Asked by MoJo
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?
Answers
Answered by
drwls
What is the question? Are they asking for the probabilities of 0, 1, 2, 3 etc per year?
Answered by
MoJo
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?
Answered by
Damon
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.