A situation is binomial when four conditions hold together: a fixed number of trials, two outcomes, a constant probability of success, and independent trials. Then you write X ~ B(n, p), where X counts the successes.
This lesson is part of probability distributions. Once the model is named, calculating binomial probabilities shows how to find the values.
What are the four conditions?
Read the question and test each condition in turn.
- Fixed n. The number of trials is decided in advance.
- Two outcomes. Each trial is a success or a failure, nothing else.
- Constant p. The probability of success is the same on every trial.
- Independence. One result does not change the next.
Only after all four are confirmed should you write X ~ B(n, p). If one fails, the binomial model is the wrong tool.
Worked example: five short scenarios
Here are original scenarios. For each, decide whether X is binomial.
| Scenario | Fixed n | Two outcomes | Constant p | Independent | Binomial? |
|---|---|---|---|---|---|
| A student guesses 12 multiple-choice answers, each with 4 options. X = number correct | 12 | Correct or wrong | 0.25 | Yes | Yes, B(12, 0.25) |
| A seed tray has 20 seeds, each germinating with probability 0.9. X = number germinating | 20 | Yes or no | 0.9 | Yes | Yes, B(20, 0.9) |
| A die is rolled until a six appears. X = number of rolls | Not fixed | Six or not | 1/6 | Yes | No |
| A player takes 15 shots, and her chance rises after each hit. X = hits | 15 | Hit or miss | Changes | No | No |
| A die is rolled 10 times. X = the total of the numbers | 10 | Six outcomes | Not applicable | Yes | No |
Three rows fail. The third has no fixed n, the fourth has a changing p, and the last has more than two outcomes for each trial.
What counts as success?
Success is the event that X counts, not the outcome you would prefer. This decides the value of p.
Consider an original case. A pack holds 25 bulbs, and each bulb is faulty with probability 0.04, independently. Let X be the number of working bulbs.
Each bulb works with probability 1 − 0.04 = 0.96. So X ~ B(25, 0.96), not B(25, 0.04).
If instead Y is the number of faulty bulbs, then Y ~ B(25, 0.04). The pack and the bulbs are the same. Only the thing being counted has changed, and p changes with it.
The mistake that costs marks
The common slip is to copy the probability given in the question into B(n, p) without checking what X counts.
| Step | Wrong | Right |
|---|---|---|
| What does X count? | (skipped) | Working bulbs |
| p for one trial | 0.04 (the faulty probability) | 0.96 (probability of working) |
| Distribution | X ~ B(25, 0.04) | X ~ B(25, 0.96) |
The fix is to write one sentence before the notation: “Let X be the number of …, so success means …”. The sentence forces p to match the count.
Check yourself
A coin is biased so that the probability of a head is 0.6. It is tossed 8 times, and the tosses are independent. Let X be the number of tails. Write the distribution of X and give P(success).
Answer
X counts tails, so success means a tail. P(tail) = 1 − 0.6 = 0.4.
The tosses are fixed at 8, each has two outcomes, p is constant and the tosses are independent.
So X ~ B(8, 0.4).
What to study next
When a repeated event does not meet the conditions, a different method is needed. See why repeated events do not always form a binomial model, then practise writing out probabilities in calculating binomial probabilities.
If you want a teacher to give you scenarios and ask you to justify each condition aloud, see online one-to-one Additional Mathematics tuition.