Probability
Exactly 1 Heads in 3 Coin Tosses: Count the Sequences
For 3 independent fair coin tosses, exactly 1 heads has probability 3/8 = 3/8. Compare one sequence with all possible arrangements.
Number JD Editorial Team · Published 2026-10-13
State the event precisely
“Exactly 1 heads in 3 tosses” means 1 heads and 2 tails, in any order. For independent fair tosses, the probability is 3/8 = 3/8, or 37.5%. There are 8 equally likely full sequences.
Exactly is different from at least. A sequence with more than the specified number of heads does not count here. Write the event in words before using a formula.
Count arrangements, not just one order
Choose which 1 of the 3 positions contain heads. The combination count C(3, 1) is 3. Every selected set of positions gives one complete sequence because the remaining positions must be tails.
Each particular sequence has probability (1/2)^3 = 1/8. Multiply by the 3 successful sequences to obtain 3/8. The choice of positions is the reason the answer may be larger than the probability of one named sequence.
Compare with all possible head counts
For 3 tosses, the counts of successful sequences for 0 through 3 heads are 1, 3, 3, 1. They add to 8. These cases cannot overlap, and together they include every complete sequence.
Exactly 2 heads has the same probability as exactly 1 heads for a fair coin. Swapping every H with T pairs the two groups. This symmetry does not hold in the same form if heads and tails have different probabilities.
Interpret repeated experiments
If you repeat the entire 3-toss experiment 160 times, the expected number of experiments with exactly 1 heads is 60. This is an average, not a guaranteed result. Keep the experiment count separate from the individual toss count.
A streak in earlier experiments does not change the next fair independent toss. If physical conditions make independence doubtful, the simple model should be reconsidered rather than patched with a “due” outcome.
How the formula changes for an unfair coin
If the chance of heads is p rather than 1/2, the expression is C(3, 1) × p^1 × (1 − p)^2. The arrangement count stays 3; only the probability of each arrangement changes.
The model needs a stated value of p. Do not substitute an observed fraction from a tiny sample as if it were an exact physical law. Here the answer follows specifically from the fair-coin assumption.
Explore the examples
Check the mathematical definitions
For fractions, percentages and worked arithmetic, see OpenStax Prealgebra. For number definitions, see OpenStax on prime and composite numbers and Wolfram MathWorld on digital roots. Worked examples on this page are calculated for this guide.
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Exactly 2 Heads in 3 Coin Tosses: Count the Sequences
For 3 independent fair coin tosses, exactly 2 heads has probability 3/8 = 3/8. Compare one sequence with all possible arrangements.
Exactly 3 Heads in 3 Coin Tosses: Count the Sequences
For 3 independent fair coin tosses, exactly 3 heads has probability 1/8 = 1/8. Compare one sequence with all possible arrangements.
Exactly 0 Heads in 4 Coin Tosses: Count the Sequences
For 4 independent fair coin tosses, exactly 0 heads has probability 1/16 = 1/16. Compare one sequence with all possible arrangements.
Exactly 1 Heads in 4 Coin Tosses: Count the Sequences
For 4 independent fair coin tosses, exactly 1 heads has probability 4/16 = 1/4. Compare one sequence with all possible arrangements.