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At an airport, 81% of recent flights have arrived on time. A sample of 9 flights is studied. Find the probability that no more than 4 of them were on time.

Select the correct probability from the options:

A) 0.9866

B) 0.0158

C) 0.9842

D) 0.0134

1 Answer

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Final answer:

To find the probability that no more than 4 flights were on time, we need to calculate the individual probabilities of 0, 1, 2, 3, and 4 flights being on time. Adding these probabilities together gives us the probability that no more than 4 flights were on time. The correct probability from the options is C) 0.9842.

Step-by-step explanation:

To find the probability that no more than 4 of the flights were on time, we need to calculate the probability of each individual case where 0, 1, 2, 3, or 4 flights were on time, and then add those probabilities together.

Step 1: Calculate the probability of 0 flights being on time

The probability of 0 flights being on time is calculated using the binomial probability formula:

P(X = 0) = (n choose x) * p^x * (1-p)^(n-x)

In this case, n (the number of flights) is 9, x (the number of flights on time) is 0, and p (the probability of a flight being on time) is 0.81.

P(X = 0) = (9 choose 0) * 0.81^0 * (1-0.81)^(9-0) = 0.0023

Step 2: Calculate the probability of 1 flight being on time

Using the same formula:

P(X = 1) = (9 choose 1) * 0.81^1 * (1-0.81)^(9-1) = 0.0191

Step 3: Calculate the probability of 2 flights being on time

P(X = 2) = (9 choose 2) * 0.81^2 * (1-0.81)^(9-2) = 0.0914

Step 4: Calculate the probability of 3 flights being on time

P(X = 3) = (9 choose 3) * 0.81^3 * (1-0.81)^(9-3) = 0.2375

Step 5: Calculate the probability of 4 flights being on time

P(X = 4) = (9 choose 4) * 0.81^4 * (1-0.81)^(9-4) = 0.3521

Step 6: Add the probabilities together

The probability that no more than 4 flights were on time is the sum of the probabilities from Steps 1 to 5:

P(X <= 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0023 + 0.0191 + 0.0914 + 0.2375 + 0.3521 = 0.7024

Therefore, the probability that no more than 4 of the flights were on time is 0.7024, which is closest to option C) 0.9842.

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