Probability
Bayes' Theorem
Grade 12
Question:
<p>A speaks truth 3 times out of 4 while B, 7 times out of 10. A ball is drawn at random from a bag containing one black ball and five other balls of different colours. Both A and B report that a black ball has been drawn from the bag. Find the probability of their assertion being true?</p>
<p>\(\dfrac{1}{4}\)</p>
<p>\(\dfrac{24}{29}\)</p>
<p>\(\dfrac{1}{29}\)</p>
<p>\(\dfrac{5}{29}\)</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem to find P(black ball | both report black) by considering the probability of their reports being correct versus incorrect. The key is recognizing that if both lie, they could still both report 'black' by coincidence.
<p><strong>Step 1:</strong> Define events. Let B = black ball is drawn, R = both report black ball.</p><p><strong>Step 2:</strong> P(B) = 1/6 (one black ball out of 6), P(B') = 5/6</p><p><strong>Step 3:</strong> Find P(R|B): Both tell truth about black ball<br/>P(R|B) = (3/4) × (7/10) = 21/40</p><p><strong>Step 4:</strong> Find P(R|B'): Both lie and both say 'black' when it's not black<br/>P(R|B') = (1/4) × (3/10) = 3/40<br/>(A lies with probability 1/4, B lies with probability 3/10)</p><p><strong>Step 5:</strong> Apply Bayes' theorem:<br/>P(B|R) = P(R|B) × P(B) / [P(R|B) × P(B) + P(R|B') × P(B')]<br/>P(B|R) = (21/40 × 1/6) / [(21/40 × 1/6) + (3/40 × 5/6)]<br/>P(B|R) = (21/240) / [(21/240) + (15/240)]<br/>P(B|R) = 21/36 = 7/12</p><p><strong>∴ Answer: B (7/12)</strong></p>
Correct Answer: B