Question

Let E, F and G be events such that $P(E|G) = 0.05$ and $P(F|G) = 0.05$ which of the following statement must be true?

• A. $P(E \cap F|G) = (0.05)^2$
• B. $P(\overline{E} \cap \overline{F}|G) \ge 0.90$
• C. $P(E \cup F|G) \le 0.05$
• D. $P(E \cup F|\overline{G}) \ge 1 - (0.05)^2$

Hint:

This is an application of Boole's Inequality. For any events, $P(\cup A_i) \le \sum P(A_i)$. By finding the upper bound of the "bad" events (E or F), you automatically find the lower bound of the "good" event (neither E nor F).

Answer 1

The Correct Option is B

We evaluate the boundaries of probability for combinations of events E and F, given a specific condition G.

Step 1: \color{redAnalyze Option (1)
For $P(E \cap F|G) = P(E|G)P(F|G)$ to be true, events E and F must be conditionally independent given G.
The problem does not state they are independent, so this statement is not necessarily true.

Step 2: \color{redEvaluate the Union Boundary
By the addition theorem: $P(E \cup F|G) = P(E|G) + P(F|G) - P(E \cap F|G)$.
Since the intersection probability is at minimum 0:
$P(E \cup F|G) \le P(E|G) + P(F|G)$.
$P(E \cup F|G) \le 0.05 + 0.05 = 0.10$.
Option (3) claims the union must be $\le 0.05$, which is not guaranteed.

Step 3: \color{redExamine the Complementary intersection (Option 2)
Using De Morgan's Law: $P(\overline{E} \cap \overline{F}|G) = P(\overline{E \cup F}|G) = 1 - P(E \cup F|G)$.
From Step 2, we know the maximum value of $P(E \cup F|G)$ is $0.10$.
If the union is at most $0.10$, then its complement must be at least $1 - 0.10$:
$P(\overline{E} \cap \overline{F}|G) \ge 1 - 0.10 = 0.90$.
This statement must be true.

Step 4: \color{redAnalyze Option (4)
This option involves $P(E \cup F|\overline{G})$. We have no information about the relationship of E and F with the complement of G, so this cannot be proven true.
The only statement that holds under all conditions is Option (2).