Probability
Bayes' Theorem
Grade 12
Question:
<p>A box contains <em>N</em> coins, <em>m</em> of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is 1/2, while it is 2/3 when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. What is the probability that the coin drawn is fair?</p>
<p>\(\dfrac{9m}{m + 8N}\)</p>
<p>\(\dfrac{9m}{m + 4N}\)</p>
<p>\(\dfrac{4m}{m + 8N}\)</p>
<p>\(\dfrac{9m}{2m + 8N}\)</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem by finding P(Fair|HT) = P(HT|Fair)×P(Fair) / P(HT). Calculate P(HT) using the law of total probability considering both fair and biased coins.
<p><strong>Step 1:</strong> Define prior probabilities.</p><p>P(Fair) = m/N, P(Biased) = (N-m)/N</p><p><strong>Step 2:</strong> Calculate likelihood of getting HT (Head then Tail).</p><p>P(HT|Fair) = (1/2) × (1/2) = 1/4</p><p>P(HT|Biased) = (2/3) × (1/3) = 2/9</p><p><strong>Step 3:</strong> Apply law of total probability.</p><p>P(HT) = P(HT|Fair)×P(Fair) + P(HT|Biased)×P(Biased)</p><p>P(HT) = (1/4)×(m/N) + (2/9)×((N-m)/N)</p><p>P(HT) = [1/(4N)]×m + [2/(9N)]×(N-m)</p><p>P(HT) = (1/N)[9m + 8(N-m)]/36 = (1/N)(m + 8N)/36</p><p><strong>Step 4:</strong> Apply Bayes' theorem.</p><p>P(Fair|HT) = [P(HT|Fair)×P(Fair)] / P(HT)</p><p>P(Fair|HT) = [(1/4)×(m/N)] / [(m + 8N)/(36N)]</p><p>P(Fair|HT) = [(m/4N) × 36N] / (m + 8N)</p><p>P(Fair|HT) = 9m / (m + 8N)</p><p>∴ Answer: A</p>
Correct Answer: A