Probability
Conditional Probability and Independence
Grade 12
Question:
<p>Let \(E_C\) denote the complement of an event E. Let \(E_1, E_2\) and \(E_3\) be any pairwise independent events with \(P(E_1) > 0\) and \(P(E_1 \cap E_2 \cap E_3) = 0\). Then \(P(E_2^C \cap E_3^C | E_1)\) is equal to</p>
<p>(a) \(P(E_2^C) + P(E_3)\)</p>
<p>(b) \(P(E_3^C) - P(E_2^C)\)</p>
<p>(c) \(P(E_3) - P(E_2^C)\)</p>
<p>(d) \(P(E_3) - P(E_2)\)</p>
Step-by-Step Solution
Key Concept: Apply De Morgan's law to convert complement intersection to complement union, then use conditional probability and pairwise independence.
<p><strong>Solution:</strong> Using De Morgan's law and conditional probability:
$$P(E_2^C \cap E_3^C | E_1) = 1 - P((E_2 \cup E_3) | E_1)$$
$$= 1 - \frac{P((E_2 \cup E_3) \cap E_1)}{P(E_1)}$$
$$= 1 - \frac{P(E_2 \cap E_1) + P(E_3 \cap E_1) - P(E_1 \cap E_2 \cap E_3)}{P(E_1)}$$
Using pairwise independence and the fact that $P(E_1 \cap E_2 \cap E_3) = 0$:
$$= 1 - \frac{P(E_2)P(E_1) + P(E_3)P(E_1) - 0}{P(E_1)}$$
$$= 1 - P(E_2) - P(E_3)$$
$$= P(E_3) - P(E_2)$$
∴ Answer is (d).</p>
Correct Answer: d