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
