Three coins are tossed. The probability of getting at least one head is----?

For n independent tosses of a fair coin, the sample space has 2n equally likely outcomes. The event ‘at least one head’ is the complement of the single event ‘no head at all’, so the fastest route is the complement rule: P(at least one head) = 1 − P(no head).

1. Total outcomes: with 3 coins, n(S) = 23 = 8.

2. The complement event ‘no head at all’ has exactly one outcome, (T, T, T), so P(no head) = 1/8.

3. By the complement rule, P(at least one head) = 1 − P(no head) = 1 − 1/8 = 7/8.

Cross-check by direct enumeration: the 8 equally likely outcomes of tossing three coins are:

• HHH, HHT, HTH, THH, HTT, THT, TTH — each of these seven sequences contains at least one head.

• TTT — the only sequence with zero heads.

This matches the complement calculation: 7 of the 8 outcomes have at least one head, so the required probability is 7/8.