---
title: "Conditional Probability and Independence"
book: "High School Mathematics"
subject: math
language: en
chapter: 32
exercises: 8
source: https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence
---

# Chapter 32 — Conditional Probability and Independence

[Probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) quantifies uncertainty; [conditional probability](#def-g12-condprob-cond) quantifies how information changes it. Learning that an [event](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) $B$ occurred reshapes the probabilities of all other [events](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) — a mechanism formalized by Bayes and misused daily in courtrooms and newspapers. This chapter sets up the rules of conditioning and the exact meaning of [independence](#def-g12-condprob-indep).

## 32.1 Probability spaces (reminder)

An experiment with finitely many outcomes is modeled by a *[sample space](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model)* $\Omega$ (the set of outcomes) and a [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\P$ assigning to each [event](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) $A \subseteq \Omega$ a number $\P(A) \in \intcc{0}{1}$, additive over disjoint [unions](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) and with $\P(\Omega) = 1$. Recall the basic rules:

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

When all outcomes are equally likely, $\P(A) = \frac{\abs A}{\abs\Omega}$ — and computing probabilities reduces to the counting techniques of [Chapter 27](https://one-course.com/books/math/2/en/chapter/27-combinatorics-and-counting#ch-g12-comb).

## 32.2 Conditional probability

**Definition 32.1 (Conditional probability).**

Let $B$ be an [event](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) with $\P(B) > 0$. The *probability of $A$ given $B$* is

$$
\pcond{B}{A} = \frac{\P(A \cap B)}{\P(B)} .
$$

The map $A \mapsto \pcond BA$ is itself a [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) (all rules apply); it represents the new state of knowledge of someone who has learned that $B$ occurred.

**Proposition 32.2 (Multiplication rule).**

For [events](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) with nonzero probabilities:

$$
\P(A \cap B) = \P(B)\,\pcond{B}{A} = \P(A)\,\pcond{A}{B},
$$

and more generally $\P(A_1 \cap A_2 \cap A_3) = \P(A_1)\,\pcond{A_1}{A_2}\,
\pcond{A_1 \cap A_2}{A_3}$, etc.

**Proof.** Rearrange the definition; the chain formula follows by iterating. ∎

**Theorem 32.3 (Law of total probability).**

Let $B_1, \dots, B_n$ be a partition of $\Omega$ into [events](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) of nonzero [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution). For every [event](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) $A$:

$$
\P(A) = \sum_{i=1}^{n} \P(B_i)\,\pcond{B_i}{A}.
$$

**Proof.** The sets $A \cap B_i$ are pairwise disjoint with [union](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) $A$, so $\P(A) = \sum_i \P(A \cap B_i) = \sum_i \P(B_i)\pcond{B_i}{A}$ by the multiplication rule. ∎

**Method 32.4 (Probability trees).**

A [tree diagram](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#met-g10-proba-tree) organizes conditional probabilities: each branch carries the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) of the next [event](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) *given* the path so far.

- The [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) of a leaf (a complete path) is the *product* of the probabilities along its branches (multiplication rule).
- The [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) of an [event](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) is the *sum* of the probabilities of the leaves realizing it (total [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) ).
- Probabilities on branches leaving one node add up to $1$ .

![A two-level tree: multiply along a path, add over the leaves. For instance (A) is the sum of the first and third leaf probabilities.](https://one-course.com/images/onecourse/chapters/math-2/g12-condprob/fig-3105e44904d1.svg)

*A two-level [tree](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#met-g10-proba-tree): multiply along a path, add over the leaves. For instance $\P(A)$ is the sum of the first and third leaf probabilities.*

**Theorem 32.5 (Bayes’ formula).**

Let $B_1, \dots, B_n$ be a partition of $\Omega$ as above and $A$ an [event](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) with $\P(A) > 0$. Then

$$
\pcond{A}{B_j} =
\frac{\P(B_j)\,\pcond{B_j}{A}}{\sum_{i=1}^{n} \P(B_i)\,\pcond{B_i}{A}} .
$$

**Proof.** $\pcond{A}{B_j} = \frac{\P(A \cap B_j)}{\P(A)}$; expand the numerator by the multiplication rule and the denominator by total [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution). ∎

**Example 32.6 (Screening test).**

A disease affects $1\%$ of a population. A test detects it with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $0.99$ (sensitivity) and gives a false positive with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $0.05$. Given a positive test, the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) of actually having the disease is

$$
\frac{0.01 \times 0.99}{0.01\times0.99 + 0.99\times0.05}
= \frac{0.0099}{0.0099 + 0.0495} = \frac{1}{6} \approx 0.17 .
$$

Despite the accurate test, five positives out of six are false — because the disease is rare. Confusing $\pcond{A}{B}$ with $\pcond{B}{A}$ is the *prosecutor’s fallacy*.

![The screening test as a tree (D: diseased, T: positive test). The two positive leaves (red) have total weight 0.0594, of which the false positives contribute five sixths.](https://one-course.com/images/onecourse/chapters/math-2/g12-condprob/fig-626ad692afeb.svg)

*The screening test as a [tree](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#met-g10-proba-tree) ($D$: diseased, $T$: positive test). The two positive leaves (red) have total weight $0.0594$, of which the false positives contribute five sixths.*

## 32.3 Independence

**Definition 32.7 (Independent events).**

Two [events](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) $A$ and $B$ are *independent* if

$$
\P(A \cap B) = \P(A)\,\P(B) .
$$

When $\P(B) > 0$, this is equivalent to $\pcond{B}{A} = \P(A)$: knowing $B$ does not change the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) of $A$.

**Proposition 32.8.**

If $A$ and $B$ are [independent](#def-g12-condprob-indep), so are $A$ and $\bar B$ (and $\bar A$ and $\bar B$).

**Proof.** $\P(A \cap \bar B) = \P(A) - \P(A \cap B) = \P(A) - \P(A)\P(B)
= \P(A)\bigl(1 - \P(B)\bigr) = \P(A)\,\P(\bar B)$. ∎

**Remark 32.9.**

Do not confuse *[independent](#def-g12-condprob-indep)* ($\P(A\cap B) = \P(A)\P(B)$) with *[incompatible](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-operations)* ($A \cap B = \varnothing$). Two [incompatible](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-operations) [events](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) of nonzero [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) are never [independent](#def-g12-condprob-indep): knowing one occurred guarantees the other did not.

**Definition 32.10 (Independent repetitions).**

When an experiment is repeated $n$ times so that the outcome of each trial does not influence the others, the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) of a specified [sequence](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-sequence) of outcomes is the product of the individual probabilities. This is the model underlying the binomial [distribution](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-rv) ([Chapter 33](https://one-course.com/books/math/2/en/chapter/33-random-variables-and-the-binomial-distribution#ch-g12-randvar)).

## 32.4 Exercises

**Exercise 32.1 ★.**

A card is drawn from a standard 52-card deck. Compute the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) that it is a king, given that it is a face card (jack, queen or king). Are the [events](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) “king” and “heart” [independent](#def-g12-condprob-indep)?

**Solution of Exercise 32.1.**

There are $12$ face cards, of which $4$ are kings: $\pcond{\text{face}}{\text{king}} = \frac{4}{12} = \frac13$.

[Independence](#def-g12-condprob-indep): $\P(\text{king} \cap \text{heart}) = \frac{1}{52}$ (the king of hearts), and $\P(\text{king})\,\P(\text{heart}) = \frac{4}{52}\times\frac{13}{52}
= \frac{1}{52}$. Equal: the [events](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) are [independent](#def-g12-condprob-indep).

**Exercise 32.2 ★.**

An urn contains 5 red and 3 blue balls. Two balls are drawn in succession *without* replacement.

1. Draw the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) [tree](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#met-g10-proba-tree) .
2. Compute the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) that both are red, and the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) that the second is red.

**Solution of Exercise 32.2.**

*1.* First branch: red $\frac58$, blue $\frac38$; second branches (without replacement): after red, red $\frac47$ / blue $\frac37$; after blue, red $\frac57$ / blue $\frac27$.

*2.* $\P(RR) = \frac58 \times \frac47 = \frac{5}{14}$. By total [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution),

$$
\P(\text{2nd red}) = \frac58\cdot\frac47 + \frac38\cdot\frac57
= \frac{20 + 15}{56} = \frac58 .
$$

(The same as the first draw — by symmetry, the second ball is a uniformly random ball of the urn.)

**Exercise 32.3 ★.**

Two fair dice are rolled. Let $A$ = “the sum is $7$”, $B$ = “the first die shows $3$”, $C$ = “the sum is $6$”. Determine whether $A$ and $B$ are [independent](#def-g12-condprob-indep), then whether $B$ and $C$ are.

**Solution of Exercise 32.3.**

$\P(A) = \frac{6}{36} = \frac16$ (six pairs sum to $7$), $\P(B) = \frac16$, and $A \cap B = \{(3,4)\}$ has [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac{1}{36} = \P(A)\P(B)$: $A$ and $B$ are [independent](#def-g12-condprob-indep). (The sum $7$ is special: whatever the first die, exactly one value of the second gives it.)

$\P(C) = \frac{5}{36}$ and $B \cap C = \{(3,3)\}$: $\P(B \cap C) = \frac{1}{36} \neq \frac16 \times \frac{5}{36}
= \frac{5}{216}$. Not [independent](#def-g12-condprob-indep).

**Exercise 32.4 ★★.**

A factory has three machines producing respectively $50\%$, $30\%$ and $20\%$ of total output, with defect rates $1\%$, $2\%$ and $4\%$.

1. What proportion of the production is defective?
2. A randomly chosen item is defective. What is the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) it came from the third machine?

**Solution of Exercise 32.4.**

*1.* Total [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) with the partition by machine:

$$
\P(D) = 0.5\times0.01 + 0.3\times0.02 + 0.2\times0.04
= 0.005 + 0.006 + 0.008 = 0.019 = 1.9\% .
$$

*2.* Bayes: $\pcond{D}{M_3} = \dfrac{0.008}{0.019} = \dfrac{8}{19} \approx 0.42$. The machine producing only a fifth of the output accounts for over $40\%$ of the defects.

**Exercise 32.5 ★★.**

In [Example 32.6](#ex-g12-condprob-test), at what disease prevalence $p$ (instead of $1\%$) would a positive test [mean](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean) at least a $90\%$ chance of disease? Solve the inequality and comment.

**Solution of Exercise 32.5.**

The condition is

$$
\frac{0.99\,p}{0.99\,p + 0.05\,(1-p)} \geq 0.9 .
$$

The denominator is positive, so this reads $0.99p \geq 0.891p + 0.045 - 0.045p$, *i.e.* $0.144\,p \geq 0.045$, *i.e.*

$$
p \geq \frac{0.045}{0.144} = 0.3125 .
$$

The test alone reaches $90\%$ certainty only if the disease already affects more than $31\%$ of the tested population — which is why mass screening of rare diseases requires confirmation tests.

**Exercise 32.6 ★★.**

A biased coin lands heads with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $p \in \intoo{0}{1}$. It is tossed three times, tosses being [independent](#def-g12-condprob-indep).

1. Compute the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) of the [event](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) $E$ = “exactly two heads”.
2. For which $p$ is $\P(E)$ maximal?

**Solution of Exercise 32.6.**

*1.* Three [sequences](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-sequence) realize $E$ (HHT, HTH, THH), each of [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $p^2(1-p)$ by [independence](#def-g12-condprob-indep): $\P(E) = 3p^2(1 - p)$.

*2.* $f(p) = 3p^2 - 3p^3$ has $f'(p) = 6p - 9p^2 = 3p(2 - 3p)$, positive then negative on $\intoo{0}{1}$: [maximum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) at $p = \frac23$, where $\P(E) = 3\cdot\frac49\cdot\frac13 = \frac49$.

**Exercise 32.7 ★★★.**

*(Monty Hall.)* A prize hides behind one of three doors. You pick a door; the host, who knows where the prize is, opens one of the two remaining doors, always revealing an empty one (choosing at random if both are empty), and offers you to switch to the other closed door. Using Bayes’ formula, compute the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) of winning if you switch, and if you do not.

**Solution of Exercise 32.7.**

Say you picked door 1 and the host opened door 3 ([event](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) $H_3$). With $B_i$ = “prize behind door $i$”, $\P(B_i) = \frac13$ and

$$
\pcond{B_1}{H_3} = \frac12, \qquad
\pcond{B_2}{H_3} = 1, \qquad
\pcond{B_3}{H_3} = 0
$$

(if the prize is behind your door, the host chooses between doors 2 and 3 at random; if it is behind door 2, he is forced to open 3). Bayes:

$$
\pcond{H_3}{B_1}
= \frac{\frac13\cdot\frac12}{\frac13\cdot\frac12 + \frac13\cdot1 + 0}
= \frac{1/6}{1/2} = \frac13,
\qquad
\pcond{H_3}{B_2} = \frac{\frac13}{\frac12} = \frac23 .
$$

Staying wins with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac13$, switching with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac23$.

**Exercise 32.8 ★★★.**

An information channel transmits bits. Each bit is flipped by noise with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\varepsilon = 0.1$, independently. To protect a bit, it is sent three times and decoded by majority.

1. Compute the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) that the decoded bit is wrong.
2. The received word is $101$ . What is the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) that the sent bit was $1$ ? (Assume $0$ and $1$ are sent with equal [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) .)

**Solution of Exercise 32.8.**

*1.* The majority is wrong when at least two of the three copies are flipped:

$$
3\varepsilon^2(1-\varepsilon) + \varepsilon^3
= 3(0.01)(0.9) + 0.001 = 0.028,
$$

much smaller than $\varepsilon = 0.1$: repetition coding works.

*2.* If $1$ was sent (as $111$), receiving $101$ requires exactly one flip: [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\varepsilon(1-\varepsilon)^2 = 0.081$. If $0$ was sent ($000$), two flips are needed: $\varepsilon^2(1-\varepsilon) = 0.009$. Bayes with equal priors:

$$
\pcond{101}{\,\text{sent }1} = \frac{0.081}{0.081 + 0.009} = 0.9 .
$$

The majority decoding ($101 \mapsto 1$) is indeed the more probable guess.

## 32.5 Problem: The doctor, the judge and the spam filter

**Problem 32.1.**

Weekend problem — Bayes’ formula against the base-rate fallacy: why a $99\,\%$ test can mean $2\,\%$, why a one-in-a-million match convicts nobody, and how your mailbox thinks

A test that catches $99\,\%$ of cases comes back positive — and the true [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) of disease is below $2\,\%$. A DNA profile matching one person in a million is found — and the suspect is probably innocent. Both statements are correct, both have been misjudged by doctors and courts, and both are settled by one line of this chapter: Bayes’ formula ([Theorem 32.5](#thm-g12-condprob-bayes)). This problem trains it on urns, then in the clinic, the courtroom and the inbox.

**Part I — Fluency.**

1. A fair coin chooses urn I ( $2$ red, $1$ blue) or urn II ( $1$ red, $3$ blue), then one ball is drawn. Compute $\P(\text{red})$ with the law of total [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) ( [Theorem 32.3](#thm-g12-condprob-total) ).
2. The ball is red: compute $\P(\text{urn I} \mid  \text{red})$ .
3. One die roll: are the [events](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) “even” and “ $\leq 4$ ” [independent](#def-g12-condprob-indep) ( [Definition 32.7](#def-g12-condprob-indep) )? And “even” and “ $\leq 3$ ”?
4. Two [independent](#def-g12-condprob-indep) smoke alarms each trigger with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $0.9$ in a fire. Using [Proposition 32.8](#prop-g12-condprob-indepcompl) , compute the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) that at least one triggers.
5. Two cards without replacement: compute $\P(\text{two kings})$ with the multiplication rule ( [Proposition 32.2](#prop-g12-condprob-product) ).

**Part II — The doctor.** A disease affects $1$ person in $1\,000$. The test detects $99\,\%$ of the sick (sensitivity) but also fires on $5\,\%$ of the healthy (false positives).

6. Natural frequencies: imagine $100\,000$ people. How many are sick, how many healthy? How many positives does each group produce, and how many positives in all?
7. From the counts, compute $\P(\text{sick} \mid \text{positive})$ . Absorb the shock.
8. Recompute it with Bayes’ formula and check the agreement.
9. In surveys, most physicians answer “about $99\,\%$ ”. Name the two conditional probabilities being confused, and explain why the tiny base rate drives the true answer.
10. A second, [independent](#def-g12-condprob-indep) test also comes back positive. Update: taking $1.9\,\%$ as the new prior, compute $\P(\text{sick} \mid \text{two positives})$ . What is evidence *accumulation* doing?
11. Compute $\P(\text{sick} \mid \text{positive})$ if the same test is used only on a high-risk group with prevalence $5\,\%$ . Deduce the public-health moral about mass screening of rare conditions.

**Part III — The judge.**

12. A crime-scene DNA profile matches one person in a million. The suspect, found by searching a database of the city’s $10$ million inhabitants, matches. Expected number of innocent matches in the city? Given only the match, what is $\P(\text{innocent} \mid  \text{match})$ , roughly?
13. Name the fallacy — which [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) did the prosecutor quote, which one did the court need? — and its exact parallel with question 9.
14. The [independence](#def-g12-condprob-indep) trap: two rare tragedies in one family were once squared into “one in $73$ million” by treating them as [independent](#def-g12-condprob-indep) , convicting an innocent mother (the Sally Clark case). If the [events](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) share family risk factors so that the second, given the first, is $5$ times likelier than the base rate, recompute the order of magnitude — and state the lesson about multiplying probabilities.
15. The defense can cheat too: “ten innocents match, so the match proves nothing” ignores everything else. What does Bayes’ formula do with *several* pieces of evidence, and what single sentence should every juror know?

**Part IV — The spam filter.**

16. Your mail is $40\,\%$ spam. The word “lottery” appears in $15\,\%$ of spam and $0.5\,\%$ of honest mail. A message contains it: compute $\P(\text{spam} \mid \text{lottery})$ .
17. Error-correcting repetition ( [Exercise 32.8](#exo-g12-condprob-8) continued): with flip [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\varepsilon = 0.1$ , majority decoding of *five* copies fails with what [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) ? Compare with the three-copy code.
18. A bit crosses $4$ [independent](#def-g12-condprob-indep) noisy relays, each flipping it with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $0.1$ : it arrives correct exactly when the number of flips is even. Compute that [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) twice: by summing the binomial terms, and by the elegant formula $\frac{1 + (1 - 2\varepsilon)^4}{2}$ .
19. Real spam filters multiply the evidence of many words as if [independent](#def-g12-condprob-indep) (“naive” Bayes). Why is that assumption false for words like “win” and “lottery” — and why does the filter work well anyway? (One or two sentences.)
20. Finale — Bayes as the arithmetic of belief: prior $\times$ likelihood, renormalize, posterior. Replay the three verdicts (clinic, courtroom, inbox) in one sentence each, then list the three traps this problem exposed: forgotten base rates, fake [independence](#def-g12-condprob-indep) , and evidence read one piece at a time.

**Solution of Problem 32.1.**

**1.** $\P(\text{red}) = \frac12 \cdot \frac23 + \frac12
\cdot \frac14 = \frac13 + \frac18 = \frac{11}{24}$.

**2.** $\P(\text{I} \mid \text{red}) =
\dfrac{\frac12 \cdot \frac23}{\frac{11}{24}} =
\dfrac{1/3}{11/24} = \dfrac{8}{11}$.

**3.** $\P(\text{even} \cap \leq 4) = \P(\{2, 4\}) =
\frac13$ and $\P(\text{even})\P(\leq 4) = \frac12 \cdot
\frac23 = \frac13$: [independent](#def-g12-condprob-indep) (unexpectedly!). For $\leq 3$: $\P(\{2\}) = \frac16$ against $\frac12 \cdot \frac12 = \frac14$: dependent.

**4.** The [complements](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-operations) of [independent](#def-g12-condprob-indep) [events](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#def-g11-prob-model) are themselves [independent](#def-g12-condprob-indep), so

$$
\P(\text{none}) = 0.1 \times 0.1 = 0.01,
\qquad
\P(\text{at least one}) = 0.99 .
$$

**5.** $\frac{4}{52} \times \frac{3}{51} =
\frac{1}{221}$.

**6.** Sick: $100$; healthy: $99\,900$. Positives: $99$ true (99 % of $100$) and $4\,995$ false (5 % of $99\,900$): $5\,094$ positives in all.

**7.** $\P(\text{sick} \mid +) = \frac{99}{5\,094}
\approx 1.9\,\%$: of the people the test flags, fewer than one in fifty is actually sick.

**8.** $\dfrac{0.001 \times 0.99}{0.001 \times 0.99 +
0.999 \times 0.05} = \dfrac{0.00099}{0.05094} \approx 0.019$: the counts and the formula agree.

**9.** Confused: $\P(+ \mid \text{sick}) = 99\,\%$ (the test’s advertised quality) with $\P(\text{sick} \mid +)$ (the patient’s actual question). With a base rate of $\frac{1}{1000}$, the sick are so rare that even a $5\,\%$ leak from the huge healthy majority swamps the true positives fifty to one.

**10.** $\dfrac{0.019 \times 0.99}{0.019 \times 0.99 +
0.981 \times 0.05} \approx 28\,\%$. Each [independent](#def-g12-condprob-indep) positive multiplies the odds by the same factor: evidence accumulates, and a third positive would push past $90\,\%$ — which is why confirmatory tests exist.

**11.** $\dfrac{0.05 \times 0.99}{0.05 \times 0.99 +
0.95 \times 0.05} \approx 51\,\%$: on a high-risk group the same test is informative. Moral: screen where the base rate is substantial; mass-screening a rare condition manufactures false alarms by the thousand.

**12.** Expected innocent matches: $10^7 \times 10^{-6} = 10$. Among the roughly $11$ matching people ($10$ innocent, $1$ guilty — if the culprit is in the city), the suspect found *by the match alone* is innocent with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) about $\frac{10}{11} \approx
91\,\%$.

**13.** The prosecutor quoted $\P(\text{match} \mid \text{innocent}) = 10^{-6}$; the court needed $\P(\text{innocent} \mid \text{match})$ — question 9’s confusion under oath. (And in the real case that popularized these errors, a second blunder compounded it: see question 14.)

**14.** Naive squaring: $\left(\frac{1}{8500}\right)^2
\approx \frac{1}{72 \text{ million}}$. With the second tragedy five times likelier given the first: $\frac{1}{8500} \times \frac{5}{8500} \approx
\frac{1}{14.5 \text{ million}}$ — five times larger, and still ignoring that the *alternative* hypothesis must be weighed by Bayes too. Lesson: multiplying probabilities requires [independence](#def-g12-condprob-indep), and shared causes (genetics, environment) destroy it.

**15.** Bayes multiplies the likelihood ratios of *all* the evidence into one posterior: the match, the alibi, the fibers, each shifting the odds. The juror’s sentence: *ask how probable the evidence is under both hypotheses, and never mistake the one conditional for its reverse*.

**16.** $\dfrac{0.4 \times 0.15}{0.4 \times 0.15 +
0.6 \times 0.005} = \dfrac{0.060}{0.063} \approx 95\,\%$ spam: one word nearly settles it.

**17.** Five copies fail when $3$, $4$ or $5$ flip: $\binom53 \varepsilon^3(1-\varepsilon)^2 + \binom54
\varepsilon^4 (1 - \varepsilon) + \varepsilon^5 \approx
0.0086$ — against $0.028$ for three copies: repetition buys reliability at the price of bandwidth.

**18.** Even flips: $\binom40 0.9^4 + \binom42 0.9^2
0.1^2 + \binom44 0.1^4 = 0.6561 + 0.0486 + 0.0001 = 0.7048$; and $\frac{1 + 0.8^4}{2} = \frac{1 + 0.4096}{2} = 0.7048$: the trick formula tracks the bias $(1 - 2\varepsilon)$ through each relay.

**19.** Spammy words travel in packs — a message with “lottery” very likely contains “win”, so their joint evidence is weaker than the product pretends. The filter still works because the exaggeration usually pushes in the *right direction*: classification needs the correct side of $50\,\%$, not the correct posterior.

**20.** Clinic: a rare condition turns a strong test into a weak verdict — update from the base rate, then accumulate. Courtroom: a rare match in a large city indicts the city, not the man — likelihoods are not verdicts. Inbox: many weak clues, multiplied, sort the mail — even naively. The three traps: forget the prior and be fooled; fake [independence](#def-g12-condprob-indep) and be very fooled; read evidence piecemeal and never know what you know.