<p>A pack of playing cards was found to contain only 51 cards. If the first 13 cards, which are examined, are all red, what is the probability that the missing card is black?</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem by conditioning on the observed event (13 red cards drawn). The missing card is black if and only if the 51 remaining cards contain all 26 red cards and 25 black cards.
<p><strong>Step 1: Set up the problem using Bayes' theorem.</strong></p><p>Let B = missing card is black, and E = first 13 cards examined are all red.</p><p>We need: P(B|E) = P(E|B)·P(B) / P(E)</p><p><strong>Step 2: Calculate P(B) and P(E|B).</strong></p><p>P(B) = 26/52 = 1/2 (originally, 26 black cards out of 52)</p><p>If missing card is black: 26 red cards and 25 black cards remain in the pack of 51.</p><p>P(E|B) = C(26,13)·C(25,0) / C(51,13) = C(26,13) / C(51,13)</p><p><strong>Step 3: Calculate P(E|R) where R = missing card is red.</strong></p><p>P(R) = 1/2</p><p>If missing card is red: 25 red cards and 26 black cards remain in the pack of 51.</p><p>P(E|R) = C(25,13)·C(26,0) / C(51,13) = C(25,13) / C(51,13)</p><p><strong>Step 4: Apply Bayes' theorem.</strong></p><p>P(E) = P(E|B)·P(B) + P(E|R)·P(R) = [C(26,13) + C(25,13)] / (2·C(51,13))</p><p>P(B|E) = [C(26,13)/(2·C(51,13))] / {[C(26,13) + C(25,13)] / (2·C(51,13))}</p><p>P(B|E) = C(26,13) / [C(26,13) + C(25,13)]</p><p><strong>Step 5: Simplify using the identity C(26,13) = (26/13)·C(25,13).</strong></p><p>P(B|E) = (26/13)·C(25,13) / [(26/13)·C(25,13) + C(25,13)] = (26/13) / (26/13 + 1) = (26/13) / (39/13) = 26/39 = 2/3</p><p>∴ Answer: B (2/3)</p>
Correct Answer: B