Probability of Minor Equipment Failure

Given that 1 percent of an airline's flights suffer a minor equipment failure in an aircraft, we need to determine the probability of having exactly two such failures in the next 100 flights.

Understanding the Problem

To solve this problem, we can model it as a binomial distribution. A binomial distribution is used when there are only two possible outcomes (success or failure) and each trial is independent of the others. In this case, the success is a minor equipment failure, and the failure is the absence of any equipment failure.

Calculating the Probability

To calculate the probability of having exactly two minor equipment failures in the next 100 flights, we can use the binomial probability formula:

P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)

where:
- P(X = k) is the probability of having exactly k successes
- n is the total number of trials (in this case, 100 flights)
- k is the number of desired successes (in this case, 2 failures)
- p is the probability of success in a single trial (in this case, 0.01 or 1%)

Applying the Formula

Let's plug in the values into the formula:

P(X = 2) = C(100, 2) * (0.01)^2 * (1 - 0.01)^(100 - 2)

Calculating the binomial coefficient C(100, 2):

C(100, 2) = 100! / (2! * (100 - 2)!)
= 100! / (2! * 98!)
= (100 * 99) / (2 * 1)
= 4950

Now, substituting the values into the formula:

P(X = 2) = 4950 * (0.01)^2 * (0.99)^98
≈ 0.1207

Interpreting the Result

Therefore, the probability of having exactly two minor equipment failures in the next 100 flights is approximately 0.1207, or 12.07%.

It's important to note that this probability only considers the minor equipment failures and assumes each flight is independent. It does not consider other factors that could influence equipment failures or any potential correlations between flights.