Probability of heads on 3 coin flips exactly twice?

This is a binomial probability problem because there is a fixed number of independent trials, each trial has two outcomes, and the probability of success is constant. Let success be flipping heads. There are 3 coin flips, and the probability of heads on each flip is 1/2. The event “exactly two heads” can occur in three ordered ways: HHT, HTH, and THH. Each sequence has probability (1/2)(1/2)(1/2) = 1/8. Since there are three favorable sequences, the total probability is 3 × 1/8 = 3/8. Equivalently, use the binomial formula: C(3,2)(1/2)^2(1/2)^1 = 3 × 1/8 = 3/8. Option B would be too large, and option D counts only one favorable sequence rather than all three. Study Guide references/topics: binomial probability, combinations, independent trials, coin-flip sample spaces.

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