Fortunately, plane crashes are rare events. Suppose that commercial crashes occur on an average (mean) of 1.1 per year (Hint: Poisson).

a. What is the probability that this year will be crash free?
b. If there is a crash, what is the probability that there will be more than two?

What is the question? Are they asking for the probabilities of 0, 1, 2, 3 etc per year?

a. What is the probability that this year will be crash free?

b. If there is a crash, what is the probability that there will be more than 2?

Probability of zero = 1.1^0 e^-1.1 / 0!

= (1/1 )e^-1.1 = .332
of 1 = 1.1^1 (.332)/1! = .365
of 2 = 1.1^2 (.332)/2! = .201

The sum of the probabilities of zero, 1 and 2 is
.898
SO, the probability of more than two is
1 - .898 = .102
ten percent. That is scary.

To solve these questions, we can use the Poisson distribution since we know the average rate at which plane crashes occur. The Poisson distribution is often used to model rare events, such as accidents or crashes, occurring randomly over time or space.

The probability mass function (PMF) of the Poisson distribution is given by:

P(X = k) = (e^(-λ) * λ^k) / k!

where:
- X is the random variable representing the number of plane crashes
- λ is the average rate at which plane crashes occur (given as 1.1 per year in this case)
- e is the mathematical constant approximately equal to 2.71828
- k is the number of plane crashes we are interested in calculating the probability for

Let's proceed with solving the questions:

a. What is the probability that this year will be crash-free?

To find the probability of having zero crashes in a year, we substitute k = 0 into the Poisson PMF equation:

P(X = 0) = (e^(-1.1) * 1.1^0) / 0!

Simplifying further:

P(X = 0) = e^(-1.1)

Using a calculator, we find that P(X = 0) is approximately 0.3329 or 33.29%.

Therefore, the probability that this year will be crash-free is approximately 0.3329 or 33.29%.

b. If there is a crash, what is the probability that there will be more than two?

To find the probability of having more than two crashes given that at least one crash occurs, we need to calculate the conditional probability.

First, we need to find the probability of having at least one crash:

P(X ≥ 1) = 1 - P(X = 0)

P(X ≥ 1) = 1 - e^(-1.1)

Using a calculator, we find that P(X ≥ 1) is approximately 0.6671 or 66.71%.

Now, we can find the probability of having more than two crashes given that at least one crash occurs:

P(X > 2 | X ≥ 1) = P(X > 2) / P(X ≥ 1)

To find P(X > 2), we can subtract the probability of having zero, one, and two crashes from 1:

P(X > 2) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))

Using the Poisson PMF equation with k = 1 and k = 2:

P(X > 2) ≈ 1 - (e^(-1.1) + (e^(-1.1) * 1.1) + (e^(-1.1) * (1.1^2) / 2))

Using a calculator, we find that P(X > 2) is approximately 0.284 or 28.4%.

Now, we can calculate the conditional probability:

P(X > 2 | X ≥ 1) = (0.284) / (0.6671)

P(X > 2 | X ≥ 1) is approximately 0.426 or 42.6%.

Therefore, if there is a crash, the probability of having more than two crashes is approximately 0.426 or 42.6%.