Question

- Poisson Distribution Q.

Answer 1

Q. The number of gamma rays emitted per second by a certain radioactive substance is a random variable, having the Poisson distribution with Lamda = 5.8. If a recording instrument becomes inoperative when there are more than 12 rays per second, what is the probability that this instrument becomes inoperative during any given second?

Answer 2

Let \(X\) denote the random variable. Then,\[P(\text{inoperative})=P(X>12)=1-P(X\le12)=1-\sum_{x=0}^{12}\frac{e^{-5.8}5.8^x}{x!}\]
Of course, this is only true if "more than 12" translates to "strictly greater than 12."

Also, you have to be careful about the units of the given \(\lambda\). Make sure it's 5.8 rays per second.