A binomial model needs more than a repeated event. The trials must be a fixed number, each with two outcomes, the same probability every time, and independent of each other.
This lesson is part of choosing the right probability model. If the conditions themselves are new to you, read recognising binomial conditions first.
How can the same draw give two different answers?
Here is an original example. A box holds 12 sweets, of which 5 are red. Three sweets are taken one at a time, and X is the number of red sweets.
Case A: with replacement. Each sweet is put back before the next draw. The probability of red is 5/12 every time, and the draws are independent, so X ~ B(3, 5/12).
P(X = 3) = (5/12)³ = 125/1728 ≈ 0.0723
Case B: without replacement. The sweets are kept out. After one red sweet is taken, only 4 red remain among 11.
P(X = 3) = (5/12) × (4/11) × (3/10) = 60/1320 = 1/22 ≈ 0.0455
The answers differ because the probability of red falls with each red sweet removed. Case B fails the constant-probability condition, so the binomial formula would overstate the answer.
What are the ways a repeated event can fail?
Checking the conditions in order catches the failures quickly.
| Condition | Question to ask | Example that fails |
|---|---|---|
| Fixed number of trials | Is n stated before we start? | Roll a die until the first six appears |
| Two outcomes | Is each trial a success or a failure? | Record the score on each roll (six outcomes) |
| Constant probability | Is p the same every time? | Sweets drawn without replacement |
| Independence | Does one result affect the next? | Two shots by a player who gains confidence after a hit |
The first row is worth a second look. “Roll until the first six” repeats an event, but n is not fixed, so the number of rolls is not binomial.
The mistake that costs marks
The common slip is to see “three sweets” and “5 red out of 12” and write X ~ B(3, 5/12) without reading how the sweets were taken. The working looks neat, so nothing seems wrong.
| Step | Wrong | Right |
|---|---|---|
| Read the sampling | (skipped) | “one at a time, not replaced” |
| Test constant p | Assumed | 5/12, then 4/11, then 3/10: not constant |
| Model | B(3, 5/12) | Multiply conditional probabilities |
| P(X = 3) | 0.0723 | 0.0455 |
The fix is to underline the phrase that tells you how the trials happen, such as “replaced”, “at random from a large batch” or “independently”. That phrase decides whether the model fits.
Check yourself
A box holds 10 pens, of which 4 are blue. Two pens are taken one after the other and not replaced. Let X be the number of blue pens. Explain why X is not binomial and find P(X = 2).
Answer
The pens are not replaced, so the probability of blue changes from 4/10 to 3/9 on the second draw. The draws are not independent, so X is not binomial.
P(X = 2) = (4/10) × (3/9) = 12/90 = 2/15 ≈ 0.133.
For comparison, a wrong binomial calculation would give (4/10)² = 0.16.
What to study next
Once the model is right, a common shortcut is worth knowing. Continue with using the complement to simplify an at-least probability, then try the practice set for this cluster.
If you want a teacher to give you questions that hide the failed condition, see online one-to-one Additional Mathematics tuition.