For an “at least” probability, subtract the short complement from 1 instead of adding a long list. For P(X ≥ 1), the complement is just P(X = 0).
This lesson is part of choosing the right probability model. It assumes you can already write a binomial term, as in calculating binomial probabilities.
Which outcomes does the complement remove?
All probabilities add to 1, so the outcomes you want and the outcomes you do not want share that total. When the unwanted group is small, count it and subtract.
For X ~ B(n, p), P(X ≥ 1) = 1 − P(X = 0) = 1 − (1 − p)ⁿ. There is only one “not wanted” outcome, which is no successes.
Worked example: long route against short route
Here is an original question. A seed packet has 6 seeds, each germinating with probability 0.2 independently. Find the probability that at least one seed germinates.
Let X ~ B(6, 0.2).
Long route. P(X ≥ 1) = P(1) + P(2) + P(3) + P(4) + P(5) + P(6). That is six terms, each needing its own ⁶Cᵣ and two powers.
Short route. P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.8⁶.
0.8⁶ = 0.262 144, so P(X ≥ 1) = 1 − 0.262 144 = 0.7379 (to 4 decimal places).
The short route has one power and one subtraction. It is also easier to check, since a probability of 0.74 for “at least one of six” feels reasonable.
A second type: at least two
The complement still works when the boundary moves, but now two terms are removed. Let X ~ B(5, 0.3) and find P(X ≥ 2).
P(X ≥ 2) = 1 − P(X = 0) − P(X = 1)
P(X = 0) = 0.7⁵ = 0.168 07
P(X = 1) = 5 × 0.3 × 0.7⁴ = 5 × 0.3 × 0.2401 = 0.360 15
P(X ≥ 2) = 1 − 0.168 07 − 0.360 15 = 1 − 0.528 22 = 0.4718
The mistake that costs marks
The common slip is to write 1 − P(X = 1) for “at least one”. It looks like a complement, so it passes a quick read.
| Step | Wrong | Right |
|---|---|---|
| Unwanted outcome for X ≥ 1 | X = 1 | X = 0 |
| Working for B(6, 0.2) | 1 − 6(0.2)(0.8⁵) | 1 − 0.8⁶ |
| Result | 0.6068 | 0.7379 |
The wrong version removes only the case of exactly one success, so it keeps “no success” inside the answer. Ask which outcomes you do not want, then subtract those from 1.
Finding the smallest n
Some questions ask how many trials are needed. Each attempt at a target succeeds with probability 0.2 independently. How many attempts n are needed so that P(at least one success) > 0.9?
- Write 1 − 0.8ⁿ > 0.9.
- Rearrange: 0.8ⁿ < 0.1.
- Take logarithms: n log 0.8 < log 0.1. Since log 0.8 is negative, the inequality reverses: n > log 0.1 ÷ log 0.8 = 10.32.
- So n = 11. Check: 0.8¹⁰ = 0.1074, which is not below 0.1, and 0.8¹¹ = 0.0859, which is.
The check in step 4 matters, because rounding 10.32 down would give the wrong answer. If logarithms are still slow, the algebra step repair trainer can rebuild that step.
Check yourself
X ~ B(8, 0.1). Find P(X ≥ 1) to 4 decimal places.
Answer
P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.9⁸.
0.9⁸ = 0.430 467 21, so P(X ≥ 1) = 1 − 0.430 467 = 0.5695.
Sense check: a 10% chance on each of 8 tries should give a little more than one chance in two of at least one success, and 0.57 does.
What to study next
The next lesson moves from counting to measuring. Continue with recovering a raw score from a stated normal percentile, or test the set with the practice questions.
For a teacher to check each line of your at-least working, see online one-to-one Additional Mathematics tuition.