Probability
Bayes' Theorem
Grade 12
Question:
<p>Die \(A\) has 4 red and 2 white faces, whereas die \(B\) has 2 red and 4 white faces. A coin is flipped once. If it shows a head, the game continues by throwing die \(A\); if it shows tail, then die \(B\) is to be used. If the probability that die \(A\) is used is 32/33 when it is given that red turns up every time in first \(n\) throws, then find the value of \(n\).</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem to relate the posterior probability of die A being used to the likelihood of observing n reds from each die. The conditional probability P(A|n reds) = P(n reds|A)·P(A) / P(n reds) must equal 32/33.
<p><strong>Step 1:</strong> Set up Bayes' theorem. Given n reds observed, probability die A was used is:</p><p>P(A|n reds) = P(n reds|A)·P(A) / P(n reds)</p><p><strong>Step 2:</strong> Calculate individual probabilities.</p><p>• P(A) = P(B) = 1/2 (coin flip)</p><p>• P(n reds|A) = (4/6)ⁿ = (2/3)ⁿ (die A has 4 red faces out of 6)</p><p>• P(n reds|B) = (2/6)ⁿ = (1/3)ⁿ (die B has 2 red faces out of 6)</p><p><strong>Step 3:</strong> Find total probability of n reds using law of total probability:</p><p>P(n reds) = P(n reds|A)·P(A) + P(n reds|B)·P(B)</p><p>P(n reds) = (2/3)ⁿ · (1/2) + (1/3)ⁿ · (1/2) = (1/2)[(2/3)ⁿ + (1/3)ⁿ]</p><p><strong>Step 4:</strong> Apply Bayes' formula:</p><p>P(A|n reds) = [(2/3)ⁿ · (1/2)] / {(1/2)[(2/3)ⁿ + (1/3)ⁿ]} = (2/3)ⁿ / [(2/3)ⁿ + (1/3)ⁿ]</p><p><strong>Step 5:</strong> Set equal to 32/33 and solve:</p><p>(2/3)ⁿ / [(2/3)ⁿ + (1/3)ⁿ] = 32/33</p><p>33(2/3)ⁿ = 32[(2/3)ⁿ + (1/3)ⁿ]</p><p>33(2/3)ⁿ = 32(2/3)ⁿ + 32(1/3)ⁿ</p><p>(2/3)ⁿ = 32(1/3)ⁿ</p><p>[(2/3)ⁿ] / [(1/3)ⁿ] = 32</p><p>2ⁿ = 32 = 2⁵</p><p><strong>∴ Answer: n = 5</strong></p>
Correct Answer: 5