Repeat the same independent trial a fixed number of times and count how often it succeeds: that count follows a binomial distribution. One formula gives every probability, and two more give its mean and variance.
If \(X \sim \text{B}(n, p)\), then \(\text{P}(X = r) = \dbinom{n}{r} p^{r} (1 - p)^{n - r}\) for \(r = 0, 1, \dots, n\), with \(\text{E}(X) = np\) and \(\text{Var}(X) = np(1 - p)\).
Fig. 1: X ~ B(10, 0.3) drawn to scale. The three shaded bars add to P(X ≤ 2) = 0.0282 + 0.1211 + 0.2335 = 0.383
Section 01
When is it binomial?
Check the four conditions every time, and say them in the context of the question when asked to justify the model.
A fixed number of trials, \(n\).
Two outcomes on each trial: success or failure.
The same probability of success, \(p\), on every trial.
The trials are independent of each other.
Then \(X\), the number of successes, satisfies \(X \sim \text{B}(n, p)\). Drawing without replacement breaks 'same \(p\)' and 'independent', so it is not binomial.
Leaving out \(\dbinom{n}{r}\). \(0.3^3 \times 0.7^7\) is the probability of ONE particular order, such as SSSFFFFFFF; there are 120 orders.
Section 02
At most, at least, between.
Cumulative questions mean adding several terms. Translate the words into an inequality on whole numbers first, and use \(1 - \text{P}(\dots)\) when that means fewer terms.
The mean \(np\) and variance \(np(1-p)\) can be used forwards, or backwards to find \(n\) and \(p\). Questions asking for the least \(n\) so that a probability passes a target are solved by trying values of \(n\) (logarithms are not needed).
Worked example Mean and variance
\(X \sim \text{B}(20, 0.15)\). Find \(\text{E}(X)\) and \(\text{Var}(X)\).
\(X \sim \text{B}(n, p)\) has mean 6 and variance 4.2. Find \(n\) and \(p\).
\(np = 6\) and \(np(1-p) = 4.2\).
Divide: \(1 - p = \dfrac{4.2}{6} = 0.7\), so \(p = 0.3\).
\(n = 6 \div 0.3 = 20\).
Answer: \(n = 20\), \(p = 0.3\)
Worked example Least number of trials
Each ticket wins with probability 0.2, independently. Find the least number of tickets needed for the probability of at least one win to exceed 0.99.
\(\text{P}(\text{at least one}) = 1 - 0.8^{n} > 0.99\), so we need \(0.8^{n} < 0.01\).
Try values: \(0.8^{20} = 0.0115\), which is too big.
\(0.8^{21} = 0.009\,22\), which is less than 0.01.
Answer: \(n = 21\)
Examiner trap
Stopping at \(n = 20\) because \(0.0115\) 'is about 0.01'. The inequality must actually hold, so check the value either side.
Section 04
Binomial in context.
Longer questions often use one binomial answer as the \(p\) of a second binomial: first find the probability that one pack, week or group 'succeeds', then count how many packs, weeks or groups succeed.
Worked example Two layers
Each seed germinates with probability 0.85, independently. Seeds are sold in packs of 12. (a) Find the probability that at least 10 seeds in a pack germinate. (b) Find the probability that, out of 5 packs, at least 4 packs have at least 10 seeds germinating.