<p>A coin is tossed three times in succession. If \(E\) is the event that there are at least two heads and \(F\) is the event in which first throw is a head, then find \(P(E/F)\).</p>
Step-by-Step Solution
Key Concept: Conditional probability P(E/F) = P(E∩F)/P(F) requires finding outcomes where BOTH at least two heads occur AND first toss is heads, then dividing by all outcomes where first toss is heads.
<p><strong>Step 1: Find sample space.</strong> Three coin tosses give 8 outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT</p><p><strong>Step 2: Identify event F (first toss is head).</strong> F = {HHH, HHT, HTH, HTT}, so |F| = 4 and P(F) = 4/8 = 1/2</p><p><strong>Step 3: Find E∩F (at least two heads AND first toss is head).</strong> From F's outcomes, those with at least 2 heads total: HHH (3 heads ✓), HHT (2 heads ✓), HTH (2 heads ✓), HTT (1 head ✗). So E∩F = {HHH, HHT, HTH}, with |E∩F| = 3</p><p><strong>Step 4: Calculate P(E∩F).</strong> P(E∩F) = 3/8</p><p><strong>Step 5: Apply conditional probability formula.</strong> P(E/F) = P(E∩F)/P(F) = (3/8)/(1/2) = 3/8 × 2/1 = 3/4</p><p>∴ Answer: <strong>3/4</strong></p>
Correct Answer: 3/4