Count the successes in a fixed number of independent tries: heads in 10 flips, made free throws in 5 shots, defective bulbs in a box. This situation comes up so often that its distribution has a name and a formula, built from combinations and the expected value.
Key idea
If there are $n$ independent trials, each a success with the same probability $p$, the number of successes $X$ is binomial:
$$P(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}, \quad k = 0, 1, \ldots, n.$$
Its mean is $\mu = np$ and its standard deviation is $\sigma = \sqrt{np(1 - p)}$.
Why it works
Take one particular sequence with $k$ successes and $n - k$ failures, such as SSFSF. By independence, its probability is the product $$p \cdot p \cdot (1 - p) \cdot p \cdot (1 - p) = p^k(1 - p)^{n - k}$$ and every such sequence has that same probability. How many sequences are there? Choose which $k$ of the $n$ positions are successes: $\binom{n}{k}$. Multiply. For the mean, X is a sum of $n$ trials that are each 1 (success) or 0, each with mean $p$, so $\mu = np$; their variances $p(1 - p)$ add because the trials are independent.
- Binary: each trial is a success or a failure.
- Independent trials.
- Number of trials fixed in advance.
- Same probability of success each time.
Worked example: Free throws
A player makes 70% of free throws, independently. In 5 shots, find $P(\text{exactly 4 made})$.
- Binomial with $n = 5$, $p = 0.7$.
- $$P(X = 4) = \binom{5}{4}(0.7)^4(0.3)^1 = 5 \times 0.2401 \times 0.3$$
- $= 0.36015$.
Answer: About $0.360$.
Worked example: At least one
Roll a fair die 4 times. Find the probability of at least one six.
- "At least one" is the complement of "none."
- $P(X = 0) = \left(\frac{5}{6}\right)^4 = \frac{625}{1296}$.
- $$P(X \ge 1) = 1 - \frac{625}{1296} = \frac{671}{1296} \approx 0.518$$
Answer: About $0.518$.
Worked example: Mean and spread
In a random sample of 100 voters, where 20% of all voters support a measure, find the mean and SD of the number of supporters.
- $\mu = np = 100(0.2) = 20$.
- $\sigma = \sqrt{100(0.2)(0.8)} = \sqrt{16} = 4$.
Answer: Mean 20, SD 4.
Common mistake
Using the binomial model when the trials are not independent or n is not fixed. Drawing 5 cards without replacement and counting hearts is not binomial (the chance changes after each draw); rolling until a six appears is not binomial (n is not fixed; see the geometric distribution). Check all four BINS conditions first.