# Confusion With “Nested” Random Variables

Problem

The probability that a compnay's workforce has no accidents in a given month is $0.7$. The numbers of accidents from month to month are independent. What is the probability that the third month in a year is the first month that at least one accident occurs?

My confusion lies with the use of random variables.

1. The answer is as simple as letting the random variable $X$ represent the month upon which at least one accident occurs and assuming that $X$~Geom. If we assign let $p=0.3$ represent the probability that an accident occurs in a month and $q=0.7$ is the probability that no accidents occur in the same month, then the solution to the problem when $X=3$ is $(0.7)^2(0.3)$.

2. However, this is how I initally solved the problem. If we let $Y$ represent the number of accidents that occur in a given month, then the probability that at least one accident occurs on any given month is $\sum_{y=1}^\infty (0.3)^y = \frac{0.3}{0.7}.$ Now, let $\psi = \frac{0.3}{0.7}$ represent the probability that at least one accident occurs in a given month. Thus, if $X$ is the month number on which an accident occurs, then $X$~Geom. When $X=3$, we have

$$\psi*(1-\psi)^{3-1} = \frac{0.3}{0.7} * (0.7)^2 = (0.3)(0.7)$$

which is wrong.

QUESTIONS

1. Why is that a single random variable in 1. can represent the month number on which at least one accident occurs without having to compute the probability that at least one accident occurs in the first place?

2. Why doesn't my method in 2. work? It seems legitimate, because we are given the probability that an accident, i.e. ony one accident, occurs, so I went ahead and computed the probability that at least one accident happens, and then used that as my probability of success for the geometric distribution.

3. How do you know when to use more than one random variable?

Prob(atleast one) is simply 1 - Prob(None) = 1 - 0.7 = 0.3