (the chance of a passenger not showing up), we can substitute the values into the formula:
P(n <= 118) = ∑ [(1 - 0.1)^n * 0.1^(118 - n)]
By summing up the probabilities for each value of n from 0 to 118, we can find the probability that every passenger who shows up can take the flight.
(b) To find the probability that the flight departs with empty seat(s), we need to find the probability that the number of passengers who show up (n) is less than or equal to the number of available seats minus one (117 in this case).
Using the same calculation as in (a), we can find:
P(n <= 117) = ∑ [(1 - 0.1)^n * 0.1^(118 - n)]
(c) To find the probability that the flight departs with exactly five empty seats, we need to find the probability that the number of passengers who show up (n) is equal to the number of available seats minus five (113 in this case).
Using the same calculation as in (a), we can find:
P(n = 113) = (1 - 0.1)^113 * 0.1^(118 - 113)
By substituting the values into the formulas and performing the calculations, we can find the probabilities for each scenario.