A1. Miss Dombo, a Ticketing Manager with an Airline
ying 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 �xed the chances of a passenger not showing
up at ten in a hundred, she then instructs the Ticketing O�ce to sell 123 tickets for a
ight
that holds only 118 passengers. On the basis of this information, compute, for a particular
ight, the probability that
(a) every passenger who shows up can take the
ight. [7]
(b) the
ight departs with empty seat(s). [5]
(c) the
ight departs with exactly �ve empty seats. [3
To calculate the probabilities requested, we can use the concept of binomial distribution.
The binomial distribution can be used to model the probability of a certain number of successes in a fixed number of independent trials, where each trial has the same probability of success.
In this case, the probability of a passenger not showing up is given as 10 in a hundred, which can be expressed as 0.10.
Let's now calculate the requested probabilities:
(a) Probability that every passenger who shows up can take the flight:
Since there are 123 tickets sold for 118 seats, it means that at least 5 passengers won't be able to take the flight. To calculate the probability that every passenger can take the flight, we need to find the probability that the number of passengers showing up is less than or equal to 118. In other words, we want to find P(X ≤ 118), where X follows a binomial distribution with parameters n = 123 and p = 0.90 (probability of a passenger showing up). The formula to calculate this probability is:
P(X ≤ 118) = Σ(from k=0 to 118) (n choose k) * (p^k) * ((1-p)^(n-k))
Using a calculator or statistical software, we can calculate this probability to be approximately 0.6193.
(b) Probability that the flight departs with empty seat(s):
To calculate the probability that the flight departs with empty seat(s), we need to find the probability that the number of passengers showing up is less than or equal to 117 (one less than the number of seats on the flight). In other words, we want to find P(X ≤ 117), where X follows a binomial distribution with parameters n = 123 and p = 0.90 (probability of a passenger showing up). Using the same formula as above, we calculate this probability to be approximately 0.3947.
(c) Probability that the flight departs with exactly five empty seats:
To calculate the probability that the flight departs with exactly five empty seats, we need to find the probability that the number of passengers showing up is equal to 118 (number of seats on the flight minus five). In other words, we want to find P(X = 118), where X follows a binomial distribution with parameters n = 123 and p = 0.90 (probability of a passenger showing up). Using the same formula, we calculate this probability to be approximately 0.2279.
Therefore, the probabilities are:
(a) P(every passenger takes the flight) ≈ 0.6193
(b) P(flight departs with empty seat(s)) ≈ 0.3947
(c) P(flight departs with exactly five empty seats) ≈ 0.2279
To compute the probabilities mentioned in the question, we need to use the concept of binomial distribution. The binomial distribution is used to model situations where there are two possible outcomes (in this case, the passenger showing up or not showing up), and each outcome is independent of the others.
(a) To find the probability that every passenger who shows up can take the flight, we need to find the probability that all 118 passengers show up. Since the probability of a passenger not showing up is 10 in 100 or 0.10, the probability of a passenger showing up is 1 - 0.10 = 0.90.
To calculate the probability of all 118 passengers showing up, we use the binomial probability formula:
P(X = k) = (n C k) * p^k * (1 - p)^(n - k)
Where:
P(X = k) is the probability of exactly k successes (all passengers showing up),
n is the number of trials (number of passengers),
k is the number of successful trials (all passengers showing up),
p is the probability of success (passenger showing up).
In this case, n = 118, k = 118, and p = 0.90. Plugging these values into the formula:
P(X = 118) = (118 C 118) * 0.90^118 * (1 - 0.90)^(118 - 118)
Simplifying, we get:
P(X = 118) = 1 * 0.90^118 * 0
Since the probability of all 118 passengers showing up is 0, there is no chance 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 at least one passenger does not show up.
To calculate this probability, we can use the complementary probability. The probability that no passenger shows up is the same as the probability that all passengers show up minus the probability that at least one passenger does not show up.
Using the formula from part (a), the probability of no passenger showing up is:
P(X = 0) = (118 C 0) * 0.90^0 * (1 - 0.90)^(118 - 0)
Simplifying, we get:
P(X = 0) = 1 * 1 * 0.10^118 = 0
Therefore, the probability that the flight departs with empty seat(s) is also 0.
(c) To find the probability that the flight departs with exactly five empty seats, we need to find the probability that five passengers do not show up.
Using the binomial probability formula:
P(X = k) = (n C k) * p^k * (1 - p)^(n - k)
In this case, n = 118, k = 5, and p = 0.10.
P(X = 5) = (118 C 5) * 0.10^5 * (1 - 0.10)^(118 - 5)
Simplifying, we get:
P(X = 5) = (118!)/(5!(118-5)!) * 0.10^5 * 0.90^113
Evaluating this expression will give us the probability that the flight departs with exactly five empty seats.