Three fair coins are tossed simultaneously. The probability of getting EXACTLY two heads is

1. Sample space: $|S| = 2^3 = 8$ outcomes — HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. 2. Event 'exactly 2 heads': enumerate — HHT, HTH, THH. So $|E| = 3$. 3. $P(\text{exactly 2 H}) = 3/8$. 4. Alternative via binomial probability: $P(\text{exactly}\,k\,\text{heads}) = \binom{n}{k}\,(1/2)^k\,(1/2)^{n-k} = \binom{n}{k}/2^n$. For $n = 3, k = 2$: $\binom{3}{2}/8 = 3/8$ ✓. 5. Option A is the probability of EXACTLY 0 or 3 heads. Option B is wrong count. Option D would require half of all outcomes. _Source: NCERT Class 11 Mathematics, Ch 14, Example 3 (Three-coin events), p. 4–5._