Question

If \(P(E)=0.9\) and \(P(F)=0.8\), show that \(P(E F) \geqslant 0.7\). In general, show that $$ P(E F) \geqslant P(E)+P(F)-1 $$ This is known as Bonferroni's inequality.

Short answer

In general, Bonferroni's inequality states that \( P(E ∩ F) \geq P(E) + P(F) - 1 \). Given the probabilities \( P(E) = 0.9 \) and \( P(F) = 0.8 \), by applying this inequality, we find that \( P(E ∩ F) \geq 0.9 + 0.8 - 1 = 0.7 \). Hence, we have shown that \( P(E ∩ F) \geq 0.7 \), verifying the truth of Bonferroni's inequality in this case.

Step-by-step solution

Understand the problem

We need to show that the probability of the intersection of two events is greater than or equal to the sum of their probabilities minus 1. This is known as Bonferroni's inequality.

Apply the formula for the probability of the union of two events

The probability of the union of two events E and F is given by: \( P(E ∪ F) = P(E) + P(F) - P(E ∩ F) \)

Rearrange the formula to solve for the intersection probability

We can rearrange the formula from Step 2 to solve for the probability of the intersection of events E and F: \( P(E ∩ F) = P(E) + P(F) - P(E ∪ F) \)

Apply the boundaries of the probabilities to conclude the inequality

Given that the probability of any event is between 0 and 1, we can conclude that the minimum value for P(E ∪ F) equals 0, and the maximum value equals 1. Thus with the minimum value of 0, \( P(E ∩ F) = P(E) + P(F) - 0 \) And with the maximum value of 1, \( P(E ∩ F) \geq P(E) + P(F) - 1 \)

Verify with the given probabilities P(E) and P(F)

Now we can check the given probabilities to see if the inequality holds: \( P(E) = 0.9, P(F) = 0.8 \) According to the inequality from step 4: \( P(E ∩ F) \geq 0.9 + 0.8 - 1 = 0.7 \) Thus, we have shown that the option \( P(E∩F) \geq 0.7 \) holds, and in general Bonferroni's inequality is true.