At least one but no more than two events occurs.

Question

We're given that at least one of the events $A_r , 1 \leq r \le n$, is certain to occur, but certainly no more than two occurs. If $P\left(A_r\right) = p$ and $P\left(A_r \cap A_s\right) = q$ , $r \neq s$, show that $p \geq \frac{1}{n}$ and $q \leq \frac{2}{n}$.

So clearly in the case that $P\left(\cup_{r}^{n}A_r\right) = 1, p = 1/n$. I can't figure out what the case(s) is/are when $ p\neq 1/n$.

One event is certain to occur. So the probability of two events occurring must be one multiplied by something. I can't get any further than this. I also don't see how to set up an inequality for p and q. Any help would be great.

Answer 1

First, it is important to realise that $P\left(A\cup B\right)$ is not, in general, equal to $p(A)+p(B)$.

The general result is $P\left(A\cup B\right)=p(A)+p(B)-P\left(A\cap B\right)$.

That is why $1=P\left(\cup_{1}^{n}A_r\right) \le pn $ and therefore $p \ge 1/n.$

We also have

$$1=P\left(\cup_{1}^{n}A_r\right)=\sum_r p(A_i)- \sum_{r,s} P\left(A_r\cap A_s\right)$$ $$1=np-\frac{n(n-1)}{2}q$$ $$\frac{n(n-1)}{2}q=np-1\le n-1$$ $$q \le \frac {2}{n}$$