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The binomial distribution

Probability · level 7, lesson 4 of 7

A question to start

A surprise quiz has 10 multiple-choice questions, each with 4 choices. You have not studied at all, so you guess every answer at random. The pass mark is 5 out of 10.

What is the chance you pass? Half the questions right does not sound like much. But each guess is right only a quarter of the time, so on average you will get \(10 \times \tfrac14 = 2.5\) right. Is passing a 1-in-3 chance? 1 in 10? 1 in 100? This lesson builds the formula that answers it exactly, and that same formula describes free throws, faulty parts, seeds that sprout and any other count of successes in repeated tries.

The idea

Repeated yes-or-no trials

A trial is one go at something with two possible results, which we call success and failure (the words are labels, not judgements: "success" might be a faulty part). The chance of success in one trial is written \(p\), so the chance of failure is \(1 - p\).

A random variable \(X\) is called binomial when it counts the successes in a situation with all four of these features (one way to remember them is BINS):

  • Binary: each trial has just two results, success or failure;
  • Independent: the result of one trial does not change the chances for another;
  • Number fixed: the number of trials \(n\) is decided in advance;
  • Same chance: every trial has the same success chance \(p\).

We write \(X \sim \operatorname{B}(n, p)\), read "\(X\) is binomial with \(n\) trials and success chance \(p\)". The guessing quiz is \(\operatorname{B}(10, \tfrac14)\): 10 trials (questions), each a success (right answer) with chance \(\tfrac14\), independent because each guess is made on its own.

Building the formula with a small case

Take \(n = 3\) trials. Write S for success and F for failure. There are \(2^3 = 8\) sequences. Because the trials are independent, the probability of a sequence is the product along it (the multiplication rule from level 5). For example, \(P(\text{SFS}) = p \cdot (1 - p) \cdot p = p^2 (1 - p)\).

Now look at the sequences with exactly 2 successes: SSF, SFS, FSS. Each has two factors of \(p\) and one factor of \((1 - p)\), in some order, so each has probability \(p^2(1 - p)\). There are 3 of them, so

\[P(X = 2) = 3\,p^2 (1 - p).\]

The order does not change the probability of a sequence; only the number of S's and F's matters. So every probability is (number of sequences with that many S's) times (probability of any one of them).

The general formula

With \(n\) trials and exactly \(k\) successes:

  1. How many sequences? A sequence is fixed once you choose which \(k\) of the \(n\) trial positions are the successes. That is a choice of \(k\) things from \(n\), which can be done in \(\dbinom{n}{k}\) ways, the combinations from Counting. (Read \(\dbinom{n}{k}\) as "\(n\) choose \(k\)"; \(\dbinom{n}{k} = \dfrac{n!}{k!\,(n - k)!}\).)
  2. Probability of each? \(k\) successes and \(n - k\) failures, multiplied together: \(p^k(1 - p)^{n - k}\).

So for \(X \sim \operatorname{B}(n, p)\):

\[P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}, \qquad k = 0, 1, 2, \dots, n.\]

For \(n = 3\) this gives \(1, 3, 3, 1\) times the matching powers: a row of Pascal's triangle again, which is why the coin flips in lesson 1 gave \(\tfrac18, \tfrac38, \tfrac38, \tfrac18\).

Do the probabilities add to 1? Yes, and the reason is the binomial theorem (which gives this distribution its name):

\[\sum_{k=0}^{n} \binom{n}{k} p^k (1 - p)^{n - k} = \big(p + (1 - p)\big)^n = 1^n = 1.\]

Expanding \((p + q)^n\) means choosing \(p\) or \(q\) from each of \(n\) brackets, and the coefficient \(\dbinom{n}{k}\) counts the ways to choose \(p\) exactly \(k\) times: the very same count as the sequences.

Mean and variance without any algebra

Write \(X\) as a sum of indicators: \(X = I_1 + I_2 + \dots + I_n\), where \(I_j\) is 1 if trial \(j\) succeeds and 0 if not. From the last two lessons, each indicator has mean \(p\) and variance \(p(1 - p)\), and the indicators are independent. So:

\[E(X) = np, \qquad \operatorname{Var}(X) = np(1 - p), \qquad \sigma = \sqrt{np(1 - p)}.\]

For the quiz: \(E(X) = 10 \cdot \tfrac14 = 2.5\) and \(\sigma = \sqrt{10 \cdot \tfrac14 \cdot \tfrac34} = \sqrt{1.875} \approx 1.37\). Proving \(E(X) = np\) straight from the formula needs a page of algebra with factorials; the indicator argument does it in a line. That is a good lesson in itself: the right way of looking at a problem can make it easy.

The shape

0 10 p = 0.2 0 10 p = 0.5 0 10 p = 0.8

With \(n = 10\): when \(p = 0.5\) the histogram is symmetric about 5. When \(p\) is small it piles up near 0 and has a long tail to the right (it is skewed right); when \(p\) is large it is the mirror image. As \(n\) grows, every binomial histogram becomes more and more bell-shaped; that is the subject of the normal approximation in level 8.

The most likely value (the mode) is always close to the mean \(np\): it is the whole number \(k\) with \(np - (1 - p) \le k \le np + p\). For the quiz, \(2.5 - 0.75 = 1.75 \le k \le 2.75\), so the mode is 2.

Worked examples

1. The quiz, in full. \(X \sim \operatorname{B}(10, \tfrac14)\). Since \(p = \tfrac14\) and \(1 - p = \tfrac34\),

\[P(X = k) = \binom{10}{k} \frac{3^{10 - k}}{4^{10}}, \qquad 4^{10} = 1\,048\,576.\]
\(k\) 0 1 2 3 4 5 6 7 8 9 10
\(P(X = k)\) 0.0563 0.1877 0.2816 0.2503 0.1460 0.0584 0.0162 0.0031 0.0004 0.00003 0.000001
0 0.056 1 0.188 2 0.282 3 0.250 4 0.146 5 0.058 6 0.016 7 0.003 8 9 10 number of correct guesses k

The chance of passing is \(P(X \ge 5) = 0.0584 + 0.0162 + 0.0031 + 0.0004 + 0.00003 + 0.000001 \approx 0.078\), about 1 in 13. (Exactly, it is \(\tfrac{81\,922}{1\,048\,576}\).) And the chance of getting every answer right is \(\left(\tfrac14\right)^{10}\), less than one in a million.

2. Exactly three right. \(P(X = 3) = \dbinom{10}{3} \left(\tfrac14\right)^3 \left(\tfrac34\right)^7 = 120 \cdot \tfrac{1}{64} \cdot \tfrac{2187}{16\,384} = \tfrac{262\,440}{1\,048\,576} \approx 0.250\).

3. Free throws. A basketball player makes 70% of her free throws, independently. She takes 5. What is the chance she makes at least 4?

\(X \sim \operatorname{B}(5, 0.7)\).

  • \(P(X = 4) = \dbinom54 (0.7)^4 (0.3)^1 = 5 \times 0.2401 \times 0.3 = 0.36015\).
  • \(P(X = 5) = (0.7)^5 = 0.16807\).
  • \(P(X \ge 4) = 0.36015 + 0.16807 \approx 0.528\).

So a 70% shooter makes 4 or 5 out of 5 only about half the time. (Whether real shots are independent is a real question: see the "hot hand" in level 5.)

4. Seeds. A packet says 90% of its seeds sprout. You plant 12. Chance at least 10 sprout? \(X \sim \operatorname{B}(12, 0.9)\):

  • \(P(X = 10) = \dbinom{12}{10}(0.9)^{10}(0.1)^2 = 66 \times 0.34868 \times 0.01 \approx 0.2301\);
  • \(P(X = 11) = 12 \times (0.9)^{11} \times 0.1 \approx 0.3766\);
  • \(P(X = 12) = (0.9)^{12} \approx 0.2824\).

Total \(\approx 0.889\). Mean \(np = 10.8\) seeds; \(\sigma = \sqrt{12 \times 0.9 \times 0.1} \approx 1.04\).

5. Is it binomial? Check BINS each time.

  • Roll a die 20 times, count the sixes: binomial, \(\operatorname{B}(20, \tfrac16)\).
  • Deal 5 cards, count the hearts: not binomial. The cards are not put back, so the chance of a heart changes from card to card (not independent, not the same \(p\)). That is the hypergeometric distribution of lesson 6.
  • Roll a die until a six appears, count the rolls: not binomial. The number of trials is not fixed; it is what we are counting. That is the geometric distribution of lesson 5.

Try it

  1. A fair coin is tossed 10 times. Find \(P(\text{exactly } 5 \text{ heads})\) and \(P(\text{at least } 8 \text{ heads})\).
  2. \(X \sim \operatorname{B}(20, 0.25)\). Find the mean, variance and standard deviation.
  3. Roll a die 6 times. Find the chance of exactly one six, and of at least one six.
  4. A fair coin is tossed 8 times. What is the chance of exactly 4 heads? Is it surprising that the "expected" result happens only about a quarter of the time?
  5. Which of these are binomial? (a) The number of girls among the next 5 babies born in a hospital (take \(p = 0.49\)). (b) The number of red cards when 3 cards are drawn from a deck without replacement. (c) The number of times you check your phone before a message arrives.
  6. A longer quiz has 20 questions with 4 choices each, and the pass mark is 10. Use a calculator or spreadsheet to find the chance of passing by guessing, and compare with the 10-question quiz.
  7. Write \(P(X = k + 1) \div P(X = k)\) for \(X \sim \operatorname{B}(n, p)\) in simplest form. Use it to explain why the bars rise while \(k < np - (1 - p)\) and fall after.
  8. A factory makes chips of which 2% are faulty, independently. In a box of 50, what is the chance that none is faulty? That at most one is faulty?
Answers
  1. \(P(5) = \tfrac{252}{1024} \approx 0.246\). \(P(\ge 8) = \tfrac{45 + 10 + 1}{1024} = \tfrac{56}{1024} \approx 0.0547\).
  2. Mean \(20 \times 0.25 = 5\). Variance \(20 \times 0.25 \times 0.75 = 3.75\). SD \(\sqrt{3.75} \approx 1.94\).
  3. Exactly one: \(6 \cdot \tfrac16 \cdot \left(\tfrac56\right)^5 = \left(\tfrac56\right)^5 \approx 0.402\). At least one: \(1 - \left(\tfrac56\right)^6 \approx 1 - 0.335 = 0.665\).
  4. \(\dbinom84 / 2^8 = \tfrac{70}{256} \approx 0.273\). It is not surprising once you see that there are 9 possible results (0 to 8) sharing the probability; the middle one is the most likely, and 3 or 5 heads together are more likely than exactly 4 (112 out of 256 against 70).
  5. (a) Binomial, \(\operatorname{B}(5, 0.49)\), if births are independent. (b) Not binomial: without replacement, so the chances change (hypergeometric). (c) Not binomial: the number of trials is not fixed.
  6. \(P(X \ge 10) = 1 - P(X \le 9) \approx 0.0139\), about 1 in 72. (In a spreadsheet: =1-BINOM.DIST(9,20,0.25,TRUE).) Much smaller than 0.078: with more questions, luck evens out and the score stays nearer its mean of 5, as the standard deviation \(\sqrt{np(1 - p)}\) grows more slowly than \(n\).
  7. \(\dfrac{P(k + 1)}{P(k)} = \dfrac{\binom{n}{k+1}}{\binom{n}{k}} \cdot \dfrac{p}{1 - p} = \dfrac{n - k}{k + 1} \cdot \dfrac{p}{1 - p}\). The bars rise when this is more than 1: \((n - k)p > (k + 1)(1 - p)\), which simplifies to \(k < np - (1 - p)\). After that the ratio is less than 1 and the bars fall.
  8. None: \(0.98^{50} \approx 0.364\). Exactly one: \(50 \times 0.02 \times 0.98^{49} \approx 0.372\). At most one: \(\approx 0.736\).

Common mistake

Forgetting the \(\dbinom{n}{k}\). \((0.25)^3(0.75)^7\) is the chance of one particular order, such as right, right, right, then seven wrong. There are 120 orders with three right answers, and every one counts.

Using the binomial when the trials are not independent or the chance changes. Drawing without replacement from a small group, or counting until something happens, breaks BINS. Check the four conditions before reaching for the formula.

Where it leads