Poisson Distribution practice problem I have just started to learn Poisson Distribution and I have no idea how to deal with the following practice from my textbook: Suppose the average amount of cars passing on a street per minute is Poisson (8.6). It takes a lady 5 seconds to cross a street and she waits for two cars to pass. What is the probability that she is safe to cross the street ?
The lady waits for 2 cars means:

the first 5 seconds, the lady waits because there is car passing.
the second 5 seconds, the lady waits because there is car passing.
the third 5 seconds, the lady waits because there is no car passing.

there is no car passing is P(0)=48.8% (you can do the math)
there is car passing is 1-P(0)=41.2%

so the probability is 41.2% x 41.2% x 48.8%
THe first step is to calculate the law of the number of car that pass on the street in 5s.

It can be shown that it follow a Poisson law with $\lambda = 5\times\frac{8.6}{60} \simeq 0.7166$

Then the number of car that pass on the street in 5s is 

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

So the probability that no car pass in the street in the next 5s is 

$$P(X=0) \simeq e^{-0.7166} \simeq 48.8 \%$$