---
title: "Probability and Sampling"
book: "High School Mathematics"
subject: math
language: en
chapter: 9
exercises: 9
source: https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling
---

# Chapter 9 — Probability and Sampling

[Probability](#def-g10-proba-distribution) assigns numbers to chance. Starting from experiments with equally likely outcomes — dice, coins, cards — this chapter builds the basic rules for computing probabilities of [events](#def-g10-proba-events), introduces [tree diagrams](#met-g10-proba-tree) for two-step experiments, and ends with the link between probabilities and observed frequencies. The theory continues in [Chapter 18](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#ch-g11-prob).

## 9.1 Random experiments and events

**Definition 9.1 (Outcomes and events).**

A *random experiment* is an experiment whose result cannot be predicted (rolling a die). The set of all possible results is the *sample space* $\Omega = \{\omega_1, \dots, \omega_n\}$; its elements are the *outcomes*. An *event* is a set of outcomes; an event *occurs* when the result of the experiment belongs to it.

**Definition 9.2 (Probability distribution).**

Assigning to each outcome $\omega_i$ a number $p_i \geq 0$, with $p_1 + \dots + p_n = 1$, defines a *probability*: the probability $\P(A)$ of an [event](#def-g10-proba-events) $A$ is the sum of the $p_i$ of the outcomes in $A$. In particular $\P(\Omega) = 1$ and the impossible [event](#def-g10-proba-events) $\varnothing$ has probability $0$.

**Proposition 9.3 (Equally likely outcomes).**

If the $n$ outcomes are equally likely (each $p_i = \frac1n$), then for every [event](#def-g10-proba-events) $A$:

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

**Proof.** $A$ contains some number $k$ of outcomes, each of [probability](#def-g10-proba-distribution) $\frac1n$, so $\P(A) = \frac kn$. ∎

**Example 9.4.**

Roll a fair die: $\Omega = \{1, 2, 3, 4, 5, 6\}$, each outcome having [probability](#def-g10-proba-distribution) $\frac16$. The [event](#def-g10-proba-events) $A$: “the result is even” is $A = \{2, 4, 6\}$, so $\P(A) = \frac36 = \frac12$. The [event](#def-g10-proba-events) $B$: “the result is at least $5$” is $B = \{5, 6\}$, so $\P(B) = \frac26 =
\frac13$.

## 9.2 Combining events

**Definition 9.5 (Intersection, union, complement).**

Let $A$ and $B$ be [events](#def-g10-proba-events).

- $A \cap B$ (“ $A$ and $B$ ”) occurs when both occur;
- $A \cup B$ (“ $A$ or $B$ ”) occurs when at least one occurs;
- the *complement* $\bar A$ (“not $A$ ”) occurs exactly when $A$ does not;
- $A$ and $B$ are *incompatible* when they cannot occur together: $A \cap B = \varnothing$ .

**Theorem 9.6 (Addition rules).**

For all [events](#def-g10-proba-events) $A$ and $B$:

$$
\P(A \cup B) = \P(A) + \P(B) - \P(A \cap B),
\qquad
\P(\bar A) = 1 - \P(A) .
$$

For [incompatible](#def-g10-proba-operations) [events](#def-g10-proba-events), the first rule simplifies to $\P(A \cup B) = \P(A) + \P(B)$.

**Proof.** Summing $\P(A) + \P(B)$ adds the [probability](#def-g10-proba-distribution) of every outcome of $A \cup B$ once, *except* those of $A \cap B$, which are counted twice — once in $A$ and once in $B$. Subtracting $\P(A \cap B)$ repairs the double counting. For the [complement](#def-g10-proba-operations): $A$ and $\bar A$ are [incompatible](#def-g10-proba-operations) and $A \cup \bar A = \Omega$, so $\P(A) + \P(\bar A) = \P(\Omega) = 1$. ∎

![Two overlapping events: in (A) + (B), the shaded intersection A ∩ B is counted twice, which explains the subtraction in the addition rule.](https://one-course.com/images/onecourse/chapters/math-2/g10-proba/fig-25bc6ba9d1f1.svg)

*Two overlapping [events](#def-g10-proba-events): in $\P(A) + \P(B)$, the shaded [intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) $A \cap B$ is counted twice, which explains the subtraction in the addition rule.*

**Example 9.7.**

Draw one card from a standard $52$-card deck. Let $A$: “the card is a heart” ($13$ cards) and $B$: “the card is a king” ($4$ cards). Then $A \cap B$ is “the king of hearts” ($1$ card), and

$$
\P(A \cup B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52}
= \frac{4}{13}.
$$

The [complement](#def-g10-proba-operations) rule is often the fastest route: the [probability](#def-g10-proba-distribution) that the card is *not* a heart is $1 - \frac{13}{52} = \frac34$.

## 9.3 Tree diagrams

**Method 9.8 (Two-step experiments).**

For an experiment performed in two steps:

1. draw a *tree* : one branch per possible result of the first step, then from each of these, one branch per result of the second step, writing the [probability](#def-g10-proba-distribution) on each branch (the branches leaving any node must sum to $1$ );
2. the [probability](#def-g10-proba-distribution) of a *path* (a leaf of the [tree](#met-g10-proba-tree) ) is the product of the probabilities along its branches;
3. the [probability](#def-g10-proba-distribution) of an [event](#def-g10-proba-events) is the sum of the probabilities of the paths that realize it.

**Example 9.9.**

An urn contains $3$ red balls and $2$ blue balls. Draw one ball, put it back, draw again. Each draw gives red (R) with [probability](#def-g10-proba-distribution) $\frac35$ and blue (B) with [probability](#def-g10-proba-distribution) $\frac25$.

![The tree of two draws with replacement. The four path probabilities sum to 1.](https://one-course.com/images/onecourse/chapters/math-2/g10-proba/fig-7a3dd3ce478e.svg)

*The [tree](#met-g10-proba-tree) of two draws with replacement. The four path probabilities sum to $1$.*

The [probability](#def-g10-proba-distribution) of getting two balls of the same color is $\P(\text{RR}) + \P(\text{BB}) = \frac{9}{25} + \frac{4}{25} =
\frac{13}{25}$; the [probability](#def-g10-proba-distribution) of at least one blue is $1 - \P(\text{RR}) = 1 - \frac{9}{25} = \frac{16}{25}$ ([complement](#def-g10-proba-operations) rule).

**Remark 9.10.**

Without replacement, the probabilities of the second draw depend on the first: with the urn above, after drawing a red ball, only $2$ red and $2$ blue remain, so the second branch carries $\frac24$ and $\frac24$. The [tree](#met-g10-proba-tree) method works unchanged; only the numbers on the second-level branches change.

## 9.4 Frequencies and sampling

When a [random experiment](#def-g10-proba-events) is repeated many times, the observed [frequency](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-series) of an [event](#def-g10-proba-events) approaches its [probability](#def-g10-proba-distribution) — this is why [probability](#def-g10-proba-distribution) is useful for describing the real world.

**Definition 9.11 (Sample).**

A *sample* of size $n$ is the result of repeating the same experiment $n$ times independently. The *observed [frequency](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-series)* of an [event](#def-g10-proba-events) in the sample is the number of occurrences divided by $n$.

**Example 9.12.**

A fair coin gives heads with [probability](#def-g10-proba-distribution) $0.5$. Tossing it $100$ times rarely gives exactly $50$ heads: [samples](#def-g10-proba-sample) *fluctuate*. Typical runs of $100$ tosses give frequencies like $0.46$, $0.53$, $0.49$ — close to $0.5$, but not equal. Larger [samples](#def-g10-proba-sample) fluctuate less: with $10\,000$ tosses, observed frequencies are usually within about $0.01$ of $0.5$.

**Remark 9.13 (Fluctuation interval).**

A useful rule of thumb, justified in [Chapter 19](https://one-course.com/books/math/2/en/chapter/19-the-binomial-distribution#ch-g11-binom): for a [sample](#def-g10-proba-sample) of size $n$ of an [event](#def-g10-proba-events) of [probability](#def-g10-proba-distribution) $p$ (with $n$ reasonably large), the observed [frequency](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-series) lies in the [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval)

$$
\intcc{p - \frac{1}{\sqrt n}}{\,p + \frac{1}{\sqrt n}}
$$

in roughly $95\%$ of [samples](#def-g10-proba-sample). If an observed [frequency](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-series) falls far outside this [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval), doubt the value of $p$: this is the starting point of statistical testing.

**Example 9.14.**

A factory claims that only $10\%$ of its parts are defective ($p = 0.1$). In a [sample](#def-g10-proba-sample) of $n = 400$ parts, $64$ are defective: observed [frequency](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-series) $\frac{64}{400} = 0.16$. The fluctuation [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) is $\intcc{0.1 - \frac{1}{20}}{0.1 + \frac{1}{20}} = \intcc{0.05}{0.15}$, and $0.16$ falls outside it: the [sample](#def-g10-proba-sample) casts serious doubt on the claim.

## 9.5 Exercises

**Exercise 9.1 ★.**

A fair die is rolled. Compute the [probability](#def-g10-proba-distribution) of each [event](#def-g10-proba-events): “getting a $6$”; “getting an odd number”; “getting at most $4$”; “getting a $7$”.

**Solution of Exercise 9.1.**

Each of the $6$ outcomes has [probability](#def-g10-proba-distribution) $\frac16$. “Getting a $6$”: $\frac16$. “Odd number” $= \{1, 3, 5\}$: $\frac36 = \frac12$. “At most $4$” $= \{1,2,3,4\}$: $\frac46 = \frac23$. “Getting a $7$”: impossible [event](#def-g10-proba-events), [probability](#def-g10-proba-distribution) $0$.

**Exercise 9.2 ★.**

A bag contains $5$ green, $3$ yellow and $2$ black marbles; one marble is drawn at random. Compute the [probability](#def-g10-proba-distribution) that it is green; that it is not black; that it is green or yellow.

**Solution of Exercise 9.2.**

There are $10$ equally likely marbles. $\P(\text{green}) = \frac{5}{10} = \frac12$. $\P(\text{not black}) = 1 - \frac{2}{10} = \frac{8}{10} = \frac45$. Green and yellow are [incompatible](#def-g10-proba-operations): $\P(\text{green or yellow}) = \frac{5}{10} + \frac{3}{10} = \frac{8}{10}
= \frac45$.

**Exercise 9.3 ★.**

In a class of $30$ students, $18$ study Spanish, $10$ study German, and $4$ study both. A student is picked at random. Using the addition rule, compute the [probability](#def-g10-proba-distribution) that this student studies at least one of the two languages, then the [probability](#def-g10-proba-distribution) that they study neither.

**Solution of Exercise 9.3.**

Let $S$: “studies Spanish”, $G$: “studies German”. Then $\P(S) = \frac{18}{30}$, $\P(G) = \frac{10}{30}$, $\P(S \cap G) = \frac{4}{30}$. Addition rule:

$$
\P(S \cup G) = \frac{18}{30} + \frac{10}{30} - \frac{4}{30}
= \frac{24}{30} = \frac45 .
$$

[Complement](#def-g10-proba-operations): $\P(\text{neither}) = 1 - \frac45 = \frac15$.

**Exercise 9.4 ★.**

A fair coin is tossed twice. Draw the [tree](#met-g10-proba-tree) of the experiment and compute the [probability](#def-g10-proba-distribution) of getting two heads; exactly one head; at least one head.

**Solution of Exercise 9.4.**

The [tree](#met-g10-proba-tree) has four paths HH, HT, TH, TT, each of [probability](#def-g10-proba-distribution) $\frac12 \times \frac12 = \frac14$. Two heads: $\frac14$. Exactly one head (HT or TH): $\frac24 = \frac12$. At least one head: $1 - \P(\text{TT}) = 1 - \frac14 = \frac34$.

**Exercise 9.5 ★★.**

An urn contains $4$ red and $6$ white balls. Two balls are drawn one after the other *without* replacement. Draw the [tree](#met-g10-proba-tree) with the correct probabilities on the second level, and compute the [probability](#def-g10-proba-distribution) of drawing two red balls, then of drawing two balls of different colors.

**Solution of Exercise 9.5.**

First draw: red with [probability](#def-g10-proba-distribution) $\frac{4}{10}$, white $\frac{6}{10}$. Second draw *without* replacement: after a red, $3$ red and $6$ white remain among $9$; after a white, $4$ red and $5$ white among $9$.

Two reds: $\frac{4}{10} \times \frac39 = \frac{12}{90} = \frac{2}{15}$.

Different colors: paths RW and WR:

$$
\frac{4}{10} \times \frac69 + \frac{6}{10} \times \frac49
= \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}.
$$

**Exercise 9.6 ★★.**

Two fair dice are rolled and their sum is recorded.

1. Explain why the [sample space](#def-g10-proba-events) can be taken as the $36$ equally likely pairs $(1,1), (1,2), \dots, (6,6)$ .
2. Compute the [probability](#def-g10-proba-distribution) that the sum is $7$ ; that it is $12$ ; that it is at most $4$ .
3. Which sum is the most likely?

**Solution of Exercise 9.6.**

*1.* Each die independently shows each face with [probability](#def-g10-proba-distribution) $\frac16$, so the $36$ ordered pairs are equally likely; the sum is determined by the pair.

*2.* Sum $7$: the pairs $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$, so $\P = \frac{6}{36} = \frac16$. Sum $12$: only $(6,6)$, $\P = \frac{1}{36}$. Sum at most $4$: pairs summing to $2$, $3$ or $4$: $(1,1)$; $(1,2), (2,1)$; $(1,3), (2,2), (3,1)$ — six pairs, so $\P = \frac{6}{36} = \frac16$.

*3.* The number of pairs achieving each sum grows from $1$ (sum $2$) to $6$ (sum $7$) then decreases: the most likely sum is $7$.

**Exercise 9.7 ★★.**

A multiple-choice question offers $4$ answers, one correct. A student answers $2$ independent such questions completely at random. Compute the [probability](#def-g10-proba-distribution) of answering both correctly, exactly one correctly, none correctly. Check that the three probabilities sum to $1$.

**Solution of Exercise 9.7.**

Each question is answered correctly with [probability](#def-g10-proba-distribution) $\frac14$, independently. [Tree](#met-g10-proba-tree) with two levels:

Both correct: $\frac14 \times \frac14 = \frac{1}{16}$.

Exactly one correct: paths correct–wrong and wrong–correct: $\frac14 \times \frac34 + \frac34 \times \frac14 = \frac{6}{16}
= \frac38$.

None correct: $\frac34 \times \frac34 = \frac{9}{16}$.

Check: $\frac{1}{16} + \frac{6}{16} + \frac{9}{16} = 1$.

**Exercise 9.8 ★★.**

A politician claims $50\%$ approval. In a poll of $n = 900$ randomly chosen voters, $396$ approve.

1. Compute the observed [frequency](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-series) and the fluctuation [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) around $p = 0.5$ for $n = 900$ .
2. Does the poll cast doubt on the claim?

**Solution of Exercise 9.8.**

*1.* Observed [frequency](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-series): $\frac{396}{900} = 0.44$. With $n = 900$: $\frac{1}{\sqrt{900}} = \frac{1}{30} \approx 0.033$, so the fluctuation [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) is about $\intcc{0.467}{0.533}$.

*2.* The observed $0.44$ lies outside the [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval): [samples](#def-g10-proba-sample) of $900$ voters from a population with $50\%$ approval rarely stray that far. The poll casts serious doubt on the claimed $50\%$.

**Exercise 9.9 ★★★.**

A game costs $2$ to play: you roll a fair die and win $6$ if you roll a $6$, $3$ if you roll a $5$, nothing otherwise.

1. Compute the [probability](#def-g10-proba-distribution) of each amount won ( $6$ , $3$ , $0$ ).
2. Over $600$ games, about how many of each outcome do you expect? Compute the total expected gain and loss: is the game worth playing?

**Solution of Exercise 9.9.**

*1.* $\P(\text{win } 6) = \frac16$ (roll a $6$), $\P(\text{win } 3) = \frac16$ (roll a $5$), $\P(\text{win } 0) = \frac46 = \frac23$.

*2.* Over $600$ games, expect about $100$ sixes, $100$ fives, $400$ other rolls. Winnings: about $100 \times 6 + 100 \times 3 = 900$. Cost: $600 \times 2 = 1200$. Expected net result: $900 - 1200 = -300$, i.e. a loss of about $0.5$ per game on average — the game is not worth playing.

## 9.6 Problem: The three doors that fooled a thousand mathematicians

**Problem 9.1.**

Weekend problem — the Monty Hall paradox and its cousins: how information changes probability, and how samples separate claims from wishful thinking

In 1990 the columnist Marilyn vos Savant published a puzzle about a game show host, three doors, a car and two goats — and answered it correctly. Ten thousand readers, among them hundreds of mathematics PhDs, wrote in to insist she was wrong. This problem arms you with [trees](#met-g10-proba-tree) ([Method 9.8](#met-g10-proba-tree)), [complements](#def-g10-proba-operations) and the addition rules ([Theorem 9.6](#thm-g10-proba-addition)), lets you settle the argument yourself, and finishes with the tool that arbitrates every such dispute in practice: the fluctuating [sample](#def-g10-proba-sample).

**Part I — The rules, warmed up.**

1. One card from a standard $52$ -card deck: compute $P(\text{king or heart})$ with the addition rule — and say why $\frac{4}{52} + \frac{13}{52}$ alone is wrong.
2. Three fair coin flips: compute $P(\text{at least one  head})$ by the [complement](#def-g10-proba-operations) .
3. A bag holds $3$ red and $2$ blue marbles; two are drawn without replacement. With a [tree](#met-g10-proba-tree) , compute $P(\text{two reds})$ and $P(\text{one of each color})$ .
4. Complete the distribution of question 3 with $P(\text{two blues})$ and check the three probabilities sum to $1$ .
5. Someone reports, for two *[incompatible](#def-g10-proba-operations)* [events](#def-g10-proba-events) , $P(A) = 0.6$ and $P(B) = 0.5$ . Convict them with one line of arithmetic.

**Part II — The three doors.** The game: a car hides behind one of three doors, goats behind the other two. You pick a door (say door 1). The host — who *knows* where the car is — opens one of the other two doors, always revealing a goat, and offers you the chance to switch to the remaining closed door.

6. Before computing anything: does switching help, hurt, or not matter? Write down your gut answer — honestly.
7. Now the [tree](#met-g10-proba-tree) , by cases on the car’s true position (each $\frac13$ ): if the car is behind your door, switching loses; if it is behind either other door, what must the host do, and what does switching then win? Compute $P(\text{win by switching})$ .
8. The one-line version: switching wins exactly when your first pick was wrong. Conclude again.
9. Now the crucial variant — “Monty Fall”: the host does *not* know where the car is, opens one of the other doors at random, and by luck reveals a goat. List the six equally likely scenarios (car position $\times$ door opened), discard those where the car is revealed, and compute the [probability](#def-g10-proba-distribution) that switching wins *among the surviving scenarios* . What changed, and why does the host’s knowledge matter?
10. The intuition pump: one hundred doors, you pick one, the knowing host opens $98$ goat doors. Switch? With what [probability](#def-g10-proba-distribution) of winning?
11. You face a skeptic who, like the thousand letter-writers, insists on “50–50”. Describe a card-based simulation (three cards, many rounds, a friend as host) and say what proportion of switching wins you expect after $300$ rounds — with the fluctuation [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) of the chapter as your margin.

**Part III — Cousins of the paradox.**

12. A family has two children; you learn that *at least one is a boy* . List the equally likely two-children configurations compatible with the information, and compute $P(\text{two boys})$ .
13. Same family, but the information is now: *the elder child is a boy* . Recompute. Why do the two answers differ, and what do they share with the two Montys?
14. Three fair coins: compute $P(\text{all three alike})$ . The mathematician d’Alembert argued in 1754 for another answer: “either they are all alike or they are not — one case in two.” Name the exact error, with [Proposition 9.3](#prop-g10-proba-uniform) in hand.
15. A game costs $2$ euros: roll a die, win $6$ euros on a six, $3$ euros on a five, nothing otherwise ( [Exercise 9.9](#exo-g10-proba-9) ). Over $600$ games, what average gain per game does the counting predict? An insurer plays the same mathematics: a $1\,\%$ yearly chance of a $10\,000$ -euro claim — what is the *fair* premium, and why does the real premium exceed it?

**Part IV — [Samples](#def-g10-proba-sample) as arbiters.**

16. A coin is flipped $100$ times. Using the fluctuation [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $p \pm \frac{1}{\sqrt n}$ , decide which of these outcomes should raise eyebrows: $55$ heads; $70$ heads.
17. A poll of $1\,000$ voters gives a candidate $52\,\%$ . Compute the fluctuation [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) around $50\,\%$ and say whether the poll *proves* the candidate is ahead.
18. A factory claims $4\,\%$ defective parts. A [sample](#def-g10-proba-sample) of $400$ shows $28$ defective ( $7\,\%$ ); a later [sample](#def-g10-proba-sample) of $2\,500$ shows $175$ ( $7\,\%$ again). Test the claim against each [sample](#def-g10-proba-sample) . Why do the verdicts differ?
19. Two error sources haunt every survey: sampling *bias* (the how-data-lies weekend problem of the Middle School volume) and sampling *fluctuation* (this chapter). Which one shrinks as $n$ grows, and which one survives any [sample](#def-g10-proba-sample) size? One sentence each.
20. Finale — the probabilist’s checklist, one line per item: outcomes equally likely before counting (question 15); [trees](#met-g10-proba-tree) and [complements](#def-g10-proba-operations) for multi-step and “at least” (questions 2–3); *information changes probabilities, and how it was obtained matters* (questions 7–14); [samples](#def-g10-proba-sample) fluctuate by about $\frac{1}{\sqrt n}$ , and simulation arbitrates (questions 11, 16–18).

**Solution of Problem 9.1.**

**1.** $P(\text{king or heart}) = \frac{4}{52} +
\frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$: the king of hearts sits in both [events](#def-g10-proba-events) and must not be counted twice — hence the subtraction ([Theorem 9.6](#thm-g10-proba-addition)).

**2.** $P(\text{no head}) = \left(\frac12\right)^3 =
\frac18$, so $P(\text{at least one}) = \frac78$.

**3.** $P(RR) = \frac35 \times \frac24 = \frac{3}{10}$; $P(\text{one of each}) = \frac35 \times \frac24 + \frac25
\times \frac34 = \frac{6}{20} + \frac{6}{20} = \frac35$.

**4.** $P(BB) = \frac25 \times \frac14 = \frac1{10}$; and $\frac{3}{10} + \frac{6}{10} + \frac{1}{10} = 1$: the [tree](#met-g10-proba-tree)’s leaves always total $1$.

**5.** [Incompatible](#def-g10-proba-operations) [events](#def-g10-proba-events) obey $P(A \cup B) = P(A) +
P(B) = 1.1 > 1$: impossible — some [probability](#def-g10-proba-distribution) was invented.

**6.** (Most people, including the thousand letter-writers, answer “it doesn’t matter, 50–50”. Keep your answer for question 8.)

**7.** Car behind door 1 ([probability](#def-g10-proba-distribution) $\frac13$): the host opens either goat door; switching leaves the car: *lose*. Car behind door 2 ($\frac13$): the host *must* open door 3; switching lands on door 2: *win*. Car behind door 3: symmetric: *win*. Total: $P(\text{switch wins}) = \frac13 + \frac13 = \frac23$.

**8.** Your first pick is wrong with [probability](#def-g10-proba-distribution) $\frac23$ — and precisely then, the host’s forced reveal leaves the car behind the remaining door: switching converts every initial error into a win. $\frac23$, whatever the gut said.

**9.** Six equally likely scenarios: car behind $1$, $2$ or $3$, host opens door $2$ or $3$ (never yours). The reveal shows the car in two of them (car@2/opens 2, car@3/opens 3): discarded. Among the four survivors — car@1/opens 2, car@1/opens 3, car@2/opens 3, car@3/opens 2 — switching wins in the last two: $\frac24 = \frac12$. With an ignorant host, 50–50 *is* correct: the paradox lives entirely in the host’s knowledge, which filters the scenarios unevenly.

**10.** Your pick holds the car with [probability](#def-g10-proba-distribution) $\frac{1}{100}$; the knowing host’s $98$ reveals funnel the remaining $\frac{99}{100}$ onto the one closed door. Switch, and win $99$ times in $100$.

**11.** Three cards (one ace, two jokers), a friend who peeks and always discards a joker from the two you did not pick; play $300$ rounds, always switching, and record wins. Expected [frequency](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-series) $\frac23 \approx 0.667$, with fluctuation $\pm\frac{1}{\sqrt{300}} \approx 0.058$: expect between about $61\,\%$ and $72\,\%$ switching wins — far from any “50–50”.

**12.** Equally likely configurations (elder–younger): BB, BG, GB, GG. “At least one boy” keeps BB, BG, GB: $P(\text{BB}) = \frac13$.

**13.** “The elder is a boy” keeps BB, BG: $P(\text{BB}) = \frac12$. Different information filters different scenario sets — exactly as the two Montys did: what you learn matters, and *how* you learned it matters just as much.

**14.** All alike: HHH, TTT: $\frac28 = \frac14$. D’Alembert’s error: his two “cases” are not equally likely — “not all alike” bundles six of the eight outcomes. [Proposition 9.3](#prop-g10-proba-uniform) counts only over equally likely outcomes.

**15.** Average winnings per game: $6 \times \frac16 + 3
\times \frac16 = 1.50$ euros, against a $2$-euro stake: an average loss of $0.50$ per game, about $300$ euros over $600$ games. The fair premium equalizes average payout: $1\,\% \times 10\,000 = 100$ euros; real premiums add costs and a safety margin — the insurer must survive bad years, not just average ones.

**16.** [Interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval): $0.5 \pm \frac{1}{\sqrt{100}} =
\intcc{0.4}{0.6}$. $55$ heads ($0.55$): inside, unremarkable. $70$ heads ($0.70$): far outside — doubt the coin.

**17.** $\frac{1}{\sqrt{1000}} \approx 0.032$: a fair $50\,\%$ candidate would typically poll in $\intcc{0.468}{0.532}$ — and $0.52$ sits comfortably inside. The poll *suggests* a lead but proves nothing: the margin swallows it.

**18.** First [sample](#def-g10-proba-sample): [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $0.04 \pm \frac{1}{20} =
\intcc{0}{0.09}$; the observed $0.07$ is inside: no verdict. Second: $0.04 \pm \frac{1}{50} = \intcc{0.02}{0.06}$; the observed $0.07$ falls outside: the claim is now in serious doubt. Same [frequency](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-series), sharper instrument: precision grows like $\sqrt n$.

**19.** Fluctuation shrinks like $\frac{1}{\sqrt n}$: quadruple the [sample](#def-g10-proba-sample), halve the noise. Bias — asking the wrong crowd — survives any $n$: two million badly chosen ballots stayed wrong (the how-data-lies weekend problem of the Middle School volume).

**20.** (i) Check outcomes are equally likely before counting — d’Alembert forgot. (ii) [Trees](#met-g10-proba-tree) multiply along branches; “at least one” surrenders to the [complement](#def-g10-proba-operations). (iii) Information filters scenarios, and its *source* changes the filter — knowing Monty against fallen Monty, elder-boy against some-boy. (iv) Frequencies approach probabilities at speed $\frac{1}{\sqrt n}$, so a patient simulation settles any argument — including one with ten thousand angry letter-writers.