Suppose an unbiased coin is tossed 6 times. Each coin toss is independent. Let E1 be the event that

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Question

Suppose an unbiased coin is tossed 6 times. Each coin toss is independent. Let E1 be the event that among the second, fourth, and sixth coin tosses, there are at least two heads. Let E2 be the event that among the first, second, third, and fifth coin tosses, there are equal number of heads and tails.The conditional probability P(E1 | E2) is equal to ____________. (rounded off to one decimal place)

Accepted Answer

Answer: 0.5Concept -- Conditional Probability with Independent Events: E2 involves tosses 1,2,3,5 with exactly 2 heads and 2 tails (equal number). The probability of E2 is C(4,2)*(0.5)^4 = 6/16 = 3/8. Now given E2, tosses 1,2,3,5 are fixed in terms of counts but not exact values? Actually, E2 specifies that among those four tosses, there are exactly 2 heads. However, the specific pattern is not fixed; all 6 equally likely patterns. But we can compute P(E1∩E2) and divide by P(E2). Alternatively, note that tosses 4 and 6 are independent of E2 (since E2 does not involve toss 4 and 6). Toss 2 is involved in both. However, because of symmetry, the conditional probability that among tosses 2,4,6 there are at least 2 heads given that among tosses 1,2,3,5 there are exactly 2 heads might still be 0.5. Detailed calculation: List all outcomes of 6 tosses with equal probability 1/64. Count those satisfying E2: choose which 2 of {1,2,3,5} are heads: C(4,2)=6 ways, each with probability 1/64, so 6/64 = 3/32? Wait, each specific pattern has probability 1/64, so number of outcomes with E2 = 6 * 2^2? Because tosses 4 and 6 are free? Actually E2 only constrains tosses 1,2,3,5. Tosses 4 and 6 can be anything (2^2=4 possibilities). So total outcomes satisfying E2 = 6 * 4 = 24. So P(E2)=24/64=3/8, consistent. Now among these 24 outcomes, count those where among tosses 2,4,6 at least 2 heads. Toss 2 is part of the constrained set. We can sum over the 6 patterns for (1,2,3,5) and for each, count the number of (4,6) such that condition holds. Because of symmetry, the conditional probability comes out to 0.5. Compute: For each fixed pattern of (1,2,3,5), toss 2 is either H or T. The condition on (2,4,6) at least 2 heads. For a given toss 2, the number of (4,6) pairs giving at least 2 heads: if toss2=H, then we need at least 1 head among {4,6} → 3 possibilities (HT,TH,HH). If toss2=T, then we need both 4 and 6 heads → 1 possibility (HH). Now, among the 6 patterns for (1,2,3,5), how many have toss2=H? Since exactly 2 heads among the four positions, the probability that a specific position (toss2) is head given exactly 2 heads in 4 is 2/4=0.5. So out of 6 patterns, 3 have toss2=H and 3 have toss2=T. Then total favorable outcomes = 3*3 + 3*1 = 9+3=12. So P(E1∩E2)=12/64=3/16. Then P(E1|E2)= (3/16) / (3/8) = (3/16)*(8/3)=8/16=0.5.Note: Conditional probability can be computed by reducing the sample space. Here the result is 0.5, which is intuitive because the extra tosses (4,6) are independent and toss2 is symmetric.

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