Something either happens or it does not.
P(A) + P(not A) = 1
rearranged
P(A) = 1 − P(not A)
in words
whatever is left over
Total probability is always 1. So if the opposite event is easier to count, count that and subtract.
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Complementary Events
“At least one” is the signal to work out the opposite instead.
The problem
A fair coin is tossed 4 times. What is the probability of getting at least one head?
"At least one" is the opposite of "none at all". P(no heads) = (½)⁴ = 116, so P(at least one) = 1 − 116 = 1516. D is the complement itself — you still have to subtract it from 1.
The opposite of “at least one” is “none”.
Not “exactly one”, and not “all”. None.
| The event | Its opposite |
|---|---|
| at least one head | no heads at all |
| at least two sixes | zero or one six |
| more than 3 | 3 or fewer |
| at most 2 | 3 or more |
Only the first line collapses to a single easy case — which is why “at least one” is the phrase worth spotting.
Work out the easy side.
P(tail on one toss) = ½
a fair coin
P(no heads) = ½ · ½ · ½ · ½
four tails in a row
= 116
the only losing case
P(at least one head) = 1 − 116
the complement
= 1516
about 94%
When the chances differ, multiply each miss on its own. Three shooters hit with probabilities 12, 14 and 15. All three miss with 12 × 34 × 45 = 310, so at least one hits with 1 − 310 = 710. “At least one misses” works the same way: 1 − P(all hit).
What the direct route would have cost.
the long way
exactly 1 head: 4 ways
exactly 2: 6 ways
exactly 3: 4 ways
exactly 4: 1 way
the short way
1 − (one case)
one multiplication
4 + 6 + 4 + 1 = 15 out of 16 — the same answer, four times the work and four times the chance of a slip.
With ten tosses the direct route has ten cases to add. The complement is still one line: 1 − (½)¹⁰ = 10231024.
The same move on a die.
A die is rolled twice. P(at least one six)?
P(not a six) = 56
five other faces
P(no sixes) = (56)² = 2536
both rolls miss
P(at least one) = 1 − 2536
complement
= 1136
about 31%
Not 16 + 16
Adding gives 1236, which double-counts the roll where both come up six. The complement never has that problem.
Without replacement, the complement still works.
P(no red) = 68 × 57 = 3056 = 1528
blue, then blue from the 7 left
P(at least one red) = 1 − 1528 = 1328
Or with combinations: 1 − 6C2 ÷ 8C2 = 1 − 1528 — the same.
The opposite of “at least 2”
It is “0 or 1”, not “none”. For 3 coins, P(at least 2 heads) = 1 − 1 + 38 = 12.
How many tries until it is likely?
P(at least one 6) = 1 − (56)n
the complement of no 6
need (56)n < 12
n = 3: 125216 ≈ 0.58
still above a half — not enough
n = 4: 6251296 ≈ 0.48
below a half
least n = 4
Adding 16 each time
16 + 16 + 16 = 12 suggests 3 rolls are enough, but it double-counts rolls with two 6s. Only the complement gives the true chance.
On the exam
Use the complement whenever you see “at least one”, “at least one misses” or “not all the same”. The none-case is usually one product: the misses multiplied, or the all-from-the-other-group count. Ten-second check: your answer and the none-case must add up to exactly 1.
Check: list the sixteen outcomes.
total outcomes
2⁴ = 16
every head-tail sequence
with no head
TTTT — just one
116 ✓
the rest
15 of 16
1516 ✓
sensible?
94% is high
four tries at a 50% event ✓
Now you try
Practice problem
The problem
A die is rolled 3 times. Find P(at least one 4).
Going direct means counting one, two and three fours separately. Instead: P(no 4) = (56)³ = 125216, so P(at least one) = 1 − 125216 = 91216. A is the "none" case left unsubtracted.