If two fair coins are flipped and at least one of the outcomes is known to be…

Concept: Conditional probability answers "given some information about an experiment, what's the chance of a specific outcome" by shrinking the full sample space down to only the outcomes consistent with that information, then measuring what fraction of that smaller space satisfies the event you care about — formally, P(event | condition) = P(event AND condition) / P(condition).

Application: For two fair coin flips, list every equally likely outcome, then apply the given condition (at least one flip is a head) to see which outcomes survive.

• (H, H)

• (H, T)

• (T, H)

• (T, T)

1. All four outcomes are equally likely before any information is given, so each has probability 1/4.

2. The condition "at least one flip is a head" rules out (T, T), since it has no heads at all.

3. That leaves three equally likely outcomes still possible: (H, H), (H, T), (T, H) — each now carries probability 1/3 within this reduced space.

4. Only (H, H) has both flips landing heads, so the probability that both are heads, given the condition, is 1/3.

Cross-check: Confirm with the conditional-probability formula directly.

1. P(both heads AND at least one head) = P(both heads) = 1/4, since "both heads" already satisfies "at least one head".

2. P(at least one head) = 1 − P(no heads) = 1 − 1/4 = 3/4.

3. P(both heads | at least one head) = (1/4) / (3/4) = 1/3 — matching the direct count above.

So the probability that both coins are heads is 1/3. This is different from the case where a specific coin is named as heads (e.g., "the first flip is a head") — that extra, more precise information would leave only two equally likely possibilities for the other coin, giving 1/2 instead. Here, however, neither flip is singled out; we're only told that at least one of the two landed heads, so the 1/3 result stands.