Probability
Independent Events
Grade 12
Question:
<p>Let \(E\) and \(F\) be two independent events. The probability that exactly one of them occurs is \(11/25\) and the probability of none of them occurring is \(2/25\). If \(P(T)\) denotes the probability of occurrence of the event \(T\), then</p>
<p>\(P(E) = \dfrac{4}{5},\ P(F) = \dfrac{3}{5}\)</p>
<p>\(P(E) = \dfrac{1}{5},\ P(F) = \dfrac{2}{5}\)</p>
<p>\(P(E) = \dfrac{2}{5},\ P(F) = \dfrac{1}{5}\)</p>
<p>\(P(E) = \dfrac{3}{5},\ P(F) = \dfrac{4}{5}\)</p>
Step-by-Step Solution
Key Concept: For independent events E and F, use P(exactly one occurs) = P(E)P(F^c) + P(E^c)P(F) and P(none occurs) = P(E^c)P(F^c) to set up a system of equations in P(E) and P(F).
<p><strong>Step 1:</strong> Set P(E) = p and P(F) = q. Since E and F are independent, P(E^c) = 1-p and P(F^c) = 1-q.</p><p><strong>Step 2:</strong> Exactly one occurs means: P(E∩F^c) + P(E^c∩F) = p(1-q) + (1-p)q = 11/25</p><p>Simplifying: p + q - 2pq = 11/25 ... (1)</p><p><strong>Step 3:</strong> None occurs means: P(E^c∩F^c) = (1-p)(1-q) = 2/25</p><p>Expanding: 1 - p - q + pq = 2/25</p><p>Therefore: p + q - pq = 23/25 ... (2)</p><p><strong>Step 4:</strong> From equation (1): p + q - 2pq = 11/25</p><p>From equation (2): p + q - pq = 23/25</p><p>Subtracting (1) from (2): pq = 23/25 - 11/25 = 12/25</p><p><strong>Step 5:</strong> Substituting back in (2): p + q = 23/25 + 12/25 = 35/25 = 7/5</p><p><strong>Step 6:</strong> So p and q are roots of: t² - (7/5)t + 12/25 = 0</p><p>Multiplying by 25: 25t² - 35t + 12 = 0</p><p>Using quadratic formula: t = (35 ± √(1225 - 1200))/50 = (35 ± 5)/50</p><p>Therefore: t = 4/5 or t = 3/5</p><p>∴ Answer: P(E) = 4/5, P(F) = 3/5 (or vice versa) — Both options must be correct, hence AD
Correct Answer: AD