# 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?

## 2 Answers

According to me the problem in your second solution lies in calculation of probability of atleast one accident in a given month.

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


We cannot assume that probability of k accidents shall be (0.3)^k because there is no information about the distribution and independence of intra-month accidents.

With regards to when to use more than one random variables, it depends on the problem context. For example if we are given two labelled dice and we are asked to find the expected number of tosses till we get a (H,T) then the problem can be relatively easily visualized using two random variables instead of one.

The key to this problem is noting that every month is independent. That means at the start of each year, the probability of getting an accident on the 3rd month is the same, regardless of what year it is. So what is the probability of getting into an accident on the third month of any given year?

\begin{align}P &= P(\text{no accident in first}) P(\text{no accident in second}) P(\text{accident in third}) \\&= (0.7)(0.7)(0.3) \end{align}