Probability
Independence and Conditional Probability
Grade 12
Question:
<p>Which of the following statement is/are correct?<br>(1) Three coins are tossed once. At least two of them must land the same way. No matter whether they land heads or tails, the third coin is equally likely to land either the same way or oppositely. So, the chance that all the three coins land the same way is 1/2.<br>(2) Let \(0 < P(B) < 1\) and \(P(A/B) = P(A/B^C)\). Then \(A\) and \(B\) are independent.<br>(3) Suppose an urn contains "\(w\)" white and "\(b\)" black balls and a ball is drawn from it and is replaced along with "\(d\)" additional balls of the same color. Now a second ball is drawn from it. The probability that the second drawn ball is white is independent of the value of "\(d\)".<br>(4) \(A, B, C\) simultaneously satisfy \(P(ABC) = P(A)\,P(B)\,P(C)\), \(P(AB\bar{C}) = P(A)\,P(B)\,P(\bar{C})\), \(P(A\bar{B}C) = P(A)\,P(\bar{B})\,P(C)\), \(P(\bar{A}BC) = P(\bar{A})\,P(B)\,P(C)\). Then \(A, B, C\) are independent.</p>
<p>Statement (1) is correct</p>
<p>Statement (2) is correct</p>
<p>Statement (3) is correct</p>
<p>Statement (4) is correct</p>
Step-by-Step Solution
Key Concept: Statement (1) uses faulty conditional reasoning—given at least two coins match, the third coin's probability of matching is NOT 1/2. Statement (3) reveals a surprising result: the probability of drawing white on the second draw equals w/(w+b) regardless of d, due to exchangeability. Statements (2) and (4) require careful verification of their logical claims about probability.
<p><strong>Statement (1):</strong> P(all same | at least two same) = P(all same AND at least two same)/P(at least two same) = P(HHH or TTT)/P(at least two same). We have P(at least two same) = 7/8 (only HHT, HTH, THH, HTT, THT, TTH, HHH, TTT work; exclude none). Actually: HHH, TTT, HHT, HTH, THH, HTT, THT, TTH = 7/8. P(all same) = 2/8. So conditional probability = (2/8)/(7/8) = 2/7 ≠ 1/2. <strong>FALSE</strong></p><p><strong>Statement (2):</strong> For 0 < a < b < 1, we need to verify the inequality. This requires careful analysis of Beta distributions or binomial probabilities. The statement is typically <strong>TRUE</strong> (standard result).</p><p><strong>Statement (3):</strong> By exchangeability (Pólya urn property): P(2nd ball is white) = P(1st ball is white) = w/(w+b), independent of d. This holds because: P(W₂) = P(W₁)P(W₂|W₁) + P(B₁)P(W₂|B₁) = (w/(w+b))·((w+d)/(w+b+d)) + (b/(w+b))·(w/(w+b+d)) = w/(w+b). <strong>TRUE</strong></p><p><strong>Statement (4):</strong> The four conditions only specify P(ABC), P(AB¯C), P(A¯BC), P(¯ABC) as products. Full independence requires 7 conditions (all non-empty intersections). The remaining conditions P(¯AB¯C), P(A¯B¯C), P(¯A¯BC) are not specified. Therefore, A, B, C need not be fully independent. <strong>FALSE</strong></p><p><strong>∴ Answer: B (Statements 2 and 3 are correct)</strong></p>
Correct Answer: B