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?
3 answers
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?
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.
= (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.
1 answer