---
title: "Statistics and Probability"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 71
exercises: 9
source: https://one-course.com/books/math/1/en/chapter/71-statistics-and-probability
---

# Chapter 71 — Statistics and Probability

Data and chance are everywhere: sports results, weather forecasts, games. This chapter teaches how to summarize a series of numbers by its [mean](#def-g9-statproba-indicators), [median](#def-g9-statproba-indicators) and [range](#def-g9-statproba-indicators), and how to compute the [probability](#def-g9-statproba-probability) of simple random experiments — including two-step experiments, handled with tree diagrams. Both subjects are developed much further in the High School volume of this series.

## 71.1 Summarizing data

**Definition 71.1 (Mean, median, range).**

For a series of $N$ numbers:

- the *mean* is the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of all the values divided by $N$ ;
- the *median* is a value splitting the sorted series in two halves of equal size: for $N$ [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) , the middle value; for $N$ [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) , the midpoint of the two middle values;
- the *range* is the largest value minus the smallest.

**Example 71.2.**

Marks of a student: $8,\ 12,\ 9,\ 15,\ 11$.

1. [Mean](#def-g9-statproba-indicators) : $\dfrac{8 + 12 + 9 + 15 + 11}{5} = \dfrac{55}{5} = 11$ .
2. [Median](#def-g9-statproba-indicators) : sort the series: $8, 9, 11, 12, 15$ ; the middle (3rd) value is $11$ .
3. [Range](#def-g9-statproba-indicators) : $15 - 8 = 7$ .

With a sixth mark of $17$: sorted $8, 9, 11, 12, 15, 17$, the [median](#def-g9-statproba-indicators) becomes the midpoint of $11$ and $12$, that is $11.5$, and the [mean](#def-g9-statproba-indicators) becomes $\frac{72}{6} = 12$.

**Example 71.3 (Weighted mean).**

The shoe sizes sold in a day:

| size | $38$ | $39$ | $40$ | $41$ | $42$ |
| --- | --- | --- | --- | --- | --- |
| count | $3$ | $7$ | $6$ | $3$ | $1$ |

The [mean](#def-g9-statproba-indicators) size is

$$
\frac{3 \times 38 + 7 \times 39 + 6 \times 40 + 3 \times 41 + 1 \times 42}
{3 + 7 + 6 + 3 + 1}
= \frac{114 + 273 + 240 + 123 + 42}{20}
= \frac{792}{20} = 39.6 .
$$

**Remark 71.4.**

The [mean](#def-g9-statproba-indicators) and the [median](#def-g9-statproba-indicators) can differ a lot. In the series $10, 10, 10, 10, 60$ the [mean](#def-g9-statproba-indicators) is $20$ (pulled up by the value $60$) while the [median](#def-g9-statproba-indicators) is $10$. Always ask which indicator represents the situation better.

## 71.2 Probability

**Definition 71.5 (Probability).**

A random experiment has several possible *outcomes*; an *event* is a set of outcomes. Each outcome gets a *probability*: a number between $0$ and $1$, with all the probabilities of the outcomes summing to $1$; the probability $\P(A)$ of an event $A$ is the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the probabilities of its outcomes. When all $n$ outcomes are equally likely,

$$
\P(A) = \frac{\text{number of favorable outcomes}}{n}.
$$

**Example 71.6.**

A wheel is split into $8$ equal sectors: $3$ red, $3$ blue, $2$ green. Each sector has [probability](#def-g9-statproba-probability) $\frac18$, so

$$
\P(\text{red}) = \frac38, \qquad
\P(\text{green}) = \frac28 = \frac14, \qquad
\P(\text{not green}) = 1 - \frac14 = \frac34 .
$$

**Proposition 71.7 (Complement).**

For every event $A$, the *contrary event* $\bar A$ (“$A$ does not occur”) satisfies

$$
\P(\bar A) = 1 - \P(A).
$$

**Proof.** Every outcome is in exactly one of $A$ and $\bar A$, so $\P(A) + \P(\bar A)$ [sums](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) the probabilities of all outcomes, which is $1$. ∎

## 71.3 Two-step experiments

**Method 71.8 (Tree diagrams).**

For an experiment in two steps:

1. draw one branch per outcome of the first step, then continue each branch with the outcomes of the second step, writing each [probability](#def-g9-statproba-probability) on its branch;
2. multiply the probabilities along a path to get the [probability](#def-g9-statproba-probability) of that path;
3. add the probabilities of all the paths forming an event.

**Example 71.9.**

A bag holds $2$ red and $3$ black tokens. Draw one token, note its color, *put it back*, draw again. Each draw gives red with [probability](#def-g9-statproba-probability) $\frac25$, black with [probability](#def-g9-statproba-probability) $\frac35$.

![The tree of the two draws with replacement: multiply along the branches, and check that the four paths sum to 1.](https://one-course.com/images/onecourse/chapters/math-1/g9-statproba/fig-f9a985036b5b.svg)

*The tree of the two draws with replacement: multiply along the branches, and check that the four paths sum to $1$.*

[Probability](#def-g9-statproba-probability) of two tokens of the same color: $\frac{4}{25} + \frac{9}{25} = \frac{13}{25}$. [Probability](#def-g9-statproba-probability) of at least one red: $1 - \P(\text{BB}) = 1 - \frac{9}{25} = \frac{16}{25}$.

**Remark 71.10 (Frequencies approach probabilities).**

Rolling a fair die $6000$ times gives *about* $1000$ sixes — not exactly. As the number of repetitions grows, observed frequencies get closer and closer to the probabilities: this is what makes [probability](#def-g9-statproba-probability) the right tool to model repeated experiments (a story developed in the High School volume of this series).

## 71.4 Exercises

**Exercise 71.1 ★.**

Compute the [mean](#def-g9-statproba-indicators), [median](#def-g9-statproba-indicators) and [range](#def-g9-statproba-indicators) of the series $14,\ 9,\ 17,\ 9,\ 12,\ 11,\ 12$.

**Solution of Exercise 71.1.**

[Mean](#def-g9-statproba-indicators): $\dfrac{14 + 9 + 17 + 9 + 12 + 11 + 12}{7} = \dfrac{84}{7} = 12$.

[Median](#def-g9-statproba-indicators): sorted, $9, 9, 11, 12, 12, 14, 17$; the middle ($4$th) value is $12$.

[Range](#def-g9-statproba-indicators): $17 - 9 = 8$.

**Exercise 71.2 ★.**

The temperatures at noon over a week were $19,\ 21,\ 24,\ 18,\ 22,\ 25,\
19$ (degrees). Compute the [mean](#def-g9-statproba-indicators) (to a tenth) and the [median](#def-g9-statproba-indicators).

**Solution of Exercise 71.2.**

[Mean](#def-g9-statproba-indicators): $\dfrac{19 + 21 + 24 + 18 + 22 + 25 + 19}{7} = \dfrac{148}{7}
\approx 21.1$ degrees.

[Median](#def-g9-statproba-indicators): sorted, $18, 19, 19, 21, 22, 24, 25$; the $4$th value is $21$ degrees.

**Exercise 71.3 ★.**

The number of goals scored by a team in $10$ matches:

| goals | $0$ | $1$ | $2$ | $3$ |
| --- | --- | --- | --- | --- |
| matches | $2$ | $4$ | $3$ | $1$ |

Compute the [mean](#def-g9-statproba-indicators) number of goals per match, and the [median](#def-g9-statproba-indicators).

**Solution of Exercise 71.3.**

[Mean](#def-g9-statproba-indicators):

$$
\frac{2 \times 0 + 4 \times 1 + 3 \times 2 + 1 \times 3}{10}
= \frac{0 + 4 + 6 + 3}{10} = \frac{13}{10} = 1.3
\text{ goals per match.}
$$

[Median](#def-g9-statproba-indicators): the sorted series is $0,0,1,1,1,1,2,2,2,3$; the $5$th and $6$th values are both $1$: [median](#def-g9-statproba-indicators) $1$.

**Exercise 71.4 ★.**

A fair die is rolled once. Compute the [probability](#def-g9-statproba-probability) of: “rolling a $3$”; “rolling an [even number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd)”; “rolling at least a $3$”; “not rolling a $6$”.

**Solution of Exercise 71.4.**

$\P(3) = \frac16$. $\P(\text{even}) = \frac36 = \frac12$. $\P(\text{at least } 3) = \P(\{3,4,5,6\}) = \frac46 = \frac23$. $\P(\text{not } 6) = 1 - \frac16 = \frac56$.

**Exercise 71.5 ★.**

A box contains $12$ balls: $5$ white, $4$ red and $3$ green; one ball is drawn at random. Compute $\P(\text{white})$, $\P(\text{red or green})$, and $\P(\text{not white})$. What do you notice about the last two?

**Solution of Exercise 71.5.**

$\P(\text{white}) = \frac{5}{12}$.

$\P(\text{red or green}) = \frac{4 + 3}{12} = \frac{7}{12}$.

$\P(\text{not white}) = 1 - \frac{5}{12} = \frac{7}{12}$.

The last two are equal: “not white” *is* the event “red or green”.

**Exercise 71.6 ★★.**

A coin is tossed, then a fair die is rolled. Draw the tree (or count the outcomes) and compute the [probability](#def-g9-statproba-probability) of getting heads *and* a $6$; then of getting heads *or* a $6$ (or both).

**Solution of Exercise 71.6.**

The $12$ outcomes ([face](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) of the coin, value of the die) are equally likely.

Heads and a $6$: one outcome out of $12$: $\frac{1}{12}$.

Heads or a $6$: the outcomes with heads ($6$ of them) plus the outcome (tails, $6$): $7$ outcomes, so $\frac{7}{12}$. Counting each event and subtracting the overlap gives the same: $\frac{6}{12} + \frac{2}{12} - \frac{1}{12} = \frac{7}{12}$.

**Exercise 71.7 ★★.**

In the bag of [Example 71.9](#ex-g9-statproba-tree) ($2$ red, $3$ black), the two draws are now made *without* replacement. Draw the new tree (careful: the second-level probabilities change) and compute the [probability](#def-g9-statproba-probability) of drawing two black tokens, then two tokens of different colors.

**Solution of Exercise 71.7.**

First draw: red $\frac25$, black $\frac35$. Second draw without replacement: after a red, $1$ red and $3$ black remain among $4$; after a black, $2$ red and $2$ black among $4$.

Two black: $\frac35 \times \frac24 = \frac{6}{20} = \frac{3}{10}$.

Different colors (RB or BR): $\frac25 \times \frac34 + \frac35 \times \frac24
= \frac{6}{20} + \frac{6}{20} = \frac{12}{20} = \frac35$.

**Exercise 71.8 ★★.**

After $9$ matches, a basketball player averages $14$ points per match.

1. How many points has she scored in total?
2. How many points must she score in the $10$ th match to average $15$ ?

**Solution of Exercise 71.8.**

*1.* $9 \times 14 = 126$ points.

*2.* An average of $15$ over $10$ matches means $150$ points in total: she must score $150 - 126 = 24$ points.

**Exercise 71.9 ★★★.**

A game: roll two fair dice; you win if the two results are equal (“a double”).

1. Compute the [probability](#def-g9-statproba-probability) of winning.
2. You play twice (independent games). Using a tree, compute the [probability](#def-g9-statproba-probability) of winning at least once.

**Solution of Exercise 71.9.**

*1.* Among the $36$ equally likely pairs, the doubles are $(1,1), (2,2), \dots, (6,6)$: six of them, so $\P(\text{double}) = \frac{6}{36} = \frac16$.

*2.* Two independent games: tree with win ($\frac16$) / lose ($\frac56$) at each level. [Probability](#def-g9-statproba-probability) of never winning: $\frac56 \times \frac56 = \frac{25}{36}$. So

$$
\P(\text{at least one win}) = 1 - \frac{25}{36} = \frac{11}{36}
\approx 0.31 .
$$

## 71.5 Problem: The Chevalier’s ruinous bets

**Problem 71.1.**

Weekend problem — the 1654 gambling dispute that created probability theory, and the birthday coincidence that fools every classroom

In 1654 a hard-gambling French nobleman, the Chevalier de Méré, grew rich on one dice bet and began losing on another that he believed equivalent. Baffled, he asked the mathematician Blaise Pascal; Pascal wrote to Pierre de Fermat; and their exchange of letters founded the theory of [probability](#def-g9-statproba-probability). This problem replays the whole affair with the tools of this chapter — equally likely outcomes, trees and the complement rule ([Proposition 71.7](#prop-g9-statproba-complement)) — and ends with the most counterintuitive coincidence of them all.

**Part I — Dice, honestly counted.**

1. Roll a fair die twice. Using a tree (or a $6 \times 6$ table), compute the [probability](#def-g9-statproba-probability) of *no* six in the two rolls, and deduce the [probability](#def-g9-statproba-probability) of at least one six.
2. A classmate argues: “one chance in six per roll, twice, so $\frac16 + \frac16 = \frac13$ .” Compare with question 1 and point at the exact outcomes his addition counts twice.
3. Rolling two dice and adding: compute the [probability](#def-g9-statproba-probability) of a [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of $7$ and of a [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of $2$ . Why is $7$ the gamblers’ favorite?
4. Old dispute: with two dice, is a [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of $9$ or a [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of $10$ more likely? Count the outcomes and settle it.
5. Generalize question 1: explain why the [probability](#def-g9-statproba-probability) of at least one six in $n$ rolls is $$1 - \left(\frac56\right)^{\!n} ,$$ the no-six branches of the tree multiplying from level to level.

**Part II — The two bets of the Chevalier.**

6. First bet, [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) money: *at least one six in four rolls of a die* . Compute $\left(\frac56\right)^4$ as a [fraction](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#def-g9-fractions-fraction) and as a decimal, and the Chevalier’s winning [probability](#def-g9-statproba-probability) . Was the bet good?
7. The Chevalier’s own reasoning was: “four rolls at $\frac16$ each: $\frac46$ of a chance.” It gave nearly the right answer here — but push it to $7$ rolls: what absurdity does it produce, and which error of question 2 does it repeat?
8. Second bet, believed equivalent by [proportionality](https://one-course.com/books/math/1/en/chapter/67-linear-and-affine-functions#def-g9-linfunc-linear) (“ $24$ rolls at $\frac1{36}$ each: again $\frac{24}{36} = \frac46$ ”): *at least one double-six in $24$ rolls of two dice* . Compute the true winning [probability](#def-g9-statproba-probability) $1 - \left(\frac{35}{36}\right)^{24}$ (calculator). Why was the Chevalier slowly ruined?
9. How many rolls of two dice would have restored his advantage? Test $n = 25$ with the calculator and conclude.
10. State the moral in two sentences: what is wrong, in general, with multiplying a [probability](#def-g9-statproba-probability) by the number of tries — and which correct tool replaces that temptation?

**Part III — [Means](#def-g9-statproba-indicators), [medians](#def-g9-statproba-indicators), birthdays.**

11. A small firm pays nine employees $2\,000$ euros a month and the director $20\,000$ . Compute the [mean](#def-g9-statproba-indicators) and the [median](#def-g9-statproba-indicators) salary ( [Definition 71.1](#def-g9-statproba-indicators) ). Which number should the job advertisement honestly quote?
12. Construct a list of five test marks with [mean](#def-g9-statproba-indicators) $12$ and [median](#def-g9-statproba-indicators) $15$ . What does the pair ( [mean](#def-g9-statproba-indicators) below [median](#def-g9-statproba-indicators) ) say about the shape of the marks?
13. The birthday problem: in a class of $23$ students, what is the [probability](#def-g9-statproba-probability) that at least two share a birthday? Set up the complement (all $23$ birthdays different): $$\frac{365}{365} \times \frac{364}{365} \times  \frac{363}{365} \times \dots \times \frac{343}{365} .$$ Compute the [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of the first three factors, describe how the factors evolve, and — given that the full [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) is about $0.493$ — answer the question. Most people expect “very unlikely”: what does the mathematics say?
14. The intuition repaired: how many *pairs* of students can share a birthday in a class of $23$ ( [Problem 37.1](https://one-course.com/books/math/1/en/chapter/37-whole-numbers#pb-g6-wholes-1) ’s handshake count)? Explain in one sentence why this number, not $23$ itself, drives the surprise.
15. Finale: describe an experiment to check either result — de Méré’s bet with a real die, or the birthday problem across several classes — and state carefully what you expect of the observed frequencies as the number of repetitions grows. (That expectation has a name, the law of large numbers; its proof awaits in the High School volume.)

**Solution of Problem 71.1.**

**1.** Each roll has $5$ non-six outcomes out of $6$; the $6 \times 6$ table shows $5 \times 5 = 25$ six-free outcomes out of $36$: $P(\text{no six}) = \frac{25}{36}$, so $P(\text{at least one}) = 1 - \frac{25}{36} = \frac{11}{36}
\approx 0.31$.

**2.** $\frac13 = \frac{12}{36} \neq \frac{11}{36}$. The addition counts the outcome “six then six” once in each $\frac16$: once too often. Probabilities of events that can happen *together* do not simply add.

**3.** [Sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $7$: the six outcomes $(1,6), (2,5), (3,4),
(4,3), (5,2), (6,1)$: $P = \frac{6}{36} = \frac16$. [Sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $2$: only $(1,1)$: $\frac{1}{36}$. Seven has the most ways to happen — the fat middle of the [sums](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) table.

**4.** [Sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $9$: $(3,6), (4,5), (5,4), (6,3)$ — four outcomes, $\frac{4}{36}$. [Sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $10$: $(4,6), (5,5), (6,4)$ — three, $\frac{3}{36}$. Nine is more likely: the count decides, not the number of “ways to write” the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) with unordered [faces](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def).

**5.** Each level of the tree multiplies the no-six [probability](#def-g9-statproba-probability) by $\frac56$, whatever happened before: after $n$ rolls, $P(\text{no six at all}) = \left(\frac56\right)^n$, and the complement rule gives the formula.

**6.** $\left(\frac56\right)^4 = \frac{625}{1296} \approx
0.482$, so the Chevalier won with [probability](#def-g9-statproba-probability) $1 - 0.482 = 0.518$ — a quiet, steady edge of almost $2\,\%$ per game: an excellent bet, and it paid for his carriages.

**7.** At $7$ rolls his rule gives $\frac76$ — a [probability](#def-g9-statproba-probability) greater than $1$, nonsense. It is question 2’s double-counting, compounded: the rolls’ successes overlap, and adding their chances counts the overlaps again and again.

**8.** $1 - \left(\frac{35}{36}\right)^{24} \approx
1 - 0.5086 = 0.491$: below one [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half). Betting [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) money on a $49.1\,\%$ event, the Chevalier lost about $2$ games in every $100$ on average — slowly, mysteriously (to him), and inevitably.

**9.** $1 - \left(\frac{35}{36}\right)^{25} \approx
1 - 0.494 = 0.506$: with $25$ rolls the bet turns favorable. One roll separated the Chevalier from profitability.

**10.** Multiplying a [probability](#def-g9-statproba-probability) by the number of tries counts overlapping successes several times, and eventually produces impossible answers above $1$. The correct route is always through the complement: chain the failure probabilities by multiplication, then subtract from $1$.

**11.** [Mean](#def-g9-statproba-indicators): $\frac{9 \times 2\,000 + 20\,000}{10} =
\frac{38\,000}{10} = 3\,800$ euros. [Median](#def-g9-statproba-indicators): the middle salaries are both $2\,000$: [median](#def-g9-statproba-indicators) $2\,000$ euros. The advertisement should quote the [median](#def-g9-statproba-indicators) — nine of the ten workers never see anything like $3\,800$; the [mean](#def-g9-statproba-indicators) is dragged up by one outlier ([Example 71.3](#ex-g9-statproba-weighted) warned of weights).

**12.** For instance $2, 8, 15, 17, 18$: [median](#def-g9-statproba-indicators) $15$ (middle value), [mean](#def-g9-statproba-indicators) $\frac{60}{5} = 12$. A [mean](#def-g9-statproba-indicators) below the [median](#def-g9-statproba-indicators) betrays a tail of *low* marks pulling the average down while the top [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) sits high.

**13.** First three factors: $1 \times \frac{364}{365} \times \frac{363}{365} \approx
0.992$. Each new student must dodge one more taken date, so the factors shrink: $\frac{343}{365} \approx 0.94$ for the $23$rd. The full [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) falls to about $0.493$ — so

$$
P(\text{at least one shared birthday}) \approx 0.507 :
$$

better than [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd), in a class of just $23$. Intuition says “$23$ people, $365$ days, no chance”; the mathematics says “flip a coin”.

**14.** $\frac{23 \times 22}{2} = 253$ pairs. Each pair is a fresh opportunity for a coincidence, and $253$ opportunities at $\frac{1}{365}$ each is no longer a small affair: the surprise dissolves once one counts pairs, not people.

**15.** Example protocol: roll a die in blocks of four, recording whether each block shows a six, for $100$ blocks; or collect the birthday lists of many classes of about $23$ and record the proportion with a coincidence. The observed [frequency](https://one-course.com/books/math/1/en/chapter/50-organizing-data#def-g7-stats-frequency) will wobble, but as the number of repetitions grows it should settle ever closer to the computed probabilities ($0.518$; $0.507$) — frequencies converge to probabilities: the law of large numbers, stated here as an expectation and proved in the High School volume.