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08 · Distributions you meet daily

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Probability, lesson 8

Distributions you meet daily

Binomial, Poisson and normal, and where each one shows up.

The last lesson summed up a spread in one number. A distribution gives the whole picture: the chance of every possible result. Three distributions cover most of what you'll meet, and each is built from things you know: C(n, k), independent trials and the square root rule.

By the end, you can:

  1. Find the chance of exactly k successes in n tries with the binomial.
  2. Find the chance of k rare events in a stretch of time with the Poisson.
  3. Estimate chances with the normal curve and z-scores.

Part 1. Counting successes

Find the chance of exactly k successes in n tries with the binomial.

Many questions count successes in a fixed number of tries: heads in 10 flips, sixes in 5 rolls, orders filled out of 20 sent. If the tries are independent and each succeeds with the same chance, one recipe gives the chance of every count.

Example 1.1

Two sixes in five rolls

Roll a die 5 times. What's the chance of exactly two sixes?

  1. Choose which rolls show the sixesC(5, 2) = 10

    The sixes could be rolls 1 and 2, or 1 and 3, and so on: a group of 2 from 5, as in the first lesson.

  2. Find the chance of one such sequence(1/6)² × (5/6)³ = 125/7,776

    Two sixes and three other faces, all independent, so the chances multiply.

  3. Add the 10 sequences10 × 125/7,776 = 1,250/7,776 ≈ 0.161

    Every sequence with two sixes has the same chance, and no two can happen together.

AnswerAbout 16%.

Roll a die 5 times 40% 0 40% 1 16% 2 3% 3 0.3% 4 0.01% 5 number of sixes
Roll a die 5 times 40% 0 40% 1 16% 2 3% 3 0.3% 4 5 number of sixes
Figure 1The chance of each number of sixes in 5 rolls. Two sixes, in orange, is 16%. The bars pile up at 0 and 1, because a six is rare.

Why multiply by C(5, 2)? A sequence's chance depends only on how many sixes it holds, not where they fall. So the answer is the number of places the sixes could go, times the chance of any one sequence: counting from the first lesson, independence from the fifth.

Binomial distribution

The number of successes in n independent tries, each a success with chance p, has a binomial distribution. Its mean is n × p. Each try is a 1-or-0 score with variance p × (1 − p), and variances add, so the variance is n × p × (1 − p).

P(k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ

The tries must be independent, with the same chance each time. Aces in a five-card hand break that: after one ace, the next card is an ace with chance 3/51, not 4/52. Then you count hands directly, as in the first lesson.

Your turn

10 fair flips. What's the chance of exactly 3 heads?

C(10, 3) = 120 ways to place 3 heads among 10 flips, and each sequence has chance (1/2)¹⁰ = 1/1,024. 1/8 is three heads in a row with the other seven flips forgotten.

Part 2. Rare events in a stretch of time

Find the chance of k rare events in a stretch of time with the Poisson.

Keep going with Pro

You've done the first part. Pro opens the rest of this lesson, and every other one.

Still to come
  1. Rare events in a stretch of time
  2. The bell curve
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Next lessonLarge numbers and the CLTWhy averages settle down, and why the bell curve keeps appearing. Start lesson 9

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