If $P(E) = 0.9$ and $P(F) = 0.9$, show that $P(EF) \geq 0.8$. In general, prove Bonferroni's inequality, namely that $P(EF) \geq P(E) + P(F)- 1$.

### Solution

We know that $P(E \cup F)\leq 1$, because probabilities by definition are less than or equal to 1.

` $$ \begin{aligned} P(E \cup F) & = P(E) + P(F) - P(EF)\leq 1\\ \Rightarrow & 0.9 + 0.9 - P(EF) \leq 1\\ \Rightarrow & 1.8 -P(EF) \leq 1\\ \Rightarrow & P(EF) \geq 1.8-1\\ \Rightarrow & P(EF) \geq 0.8. \end{aligned} $$ `

We know that $P(E \cup F)\leq 1$, because probabilities by definition are less than or equal to 1.

Also we know the result

`$$P(E\cup F) = P(E) + P(F) - P(EF)$$`

Using above condition,

`$$P(E \cup F) = P(E) + P(F) - P(EF)\leq 1.$$`

Hence,

`$$P(E) + P(F) - P(EF)\leq 1$$`

So,

`$$P(EF)\geq P(E) +P(F) -1.$$`

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators