Probability
Conditional Probability / Bayes Theorem
Grade 12
Question:
<p>If 3 identical cards are coloured both the sides such that the first card are colored red, both sides of the second card are colored black, and one side of the third card is colored red and the other side black. The 3 cards are mixed up, and 1 card is randomly selected and put down on the ground. If the upper side of the chosen card is colored black, what is the probability that the other side is colored red?</p>
<p>(a) \(1/2\)</p>
<p>(b) \(2/3\)</p>
<p>(c) \(1/3\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem by finding P(Red-Black card | Black face visible). The observed black face could come from either the Black-Black card or the Red-Black card, and we must weight their probabilities accordingly.
<p><strong>Step 1: Identify the three cards</strong></p><p>Card 1 (RR): Red-Red</p><p>Card 2 (BB): Black-Black</p><p>Card 3 (RB): Red-Black</p><p><strong>Step 2: Find P(Black face visible)</strong></p><p>Each card is equally likely to be selected: P(each card) = 1/3</p><p>P(Black visible) = P(Black | RR)·P(RR) + P(Black | BB)·P(BB) + P(Black | RB)·P(RB)</p><p>P(Black visible) = 0·(1/3) + 1·(1/3) + (1/2)·(1/3) = 1/3 + 1/6 = 1/2</p><p><strong>Step 3: Find P(Black visible AND other side is Red)</strong></p><p>This occurs only when Card 3 (RB) is selected and the black side is facing up:</p><p>P(Black visible AND Red on other side) = P(RB)·P(Black | RB) = (1/3)·(1/2) = 1/6</p><p><strong>Step 4: Apply conditional probability</strong></p><p>P(Other side is Red | Black visible) = P(Black visible AND Red on back) / P(Black visible)</p><p>= (1/6) / (1/2) = 1/3</p><p>∴ Answer: C (or 1/3)</p>
Correct Answer: C