The Binomial Distribution

About 7% of blood donors have type O-negative blood. If 12 donors walk into a blood drive, how likely is it that exactly 2 of them are O-negative? The count comes from a fixed number of tries, 12 donors, each of which either "hits" (is O-negative) or doesn't with the same chance, 0.07. Every count built that way has a binomial distribution.

What makes a random variable binomial

A binomial random variable $X$ counts the number of successes in a sequence of trials that meet four conditions:

1. Binary: each trial has exactly two outcomes, called success and failure. "Success" is just the outcome you are counting; an O-negative donor is a success even though nothing good happened.
2. Independent: the outcome of one trial doesn't change the probabilities for any other trial.
3. Number: the number of trials, $n$, is fixed in advance.
4. Same probability: the probability of success, $p$, is the same on every trial, so the probability of failure is $1 - p$ every time.

If all four hold, $X$ has a binomial distribution with parameters $n$ and $p$. For the blood drive, $n = 12$ donors and $p = 0.07$.

Two common ways to fail the conditions are worth knowing by sight. If you keep going until something happens (shoot free throws until the first miss), the number of trials isn't fixed, so the count is not binomial. If you draw without replacement from a small group, the success probability changes from draw to draw, so the trials are not independent: deal 5 cards from 20 trivia cards of which 8 are sports cards, and the chance the second card is a sports card is $7/19$ or $8/19$, depending on the first card, not $8/20$.

Sampling without replacement from a large population is different. Choosing 15 adults from a city of 400,000 changes the proportion left in the city by a negligible amount, so the trials are close enough to independent. AP Statistics uses the rule that the sample should be at most 10% of the population; 15 is far below 10% of 400,000.

The probability of exactly $x$ successes

Start with a small case. Suppose $n = 4$ and you want exactly 2 successes. One way is SSFF. Because the trials are independent, its probability is the product $p \cdot p \cdot (1-p) \cdot (1-p) = p^2(1-p)^2$. Every other arrangement, like SFSF or FFSS, contains the same two S's and two F's, so it has the same probability. The number of arrangements is the number of ways to choose which 2 of the 4 trials are the successes, $\binom{4}{2} = 6$. So $P(X = 2) = 6p^2(1-p)^2$.

The same reasoning works for any $n$ and $x$:

where $\binom{n}{x} = \frac{n!}{x!(n-x)!}$ (on a calculator, $n \, \text{nCr} \, x$) counts the arrangements. This is the binomial probability function.

Questions usually ask about a range of values rather than one, and the wording decides the range. With $n = 10$: "at most 3" is $P(X \le 3)$; "fewer than 3" is $P(X \le 2)$; "at least 3" is $P(X \ge 3) = 1 - P(X \le 2)$; "more than 3" is $P(X \ge 4) = 1 - P(X \le 3)$. On a TI-84, binompdf(n, p, x) gives $P(X = x)$ and binomcdf(n, p, x) gives $P(X \le x)$, the cumulative probability. The complement is what makes "at least" quick: adding many separate probabilities is slow and error-prone.

Mean and standard deviation: $np$ and $\sqrt{np(1-p)}$

Look at one trial first. It produces 1 success with probability $p$ and 0 successes with probability $1 - p$. Topic 2.9's formulas give its mean, $1 \cdot p + 0 \cdot (1-p) = p$, and its variance, $(1-p)^2 p + (0-p)^2(1-p) = p(1-p)$. A binomial count is the total of $n$ such trials. Their means add, so the mean count is $np$; because the trials are independent, their variances add too, giving $np(1-p)$, and the standard deviation is its square root.

The answers are parameters, and they need the same long-run interpretation as any mean and standard deviation. For the blood drive, $\mu_X = 12(0.07) = 0.84$ and $\sigma_X = \sqrt{12(0.07)(0.93)} \approx 0.88$: over many, many groups of 12 randomly chosen donors, the average number of O-negative donors per group is about 0.84, and the number in a group typically varies from 0.84 by about 0.88 donors.

The shape depends on $p$. When $p = 0.5$ the distribution is symmetric. When $p$ is small, most of the probability sits at low counts and there's a tail to the right; when $p$ is large, the tail runs to the left, as the figure in Example 3 shows.

Estimating binomial probabilities by simulation

Every binomial probability can be computed exactly, but a simulation (Topic 2.3) estimates it with no formula at all, and it is the method to fall back on when the setup is messy. Suppose a student guesses on all 5 questions of a quiz, each with 4 choices, so $X$ = number correct is binomial with $n = 5$ and $p = 0.25$. To simulate one quiz, generate 5 random integers from 1 to 4 (randInt(1, 4, 5) on a TI-84), call a 1 a correct answer, and count the 1's. One run of 200 simulated quizzes gave these results:

The estimate of $P(X \ge 3)$ is the fraction of simulated quizzes with 3 or more correct: $\frac{16 + 4 + 1}{200} = 0.105$. The exact value is $P(X \ge 3) = 1 - P(X \le 2) \approx 0.104$. The two agree closely because 200 repetitions is a fair number; with 20 repetitions the estimate could easily be off by 0.05 or more. Either way, a guesser gets 3 or more right only about 1 time in 10.

Worked examples

Worked exampleExample 3: an exam-style question

A seed company states that 85% of its tomato seeds germinate. A gardener plants 20 seeds randomly selected from a large shipment, in separate pots. Let $X$ be the number that germinate. Assume the company's claim is true.

(a) Calculate and interpret the mean of $X$.

(b) Calculate and interpret the standard deviation of $X$.

(c) Find the probability that fewer than 15 seeds germinate.

(d) Only 13 of the gardener's seeds germinate. Does this give reason to doubt the company's claim? Justify your answer with a probability.

Show solution

$X$ is binomial with $n = 20$ and $p = 0.85$: each seed germinates or doesn't, 20 seeds are planted, the seeds come from a large shipment (20 is less than 10% of it) and sit in separate pots, so they germinate independently, each with probability 0.85 under the claim.

(a) $\mu_X = np = 20(0.85) = 17$ seeds. If the gardener planted many, many sets of 20 of these seeds, the average number germinating per set would be about 17 seeds.

(b) $\sigma_X = \sqrt{np(1-p)} = \sqrt{20(0.85)(0.15)} = \sqrt{2.55} \approx 1.60$ seeds. The number of seeds out of 20 that germinate typically varies from the mean of 17 by about 1.60 seeds.

(c) "Fewer than 15" means 14 or fewer: $P(X \le 14) = \text{binomcdf}(n = 20, p = 0.85, x = 14) \approx 0.067$.

(d) The relevant probability is getting a result as low as 13 or lower: $P(X \le 13) = \text{binomcdf}(n = 20, p = 0.85, x = 13) \approx 0.022$. If 85% of the seeds really germinate, only about 2.2% of gardeners planting 20 seeds would see 13 or fewer germinate. That is unlikely enough to give reason to doubt the claim; the true germination rate may be lower than 85%.

Why "13 or fewer" and not "exactly 13"? Any single count is fairly unlikely (even the most likely value, 17, has probability about 0.243). The question is whether the gardener's result is unusually low, which is about the whole tail at or below 13. Topic 3.6 turns this idea into the p-value.

Where students lose points

On the exam

Binomial questions appear in multiple choice and in free response, usually with a calculator. Free-response parts typically ask you to identify the distribution with its parameters, find a probability (often "at least" or "at most"), compute or interpret the mean or standard deviation, and judge whether an observed count is surprising. Judge that last task with a tail probability like $P(X \le 13)$, not the probability of the single observed value.