8 coins are tossed. What is the probability that one and only one turns up…

Concept: For n independent fair-coin tosses, the sample space has 2n equally likely outcomes, and the number of outcomes with exactly k heads equals the binomial coefficient C(n, k). So P(exactly k heads) = C(n, k) / 2n.

1. Here n = 8 tosses, so the total number of equally likely outcomes is 28 = 256.

2. We need exactly one head (k = 1). The single head can fall on any of the 8 tosses, so the favourable outcomes are C(8, 1) = 8.

3. Probability = 8 / 256 = 1/32.

Cross-check: the probability of any one specific arrangement (say, head on the first coin and tails on the rest) is (1/2)8 = 1/256. Since the single head can occupy any of the 8 positions, multiply by 8: 8 x 1/256 = 1/32, which matches the result above.