Why is P(maximum of 3 functions >=3 ) = 1- prob. their intersection <= 3

Question

I have this question:

In a small metropolitan area, annual losses due to storm, fire, and theft are independently distributed random variables. The pdf's are:

Storm $e^{-x}$ Fire $\frac{2e^{-2x/3}}{3}$ Theft $\frac{5e^{-5x/12}}{12}$

Determine the probability that the maximum of these losses exceeds 3.

I read the solution and they have this step: $P[max\lbrace{S,F,T\rbrace} \leq 3] = P[(S\leq 3) \cap (F\leq 3) \cap (T\leq 3)]$.

So I understand that they've reworded it with the complementary probability. I just wanted to confirm that we can only make this statement of equality since we are using the complement. Else, we would need to find all the intervals where the maximum is greater than or equal to 3 and sum those.

Thanks in advance!

Answer 1

Here is a start: Give reasons for each equal sign and compute the answer.

Let $W = \max(S,F,T).$ Then $$P(W \le w) = P(S \le w,\, F \le w,\, T\le w)\\ =P(S \le w)P(F \le w)P(T \le w) \\ = (1-e^{-w})(1 - e^{-(2/3)w})(1 - e^{-(5/12)w}).$$

Note: Based on the following simulation in R, you will find $P(W \le 5) \approx 0.84.$

m = 10^6;  s = rexp(m);  f = rexp(m, 2/3);  t = rexp(m, 5/12)
w = pmax(s,f,t);  mean(w)
## 3.148749
mean(w <= 5)
## 0.838767    # aprx P(W < 5)
pexp(5,1)*pexp(5,2/3)*pexp(5,5/12)
## 0.8385649   # exact P(W < 5)

I realize this is not exactly the numerical question asked, but it is similar. I didn't want to do your assignment, but did want to show you how to check your answer when you get it.