Q. 2.11 - A First Course in Probability

Question

If $P\left(E\right) = .9$and$P\left(F\right) = .8$, show that$P\left(EF\right) \ge .7$. In general, prove Bonferroni’s inequality, namely$P\left(EF\right) \ge P\left(E\right) + P\left(F\right) - 1$.

Step-by-Step Solution

Answer

Therefore,

Transform$P(E\cup F)\le 1$.

Use the proven inequality to get$P\left(EF\right)\ge 0.7$.

Step 1 Given Information.

Given $P\left(E\right) = .9$and$P\left(F\right) = .8$.

Step 2 Explanation.

For any events $E, F$

$P\left(E\right)+P\left(F\right)-1\le P\left(EF\right)$

This is because:

$$\begin{gathered} P(E\cup F)\le 1 \\ P(E\cup F)=P\left(E\right)+P\left(F\right)-P(E F) \\ P\left(E\right)+P\left(F\right)-P(E F) \le 1 \\ P\left(E\right)+P\left(F\right)-1\le P\left(EF\right) \end{gathered}$$

So if $P\left(E\right)=0.9$and$P\left(F\right)=0.8$.

$P\left(EF\right)\ge P\left(E\right)+P\left(F\right)-1=0.9+0.8-1=0.7$