Posted by **MoJo** on Thursday, January 24, 2008 at 11:03am.

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?

- Stats -
**drwls**, Thursday, January 24, 2008 at 11:12am
What is the question? Are they asking for the probabilities of 0, 1, 2, 3 etc per year?

- Stats -
**MoJo**, Thursday, January 24, 2008 at 12:17pm
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?

- Stats -
**Damon**, Thursday, January 24, 2008 at 4:30pm
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.

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