Probability of At Least One Head in 3 Coin Tosses

easyCBSE-10CBSE-12CBSE 2024 Board Exam3 min read

Question

Three coins are tossed simultaneously. Find the probability of getting:

  1. At least one head
  2. Exactly two heads
  3. All tails

Solution — Step by Step

Step 1: Write out the sample space.

When three coins are tossed, each coin can show H or T. Total outcomes = 2 × 2 × 2 = 8.

The complete sample space: S =

All 8 outcomes are equally likely (assuming fair coins).

Part 1: P(at least one head)

Step 2: Use the complement method.

"At least one head" means 1, 2, or 3 heads. That's a lot of cases to add. The complement — "no heads at all" — is just one case.

P(at least one head) = 1 − P(no heads)

P(no heads) = P(all tails) = P(TTT) = 1/8

P(at least one head) = 1 − 1/8 = 7/8

Part 2: P(exactly two heads)

Step 3: Count outcomes with exactly 2 heads.

From the sample space: HHT, HTH, THH — that's 3 outcomes.

P(exactly two heads) = 3/8

We can also think of this as: choose which 2 of the 3 coins show heads = ³C₂ = 3 ways.

Part 3: P(all tails)

Step 4: Only one outcome has all tails: TTT.

P(all tails) = 1/8

Answers

P(at least one head) = 7/8

P(exactly two heads) = 3/8

P(all tails) = 1/8

Verification: Let's check the probabilities of all cases sum to 1.

P(0 heads) + P(1 head) + P(2 heads) + P(3 heads) = 1/8 + 3/8 + 3/8 + 1/8 = 8/8 = 1 ✓

Why This Works

The complement approach for "at least one" problems is a powerful shortcut. Instead of adding P(1 head) + P(2 heads) + P(3 heads), we use the simpler:

1 − P(0 heads) = 1 − 1/8 = 7/8

The logic: either you get at least one head OR you get no heads at all. These two events are complementary — they cover all possibilities and can't happen simultaneously.

Why the Direct Method Also Works

You could also count directly. From the sample space, outcomes with at least one head: HHH, HHT, HTH, THH, HTT, THT, TTH = 7 outcomes.

P(at least one head) = 7/8. Same answer.

For small sample spaces like this (8 outcomes), direct counting is equally fast. For larger problems (like 10 coin tosses), the complement method is much more efficient.

💡 Expert Tip

The phrase "at least one" is your signal to use the complement. "At most one" is also often faster with the complement (1 minus "two or more" cases). And "none at all" is just one calculation — P(all fail). Recognise these language patterns and save yourself from long enumeration.

Common Mistake

⚠️ Common Mistake

Mistake: Listing only 6 outcomes instead of 8 for three coins.

Some students think: since each coin can be H or T, three coins give 3 × 2 = 6 outcomes. That's wrong. The correct calculation is 2³ = 8 because each outcome is an ordered sequence of three H/T results, and each position independently has 2 choices. Underestimating the sample space inflates all your probabilities. Always compute total outcomes as 2ⁿ for n fair coins.

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