---
title: "Random Variables and the Binomial Distribution"
book: "High School Mathematics"
subject: math
language: en
chapter: 33
exercises: 8
source: https://one-course.com/books/math/2/en/chapter/33-random-variables-and-the-binomial-distribution
---

# Chapter 33 — Random Variables and the Binomial Distribution

A [random variable](#def-g12-randvar-rv) attaches a number to each outcome of an experiment: a gain, a count, a duration. Its [expectation](#def-g12-randvar-exp) is the long-run average of the values it produces, its [variance](#def-g12-randvar-exp) measures their spread. The star of this chapter is the [binomial distribution](#def-g12-randvar-binomial), which counts successes in repeated [independent](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep) trials. [Random variables](#def-g12-randvar-rv) and the [binomial distribution](#def-g12-randvar-binomial) were first met in Chapters [18](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#ch-g11-prob) and [19](https://one-course.com/books/math/2/en/chapter/19-the-binomial-distribution#ch-g11-binom); this chapter reviews and deepens them, with the combinatorial tools of [Chapter 27](https://one-course.com/books/math/2/en/chapter/27-combinatorics-and-counting#ch-g12-comb) now available.

## 33.1 Discrete random variables

**Definition 33.1 (Random variable, distribution).**

A *random variable* on a finite [sample space](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) $\Omega$ is a [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $X \colon \Omega \to \R$. Its *distribution* (or law) is the data of its possible values $x_1, \dots, x_k$ and of the probabilities

$$
p_i = \P(X = x_i), \qquad \sum_{i=1}^{k} p_i = 1 .
$$

**Definition 33.2 (Expectation, variance, standard deviation).**

The *expectation* of $X$ is

$$
\E(X) = \sum_{i=1}^{k} p_i\, x_i ,
$$

its *variance* and *[standard deviation](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-variance)* are

$$
\V(X) = \E\bigl((X - \E(X))^2\bigr) = \sum_{i=1}^k p_i\bigl(x_i - \E(X)\bigr)^2,
\qquad
\sigma(X) = \sqrt{\V(X)} .
$$

**Proposition 33.3 (König–Huygens formula).**

$\V(X) = \E(X^2) - \E(X)^2$.

**Proof.** Write $m = \E(X)$ and expand:

$$
\V(X) = \sum_i p_i (x_i^2 - 2m x_i + m^2)
= \E(X^2) - 2m\sum_i p_i x_i + m^2\sum_i p_i
= \E(X^2) - 2m^2 + m^2 . \qedhere
$$

∎

**Proposition 33.4 (Affine transformation).**

For $a, b \in \R$:

$$
\E(aX + b) = a\,\E(X) + b, \qquad
\V(aX + b) = a^2\,\V(X) .
$$

**Proof.** The first is a rearrangement of the defining sum. For the second, $aX + b$ deviates from its [mean](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean) by $a(X - \E(X))$, and squaring multiplies by $a^2$. ∎

**Example 33.5 (Fair games).**

A game costs $2$ euros; a die is rolled, and the player receives the value shown if it is at least $5$, nothing otherwise. The gain $G$ takes values $-2$ ([probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac46$), $3$ ($\frac16$), $4$ ($\frac16$):

$$
\E(G) = \frac{-8 + 3 + 4}{6} = -\frac{1}{6} < 0 .
$$

On average, the player loses $17$ cents per game: the game is unfavorable (as most real games are).

## 33.2 Bernoulli trials and the binomial distribution

**Definition 33.6 (Bernoulli distribution).**

A *Bernoulli trial* is an experiment with two outcomes, *success* ([probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $p$) and *failure* ($q = 1 - p$). The indicator $X$ of success ($X = 1$ on success, $0$ on failure) follows the *Bernoulli [distribution](#def-g12-randvar-rv)* $\mathcal B(p)$:

$$
\E(X) = p, \qquad \V(X) = p(1-p) .
$$

(Indeed $\E(X) = p$, $\E(X^2) = p$, and König–Huygens gives $\V(X) = p - p^2$.)

**Definition 33.7 (Binomial distribution).**

Repeat a [Bernoulli trial](#def-g12-randvar-bernoulli) $n$ times independently, and let $X$ be the total number of successes. The [distribution](#def-g12-randvar-rv) of $X$ is the *binomial distribution* $\mathcal B(n, p)$.

![Three Bernoulli trials: exactly 32 = 3 of the 23 paths give two successes (red), each with probability p2(1-p), so (X = 2) = 3p2(1-p).](https://one-course.com/images/onecourse/chapters/math-2/g12-randvar/fig-c70f3b626032.svg)

*Three [Bernoulli trials](#def-g12-randvar-bernoulli): exactly $\binom{3}{2} = 3$ of the $2^3$ paths give two successes (red), each with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $p^2(1-p)$, so $\P(X = 2) = 3p^2(1-p)$.*

**Theorem 33.8.**

If $X \sim \mathcal B(n, p)$, then for $0 \leq k \leq n$:

$$
\P(X = k) = \binom{n}{k} p^k (1-p)^{n-k},
$$

and

$$
\E(X) = np, \qquad \V(X) = np(1-p) .
$$

**Proof.** A specified [sequence](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-sequence) of outcomes with $k$ successes and $n-k$ failures has [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $p^k(1-p)^{n-k}$ by [independence](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep); the number of such [sequences](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-sequence) is the number of ways to place the $k$ successes among the $n$ trials, namely $\binom nk$ ([Chapter 27](https://one-course.com/books/math/2/en/chapter/27-combinatorics-and-counting#ch-g12-comb)). Summing over the [sequences](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-sequence) gives the formula — and the binomial theorem confirms $\sum_k \P(X = k) = (p + q)^n = 1$.

For the [expectation](#def-g12-randvar-exp), using $k\binom nk = n\binom{n-1}{k-1}$ ([Exercise 27.7](https://one-course.com/books/math/2/en/chapter/27-combinatorics-and-counting#exo-g12-comb-7)):

$$
\E(X) = \sum_{k=1}^{n} k\binom nk p^k q^{n-k}
= np\sum_{k=1}^{n} \binom{n-1}{k-1} p^{k-1} q^{(n-1)-(k-1)}
= np\,(p + q)^{n-1} = np .
$$

The [variance](#def-g12-randvar-exp) formula is proved similarly with the identity $k(k-1)\binom nk = n(n-1)\binom{n-2}{k-2}$, giving $\E(X(X-1)) = n(n-1)p^2$, whence $\V(X) = n(n-1)p^2 + np - (np)^2 = np(1-p)$. (A structural proof — the [variance](#def-g12-randvar-exp) of a sum of [independent](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep) variables — comes with [Theorem 34.4](https://one-course.com/books/math/2/en/chapter/34-sums-of-random-variables-and-the-law-of-large-numbers#thm-g12-sums-variance).) ∎

![The distribution B(20, 0.3): mean np = 6, standard deviation √npq 2.05.](https://one-course.com/images/onecourse/chapters/math-2/g12-randvar/fig-a340174233d1.svg)

*The [distribution](#def-g12-randvar-rv) $\mathcal B(20, 0.3)$: [mean](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean) $np = 6$, [standard deviation](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-variance) $\sqrt{npq} \approx 2.05$.*

**Method 33.9 (Recognizing a binomial situation).**

Check the three ingredients before writing $X \sim \mathcal B(n,p)$: a *fixed number* $n$ of trials; *two outcomes* per trial with the *same* success [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $p$; *[independence](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep)* of trials (sampling *with* replacement, or from a large population). Then use

$$
\P(X \geq 1) = 1 - (1-p)^n
$$

for “at least one success”, and a calculator or cumulative tables for general $\P(X \leq k)$.

**Example 33.10.**

How many times must one roll a die to have at least a $99\%$ chance of rolling a six? With $X \sim \mathcal B\left(n, \frac16\right)$: $\P(X \geq 1) = 1 - \left(\frac56\right)^n \geq 0.99$ means $\left(\frac56\right)^n \leq 0.01$, *i.e.* $n \geq \frac{\ln 0.01}{\ln(5/6)} \approx 25.3$: from $n = 26$ rolls on.

## 33.3 Exercises

**Exercise 33.1 ★.**

A [random variable](#def-g12-randvar-rv) $X$ takes the values $-1, 0, 2, 5$ with probabilities $0.3, 0.2, 0.4, 0.1$. Compute $\E(X)$, $\V(X)$ and $\sigma(X)$.

**Solution of Exercise 33.1.**

$\E(X) = -0.3 + 0 + 0.8 + 0.5 = 1$. $\E(X^2) = 0.3 \times 1 + 0 + 0.4\times4 + 0.1\times25 = 4.4$, so by König–Huygens $\V(X) = 4.4 - 1 = 3.4$ and $\sigma(X) = \sqrt{3.4} \approx 1.84$.

**Exercise 33.2 ★.**

A multiple-choice test has $10$ questions, each with $4$ choices, one of which is correct. A student answers uniformly at random, independently. Let $X$ be the number of correct answers.

1. Give the [distribution](#def-g12-randvar-rv) of $X$ , $\E(X)$ and $\sigma(X)$ .
2. Compute $\P(X = 0)$ , $\P(X = 5)$ and $\P(X \geq 1)$ .

**Solution of Exercise 33.2.**

*1.* The $10$ questions are [independent](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep) [Bernoulli trials](#def-g12-randvar-bernoulli) with $p = \frac14$: $X \sim \mathcal B\left(10, \frac14\right)$, $\E(X) = 2.5$, $\sigma(X) = \sqrt{10 \times \frac14 \times \frac34} = \sqrt{1.875}
\approx 1.37$.

*2.* $\P(X = 0) = \left(\frac34\right)^{10} \approx 0.056$;

$$
\P(X = 5) = \binom{10}{5}\left(\frac14\right)^5\left(\frac34\right)^5
\approx 0.058;
\quad
\P(X \geq 1) = 1 - \left(\tfrac34\right)^{10} \approx 0.944 .
$$

**Exercise 33.3 ★.**

An insurance company insures $n = 400$ clients; each files a claim during the year with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $p = 0.05$, independently. Let $X$ be the number of claims. Identify the [distribution](#def-g12-randvar-rv) of $X$ and compute its [expectation](#def-g12-randvar-exp) and [standard deviation](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-variance).

**Solution of Exercise 33.3.**

[Independent](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep) identical trials: $X \sim \mathcal B(400,\ 0.05)$, so $\E(X) = 20$ claims and

$$
\sigma(X) = \sqrt{400 \times 0.05 \times 0.95} = \sqrt{19} \approx 4.4 .
$$

**Exercise 33.4 ★★.**

In the game of [Example 33.5](#ex-g12-randvar-game), the organizer wants a fair game ($\E(G) = 0$) by changing the entry price $c$. Find $c$. Compute the [variance](#def-g12-randvar-exp) of the gain for this fair version; is “fair” the same as “riskless”?

**Solution of Exercise 33.4.**

The payment $P$ received satisfies $\E(P) = \frac{5 + 6}{6} = \frac{11}{6}$, so the fair price is $c = \frac{11}{6} \approx 1.83$ euros. The fair gain $G = P - \frac{11}6$ takes values $-\frac{11}{6}, \frac{19}{6}, \frac{25}{6}$ with probabilities $\frac46, \frac16, \frac16$:

$$
\V(G) = \E(G^2)
= \frac{4 \times 121 + 361 + 625}{6 \times 36} = \frac{1470}{216}
= \frac{245}{36} \approx 6.8,
\qquad \sigma(G) \approx 2.6 .
$$

A fair game has zero *average* gain but its outcomes still fluctuate: fair is not riskless.

**Exercise 33.5 ★★.**

A basketball player scores free throws with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $0.7$. She shoots $8$ times ([independent](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep) shots). Compute the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) that she scores: exactly $6$; at least $6$; at least once. What is the most probable number of scores?

**Solution of Exercise 33.5.**

$X \sim \mathcal B(8,\ 0.7)$.

$$
\P(X = 6) = \binom86 (0.7)^6(0.3)^2 \approx 0.296 ;
$$

$$
\P(X \geq 6) = \P(6) + \P(7) + \P(8)
\approx 0.296 + 8(0.7)^7(0.3) + (0.7)^8
\approx 0.296 + 0.198 + 0.058 = 0.552 ;
$$

$\P(X \geq 1) = 1 - (0.3)^8 \approx 0.99993$. Mode: $(n+1)p = 6.3$, so the most probable value is $k^* = 6$ ([Exercise 33.7](#exo-g12-randvar-7)).

**Exercise 33.6 ★★.**

An airline knows that each booked passenger shows up with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $0.9$, independently. A flight has $100$ seats and the airline sells $104$ tickets. Express, using a [binomial distribution](#def-g12-randvar-binomial), the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) that more passengers show up than there are seats, and bound it numerically using a calculator (give the exact expression).

**Solution of Exercise 33.6.**

The number of passengers showing up is $X \sim \mathcal B(104,\ 0.9)$; the flight is overbooked when $X \geq 101$:

$$
\P(X \geq 101) = \sum_{k=101}^{104}\binom{104}{k}(0.9)^k(0.1)^{104-k}
\approx 0.006 .
$$

Selling $4\%$ more tickets than seats causes an incident on only about $0.6\%$ of flights — the economics behind overbooking.

**Exercise 33.7 ★★.**

Let $X \sim \mathcal B(n, p)$. Show that

$$
\frac{\P(X = k+1)}{\P(X = k)} = \frac{n-k}{k+1}\cdot\frac{p}{1-p},
$$

and deduce that the [distribution](#def-g12-randvar-rv) increases up to $k^* = \floor{(n+1)p}$ and decreases afterwards (the *mode* of the binomial).

**Solution of Exercise 33.7.**

$$
\frac{\P(X = k+1)}{\P(X = k)}
= \frac{\binom{n}{k+1}}{\binom nk}\cdot\frac{p}{1-p}
= \frac{n - k}{k + 1}\cdot\frac{p}{1-p},
$$

using $\binom{n}{k+1} = \binom nk \frac{n-k}{k+1}$. This ratio is $\geq 1$ iff $(n-k)p \geq (k+1)(1-p)$ iff $np - k p \geq k - kp + 1 - p$ iff $k \leq (n+1)p - 1$. So the probabilities increase strictly while $k + 1 \leq (n+1)p$ and decrease afterwards: the [maximum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) is attained at $k^* = \floor{(n+1)p}$ (shared with $k^* - 1$ when $(n+1)p$ is an [integer](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets)).

**Exercise 33.8 ★★★.**

*(Saint Petersburg, tamed.)* A fair coin is tossed until heads appears, but at most $10$ times. Let $N$ be the number of tosses used, and the player receives $2^N$ euros if heads appeared, $0$ otherwise.

1. Give the [distribution](#def-g12-randvar-rv) of $N$ restricted to the winning outcomes: $\P(\text{heads first at toss } k) = 2^{-k}$ for $1 \leq k \leq 10$ , and check the total [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) of winning.
2. Compute the expected payoff. What would it become without the cap of $10$ tosses?

**Solution of Exercise 33.8.**

*1.* Heads first at toss $k$ means $k-1$ tails then heads: [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\left(\frac12\right)^{k-1}\cdot\frac12 = 2^{-k}$, for $1 \leq k \leq 10$. Total winning [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\sum_{k=1}^{10} 2^{-k} = 1 - 2^{-10} = \frac{1023}{1024}$ (the game is lost only on ten consecutive tails).

*2.* Expected payoff:

$$
\sum_{k=1}^{10} 2^k \cdot 2^{-k} = \sum_{k=1}^{10} 1 = 10 \text{ euros}.
$$

Without the cap, the sum $\sum_{k\geq1} 1$ diverges: the expected payoff is infinite, although the game almost always pays a small amount — the famous *Saint Petersburg paradox*, showing that [expectation](#def-g12-randvar-exp) alone does not measure the value of a game.

## 33.4 Problem: The overbooked flight

**Problem 33.1.**

Weekend problem — airlines sell more seats than they have, factories accept lots they barely inspected, and the binomial distribution referees both

An airline with $100$ seats happily sells $105$ tickets: about $10\,\%$ of passengers never show up, and empty seats earn nothing. How far can the airline push before bumped passengers eat the profit? The [binomial distribution](#def-g12-randvar-binomial) ([Theorem 33.8](#thm-g12-randvar-binomial)) answers to the decimal — and the same machinery inspects factory lots, prices raffles, and keeps insurers solvent. Decisions under repetition: this is the binomial’s day job.

**Part I — Fluency.**

1. Give the full [distribution](#def-g12-randvar-rv) table of the number $X$ of sixes in three rolls of a die.
2. Compute $\E(X)$ and $V(X)$ ( [Definition 33.2](#def-g12-randvar-exp) ).
3. A stall charges $1$ euro for three rolls and pays $2$ euros per six obtained. Compute the expected gain: fair game?
4. Checklist practice ( [Method 33.9](#met-g12-randvar-recognize) ): is the number of hearts in $5$ cards drawn *without* replacement binomial? With replacement? Justify.
5. For $X \sim \mathcal B(20,\ 0.3)$ : give $\E(X)$ and $\sigma(X)$ , then $\E(S)$ and $\sigma(S)$ for the score $S = 5X - 10$ ( [Proposition 33.4](#prop-g12-randvar-affine) ).

**Part II — The overbooked flight.** Seats: $100$. Tickets sold: $105$. Each ticket-holder shows up with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $0.9$, independently; let $X$ be the number who show.

6. Model: justify $X \sim \mathcal B(105,\ 0.9)$ with the checklist — and confess the model’s weakest point (are no-shows really [independent](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep) ? think groups and storms).
7. Compute $\E(X)$ and $\sigma(X)$ .
8. Bumping occurs when $X \geq 101$ : write $\P(X \geq 101)$ as an explicit sum of five binomial terms.
9. Evaluate the sum (calculator): what fraction of flights sees at least one bumped passenger?
10. Money: the five extra tickets bring $5 \times 200 = 1\,000$ euros; each bumped passenger costs $800$ euros in compensation. Compute the expected number of bumped passengers, $\sum_{k=101}^{105} (k - 100)\,\P(X = k)$ , the expected compensation, and the verdict on the policy.
11. Design: the regulator tolerates $\P(\text{bumping}) < 5\,\%$ . By testing $n = 105, 106, 107, \dots$ tickets, find the largest allowed $n$ .
12. The bell shortcut: for $n = 105$ , compute the [z-score](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#pb-g11-stat-1) of the bumping threshold, $z = \frac{100.5 - \E(X)}{\sigma(X)}$ , and compare the crude “about $2\sigma$ , so roughly $2$ – $3\,\%$ in the upper tail” estimate with your exact answer. (The curve behind this shortcut is the next chapters’ star.)

**Part III — The factory gate.** A lot of parts is accepted if a [sample](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-sample) of $20$ contains at most one defective part.

13. If the true defect rate is $2\,\%$ (an honest lot), compute the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) of acceptance.
14. If the rate is $10\,\%$ (a bad lot), compute it again.
15. Name the two risks of the procedure (the honest lot rejected; the bad lot accepted), read their values from questions 13–14, and say what single change improves both at once — at what cost.
16. Behind the improvement: show that the observed defect *[frequency](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-series)* $\frac Xn$ in a [sample](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-sample) of size $n$ has [standard deviation](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-variance) $\sqrt{\frac{p(1-p)}{n}}$ , and conclude how precision scales with [sample](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-sample) size (an old friend: the $\frac{1}{\sqrt n}$ of grade 10).

**Part IV — Games, raffles, reserves.**

17. The capped Saint Petersburg game of [Exercise 33.8](#exo-g12-randvar-8) has expected payoff $10$ euros. Contrast in one paragraph with the uncapped paradox met in [Problem 18.1](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#pb-g11-prob-1) : what exactly does the cap tame, and what price would now be fair?
18. A charity raffle sells $200$ tickets at $2$ euros; prizes: one $100$ -euro and two $50$ -euro baskets. Compute a ticket’s expected gain and the raffle’s margin; compare with European roulette’s house margin of $\frac{1}{37} \approx 2.7\,\%$ . Why does the raffle get away with it?
19. An insurer holds $1\,000$ [independent](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep) policies: claim [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $0.01$ each, claim size $10\,000$ euros, premium $130$ euros. Compute the expected annual profit *and* the [standard deviation](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-variance) of total claims. Compare the two numbers: what does the comparison force real insurers to hold?
20. Finale — the binomial as decision referee: the recognition checklist; the compass $\E \pm \sigma$ ; tail probabilities as the price of risk; and the two standing caveats ( [independence](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep) is a modeling claim, and rare tails, not means, cause ruin). One sentence each, with the pointer: the law of large numbers and the bell curve, next chapters, complete the referee’s rulebook.

**Solution of Problem 33.1.**

**1.** $X \sim \mathcal B\!\left(3, \frac16\right)$: $\P(X = 0) = \frac{125}{216}$, $\P(X = 1) = \frac{75}{216}$, $\P(X = 2) = \frac{15}{216}$, $\P(X = 3) = \frac{1}{216}$.

**2.** $\E(X) = 3 \times \frac16 = \frac12$; $V(X) = 3 \times \frac16 \times \frac56 = \frac{5}{12}$.

**3.** Expected payout $2\,\E(X) = 1$ euro against a $1$-euro stake: gain $0$ — exactly fair (a rarity).

**4.** Without replacement the draws are dependent (the second card’s chances depend on the first): not binomial — the checklist’s [independence](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep) box fails. With replacement: fixed $n = 5$, constant $p = \frac14$, [independent](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep) draws: binomial.

**5.** $\E(X) = 6$, $\sigma(X) = \sqrt{4.2} \approx
2.05$. $\E(S) = 5 \times 6 - 10 = 20$; $\sigma(S) = 5\sigma(X) \approx 10.25$.

**6.** Fixed $n = 105$ trials (the tickets), each a show/no-show with the same $p = 0.9$, assumed [independent](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep): binomial. The confession: families miss flights *together* and storms empty whole planes — [independence](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep) is the model’s leap of faith, and correlated no-shows make the tails fatter than the binomial promises.

**7.** $\E(X) = 94.5$; $\sigma(X) = \sqrt{105 \times 0.9 \times 0.1} =
\sqrt{9.45} \approx 3.07$.

**8.** $\P(X \geq 101) = \sum_{k=101}^{105}
\binom{105}{k} 0.9^k\, 0.1^{105 - k}$.

**9.** $\approx 0.0167$: about one flight in sixty bumps anyone at all.

**10.** $\E(\text{bumped}) \approx 0.023$ passengers per flight: expected compensation $\approx 19$ euros — against $1\,000$ euros of extra revenue. Overbooking by five is overwhelmingly profitable; hence every airline does it.

**11.** From the table of tail probabilities: $n = 105$: $1.7\,\%$; $n = 106$: $4.0\,\%$; $n = 107$: $8.1\,\%$. The largest compliant sale is $n = 106$ tickets.

**12.** $z = \frac{100.5 - 94.5}{3.07} \approx 1.95$: the threshold sits two [standard deviations](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-variance) above the [mean](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean), and the bell-curve rule of thumb (“upper $2\sigma$ tail $\approx 2.5\,\%$”) lands close to the exact $1.7\,\%$ — the smooth curve shadowing the binomial is the coming chapters’ protagonist.

**13.** $0.98^{20} + 20 \times 0.02 \times 0.98^{19}
\approx 0.94$: the honest lot passes $94\,\%$ of the time.

**14.** $0.9^{20} + 20 \times 0.1 \times 0.9^{19}
\approx 0.39$: the bad lot still sneaks through $39\,\%$ of the time.

**15.** Producer’s risk: a good lot rejected ($\approx 6\,\%$); consumer’s risk: a bad lot accepted ($\approx 39\,\%$). A larger [sample](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-sample) (with a proportional acceptance threshold) shrinks both — at the cost of more inspection: quality has a budget line.

**16.** $V\!\left(\frac Xn\right) = \frac{V(X)}{n^2} =
\frac{np(1-p)}{n^2} = \frac{p(1-p)}{n}$: [standard deviation](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-variance) $\sqrt{\frac{p(1-p)}{n}}$. Quadruple the [sample](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-sample), halve the noise: the $\frac{1}{\sqrt n}$ law of grade 10’s fluctuation [intervals](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval), now derived.

**17.** The cap bounds the payout at $2^{10} = 1024$ euros, so each of the ten rounds contributes exactly $1$ euro of [expectation](#def-g12-randvar-exp): $\E = 10$. The uncapped game’s infinite [expectation](#def-g12-randvar-exp) came entirely from astronomically rare, astronomically large payoffs; capping confesses that no bank pays $2^{50}$ euros. A fair ticket price for the capped game: $10$ euros — and suddenly nobody is paradoxed.

**18.** Expected prize money per ticket: $\frac{100 + 2 \times 50}{200} = 1$ euro against a $2$-euro ticket: margin $50\,\%$ — eighteen times roulette’s $2.7\,\%$. The raffle survives because its players are knowingly donating: the “loss” is the point.

**19.** Expected claims: $1\,000 \times 0.01 \times
10\,000 = 100\,000$; premiums $130\,000$: expected profit $30\,000$ euros. [Standard deviation](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-variance) of total claims: $10\,000 \times \sqrt{1\,000 \times 0.01 \times 0.99}
\approx 31\,500$ euros — one ordinary bad year devours the whole expected profit. Hence capital reserves, reinsurance, and portfolios far larger than a thousand policies: insurers live off the law of large numbers and keep reserves against its slowness.

**20.** Checklist first — fixed $n$, same $p$, [independence](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep) *claimed* — or no binomial at all. Then navigate by $\E \pm \sigma$: means locate, deviations warn. Then price the tails: bumping, bad lots, ruinous years all live beyond $2\sigma$. Caveats: [independence](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#def-g12-condprob-indep) is the modeler’s promise, not the world’s; and [expectation](#def-g12-randvar-exp) ignores exactly what destroys you. The rulebook’s missing pages — how fast frequencies settle, and what shape the fluctuations take — are the next two chapters.