The monthly worldwide average number of airplane crashes of commercial airlines is 3.5. What is the probability that there will be

(a) at least 2 such accidents in the next month;

(b) at most 1 accident in the next month?

Explain your reasoning!

The monthly

hansel

Answered by Aaron 3 years ago

These circumstances fit a Poisson distribution,

$$P(X=k)=\frac{\lambda^k e^{-\lambda}}{k!}$$

with

$$\lambda = 3.5$$

worldwide monthly crashes

**part (a)**

$$P(\text{at least 2 such accidents})=1-P(\text{0 or 1 crashes})$$

$$=1-\frac{3.5^0 e^{-3.5}}{0!}-\frac{3.5^1 e^{-3.5}}{1!}\approx0.8641$$

**part (b)**

this is just the complement of part (a) or

$$1-0.8641=0.1359$$

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Asked: 3 years ago