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A waste disposal company averages 6:5 spills of toxic waste per month. Assume spills occur randomly at a uniform rate, and independently of each other, with a negligible chance of 2 or more occurring at the same time. Find the probability there are 4 or more spills in a 2 month period

The way I read this, let $X = \{\text{Number of spills in 2 month period}\}$

We want $P(X \ge 4)$.

But I am confused and dont know how to proceed?

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    What does $6:5$ mean in this context? Beyond that, I'd say it sounds like a good situation in which to use a [Poisson distribution](https://en.wikipedia.org/wiki/Poisson_distribution) – 2017-02-10

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I assume you mean that the rate of spills per month is 6.5. Then you might model spills over a two-month period as $Y \sim \mathsf{Pois}(\lambda_2 = 13).$ With that model, you seek $P(Y \ge 4) = 1 - P(Y \le 3).$

You can find this probability using a calculator and the PDF for a Poisson distribution or using software. I got an answer very near one, which makes sense since the average number of spills in two months is $E(Y) = \lambda_2 = 13.$

The sketch below shows the PDF of $Y.$ You want the probability represented by the bars to the right of the vertical dotted line.

enter image description here

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    Still there? Can you find $P(Y=0) = e^{-13}, P(Y=1), P(Y=2),$ and $P(Y=3)?$ Back in several hours. – 2017-02-11