# SOA practice problem 302: CDF problem

An insurance company sells automobile liabiliy and collision insurance. Let $X$ denote the percentage of liability policies that will be renewed at the end of their terms and $Y$ the percentage of collision policies that will be renewed at the end of their terms. $X$ and $Y$ have a joint cumulative distribution $$F(x,y) = \frac{xy(x+y)}{2000000},\quad 0\leq x\leq 100,\quad 0\leq y\leq 100.$$

Calculate the probability that at least $80\%$ of liability policies and $80\%$ of collision policies will be renewed at the end of their terms.

$(A) \quad 0.072$

$(B) \quad 0.200$

$(C) \quad 0.280$

$(D) \quad 0.360$

$(E) \quad 0.488$

So I am having some trouble with a few things, I thought I had this solved but turns out I wasn't correct.

I thought I would do: $$P(X\geq80 \,\cap Y\geq80) = 1-P(X<80)-P(Y<80)+P(X<80\,\cup Y<80)$$ which wasn't pretty at all, so I knew something was wrong and I looked at the solution. I don't understand how $$P(X\geq80 \,\cap Y\geq80) = 1-P(X<80\,\cup\,Y<80)$$ This doesn't make sense to me, can anyone explain why this is the case? Also further on it says the following: $$1-P(X<80)-P(Y<80)+P(X<80\,\cap Y<80)$$ This I understand how they get this from the previous equation (even if I don't understand how they get to that equation) what my issue is, is they say that this equals

$$1-[F(80,100)+F(100,80)-F(80,80)]$$ How is this the case? Why is $P(X<80) = F(80,100)$ I understand why we plug in $80$ for $x$ since it's a CDF, but why do we do $100$ for $y$?

I get this is a multifaceted question, so I appreciate any help with this. I just don't understand how we can say that $P(X\geq80\cap\,Y\geq80) = 1-$ it's union with them being less than.

If $A$ and $B$ are two events, then $P(A\cap B) = 1 - P((A\cap B)^c) = 1 - P(A^c \cup B^c)$ because the complementary event of $A\cap B$ is $A^c\cup B^c$. Think of it with "and"s and "or"s. Not having $A$ and $B$ means not having $A$ or not having $B$. This explains why $P(X\geq 80\cap Y\geq 80)=1-P(X<80 \cup Y<80)$.
Now $P(X<80)=P(X<80,Y<100)$ because by definition $Y$ is always less than $100$. Thus intersecting with the event $Y<100$ is just doing nothing. Then $P(X<80,Y<100) = P(X\leq 80,Y\leq 100)$ because we have a continuous joint distribution, which implies that you can use either $<$ or $\leq$. Finally, $F(x,y)$ is $P(X\leq x,Y\leq y)$ by definition. Hence $P(X<80)=F(80,100)$.
• You mean $(A\cap B)^c=A^c\cup B^c$? This is de Morgan's law and of course you can use it. Commented Apr 12, 2017 at 21:25