Chapter 13 · Probability

Exercise 13.1 · Q11: Die with $E = \{1,3,5\}$ ⁠, $F = \{2,3\}$ ⁠, $G = \{2,3,4,5\}$ ⁠: six conditional probabilities

Exercise 13.1Conditional probability · Section 13.2

The question

A fair die is rolled. Consider events $E = \{1, 3, 5\}$ ⁠, $F = \{2, 3\}$ and $G = \{2, 3, 4, 5\}$ ⁠. Find

(i) $P(E|F)$ and $P(F|E)$ (ii) $P(E|G)$ and $P(G|E)$

(iii) $P((E\cup F)|G)$ and $P((E\cap F)|G)$

Exam answer

Here $S = \{1, 2, 3, 4, 5, 6\}$ ⁠, and each outcome has probability $\frac{1}{6}$ since the die is fair. So $P(E) = \frac{3}{6}$ ⁠, $P(F) = \frac{2}{6}$ and $P(G) = \frac{4}{6}$ ⁠.

By the definition of conditional probability:

(i) $E\cap F = \{3\}$ ⁠, so $P(E\cap F) = \frac{1}{6}$ ⁠.

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \frac{1}{6}\div\frac{2}{6} \end{aligned}

\begin{aligned} P(F|E) &= \frac{P(E\cap F)}{P(E)} \\ &= \frac{1}{6}\div\frac{3}{6} \end{aligned}

$P(E|F) = \frac{1}{2}$ ⁠, $P(F|E) = \frac{1}{3}$ ⁠.

(ii) $E\cap G = \{3, 5\}$ ⁠, so $P(E\cap G) = \frac{2}{6}$ ⁠.

\begin{aligned} P(E|G) &= \frac{P(E\cap G)}{P(G)} \\ &= \frac{2}{6}\div\frac{4}{6} \end{aligned}

\begin{aligned} P(G|E) &= \frac{P(E\cap G)}{P(E)} \\ &= \frac{2}{6}\div\frac{3}{6} \end{aligned}

$P(E|G) = \frac{1}{2}$ ⁠, $P(G|E) = \frac{2}{3}$ ⁠.

(iii) $E\cup F = \{1, 2, 3, 5\}$ ⁠, so $(E\cup F)\cap G = \{2, 3, 5\}$ and $P((E\cup F)\cap G) = \frac{3}{6}$ ⁠.

\begin{aligned} &P((E\cup F)|G) \\ &= \frac{P((E\cup F)\cap G)}{P(G)} \\ &= \frac{3}{6}\div\frac{4}{6} \end{aligned}

Also $E\cap F = \{3\}$ ⁠, so $(E\cap F)\cap G = \{3\}$ and $P((E\cap F)\cap G) = \frac{1}{6}$ ⁠.

\begin{aligned} &P((E\cap F)|G) \\ &= \frac{P((E\cap F)\cap G)}{P(G)} \\ &= \frac{1}{6}\div\frac{4}{6} \end{aligned}

$\therefore\ P((E\cup F)|G) = \dfrac{3}{4}$ and $P((E\cap F)|G) = \dfrac{1}{4}$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

All six faces are equally likely, so every conditional probability is a count: $P(A|B) = \dfrac{n(A\cap B)}{n(B)}$ ⁠. The event after the bar is the new sample space; we count how many of its faces also lie in the event before the bar. The order matters: $P(E|F)$ and $P(F|E)$ divide by different events.

Given, to find, and the plan

Given: One roll of a fair die: $S = \{1, 2, 3, 4, 5, 6\}$ ⁠, each face with probability $\frac16$ ⁠. Events $E = \{1, 3, 5\}$ (an odd number), $F = \{2, 3\}$ ⁠, $G = \{2, 3, 4, 5\}$ ⁠.
To find: (i) $P(E|F)$ ⁠, $P(F|E)$ ⁠; (ii) $P(E|G)$ ⁠, $P(G|E)$ ⁠; (iii) $P((E\cup F)|G)$ ⁠, $P((E\cap F)|G)$ ⁠.
Plan: Find $P(E)$ ⁠, $P(F)$ ⁠, $P(G)$ ⁠. For each part, find the needed intersection as a set, then divide by the probability of the event after the bar. Check (iii) with Property 2 of conditional probability.

Basics you need

1Union and intersection of sets

$A\cup B$ ⁠: the faces in $A$ or $B$ (or both); $A\cap B$ ⁠: the faces in both. Basics: events as sets

2Classical probability

With equally likely outcomes, $P(A) = \frac{n(A)}{n(S)}$ ⁠; here $n(S) = 6$ ⁠. Basics: P(E) = n(E)/n(S)

3Conditional probability

$P(A|B) = \dfrac{P(A\cap B)}{P(B)}$ ⁠, provided $P(B) \ne 0$ (Definition 1, Section 13.2). Lesson: conditional probability

4Property 2 of conditional probability

$P((A\cup B)|F)$ $= P(A|F) + P(B|F)$ $-\, P((A\cap B)|F)$ ⁠, the addition rule inside the new sample space $F$ (Section 13.2.1). Basics: add, but do not count the overlap twice

Part (i): $P(E|F)$ and $P(F|E)$

The probabilities of $E$ ⁠, $F$ ⁠, $G$
$P(E) = \frac{3}{6},\quad P(F) = \frac{2}{6},\quad P(G) = \frac{4}{6}.$
$E$ has 3 faces, $F$ has 2 and $G$ has 4, each face with probability $\frac16$ ⁠. None of them is 0, so every conditional probability asked for is defined. We keep the sixths uncancelled; they cancel in each division.

Classical probability. Basics: P(E) = n(E)/n(S)

The six faces, once for each of $E$ ⁠, $F$ ⁠, $E\cup F$ and $E\cap F$ ⁠, coloured against $G = \{2, 3, 4, 5\}$ ⁠. Blue and gold faces make up $G$ ⁠, the new sample space of parts (ii) and (iii); the blue ones are also in the event of that row. So $P(E|G) = \frac24$ ⁠, $P((E\cup F)|G) = \frac34$ and $P((E\cap F)|G) = \frac14$ ⁠.
Intersect and divide both ways

$E\cap F = \{3\}$ ⁠, the only odd number in $F$ ⁠, so $P(E\cap F) = \frac{1}{6}$ ⁠.
$P(E|F) = \frac{P(E\cap F)}{P(F)} = \frac{1/6}{2/6} = \frac{1}{2},$ $P(F|E) = \frac{P(E\cap F)}{P(E)} = \frac{1/6}{3/6} = \frac{1}{3}.$
Both use the same top, $P(E\cap F) = P(F\cap E)$ ⁠, but different bottoms. Given $F$ ⁠, the world is $\{2, 3\}$ and 1 of its 2 faces is odd. Given $E$ ⁠, the world is $\{1, 3, 5\}$ and 1 of its 3 faces is in $F$ ⁠.

Definition 1, Section 13.2. Lesson: conditional probability

Part (ii): $P(E|G)$ and $P(G|E)$

Intersect and divide both ways

$E\cap G = \{3, 5\}$ ⁠, so $P(E\cap G) = \frac{2}{6}$ ⁠.
$P(E|G) = \frac{P(E\cap G)}{P(G)} = \frac{2/6}{4/6} = \frac{1}{2},$ $P(G|E) = \frac{P(E\cap G)}{P(E)} = \frac{2/6}{3/6} = \frac{2}{3}.$
The odd faces 1, 3, 5 that are also in $\{2, 3, 4, 5\}$ are 3 and 5. Given $G$ ⁠, 2 of its 4 faces are odd; given $E$ ⁠, 2 of its 3 faces are in $G$ ⁠.

Definition 1, Section 13.2. Lesson: conditional probability

Part (iii): $P((E\cup F)|G)$ and $P((E\cap F)|G)$

The union and the intersection, cut down to $G$
$\begin{aligned} E\cup F &= \{1, 2, 3, 5\}, \\ (E\cup F)\cap G &= \{2, 3, 5\}, \\ E\cap F &= \{3\}, \\ (E\cap F)\cap G &= \{3\}. \end{aligned}$
$E\cup F$ collects every face of $E$ or $F$ ⁠: 1, 3, 5 and 2, 3, with 3 written once. Of these, 1 is not in $G$ ⁠. The single face of $E\cap F$ ⁠, 3, is in $G$ ⁠.

Union and intersection of sets. Basics: events as sets
Divide by $P(G)$
$\begin{aligned} P((E\cup F)|G) &= \frac{P((E\cup F)\cap G)}{P(G)} \\ &= \frac{3/6}{4/6} = \frac{3}{4}, \\ P((E\cap F)|G) &= \frac{P((E\cap F)\cap G)}{P(G)} \\ &= \frac{1/6}{4/6} = \frac{1}{4}. \end{aligned}$
The definition works for any event before the bar, including a union or an intersection: intersect it with $G$ ⁠, then divide by $P(G) = \frac46$ ⁠. In counts: 3 of the 4 faces of $G$ are in $E\cup F$ ⁠, and 1 of them is in $E\cap F$ ⁠.

Definition 1, Section 13.2. Lesson: conditional probability
Check the first answer with Property 2

Here $F\cap G = \{2, 3\}$ ⁠, so $P(F|G) = \frac{2/6}{4/6} = \frac{1}{2}$ ⁠. Then
$\begin{aligned} &P(E|G) + P(F|G) - P((E\cap F)|G) \\ &= \frac{1}{2} + \frac{1}{2} - \frac{1}{4} = \frac{3}{4}, \end{aligned}$
which is $P((E\cup F)|G)$ ⁠, as found above.

Property 2 is the addition rule used inside the world $G$ ⁠. Adding $P(E|G)$ and $P(F|G)$ counts face 3 twice, so its share $P((E\cap F)|G) = \frac14$ is taken off once.

Property 2, Section 13.2.1. Basics: add, but do not count the overlap twice

Result

(i) $P(E|F) = \frac{1}{2}$ ⁠, $P(F|E) = \frac{1}{3}$ (ii) $P(E|G) = \frac{1}{2}$ ⁠, $P(G|E) = \frac{2}{3}$ (iii) $P((E\cup F)|G) = \frac{3}{4}$ ⁠, $P((E\cap F)|G) = \frac{1}{4}$ ⁠.

(Textbook answer: (i) $\frac12$ ⁠, $\frac13$ (ii) $\frac12$ ⁠, $\frac23$ (iii) $\frac34$ ⁠, $\frac14$ ⁠.)

Check it with numbers

Count faces directly. (i) In $\{2, 3\}$ ⁠, 1 of 2 is odd: 0.5; in $\{1, 3, 5\}$ ⁠, 1 of 3 is in $F$ ⁠: $\frac13 \approx 0.3333$ ⁠. (ii) In $\{2, 3, 4, 5\}$ ⁠, 2 of 4 are odd: 0.5; in $\{1, 3, 5\}$ ⁠, 2 of 3 are in $G$ ⁠: $\frac23 \approx 0.6667$ ⁠. (iii) In $\{2, 3, 4, 5\}$ ⁠, 3 of 4 are in $\{1, 2, 3, 5\}$ ⁠: 0.75; and 1 of 4 is 3: 0.25.

A curious detail: $P(E) = \frac36 = 0.5$ ⁠, and also $P(E|F) = P(E|G) = 0.5$ ⁠. Knowing $F$ or $G$ leaves the chance of an odd number unchanged; Section 13.4 calls such events independent.

Common mistakes

Swapping the order. $P(E|F)$ divides by $P(F)$ ⁠, $P(F|E)$ by $P(E)$ ⁠. They are different here: $\frac12$ and $\frac13$ ⁠.
Forgetting to intersect with $G$ in (iii). $P(E\cup F) = \frac46$ ⁠, but face 1 is outside $G$ ⁠, so the top is $P((E\cup F)\cap G) = \frac36$ ⁠.
Writing face 3 twice in $E\cup F$ ⁠. A union lists each face once, so $n(E\cup F) = 4$ ⁠, not 5.
Dividing by 6. $\frac{n(E\cap F)}{6}$ is $P(E\cap F)$ ⁠, not a conditional probability.

What to take away

$P(A|B) = \dfrac{n(A\cap B)}{n(B)}$ for equally likely outcomes: the event after the bar is the new sample space.
The definition works for any event before the bar; for a union, Property 2 gives the same answer as direct counting.

Every fact used, and where it comes from

Fact	Used in	Source
A fair die: 6 equally likely faces, $P(A) = \frac{n(A)}{6}$	All parts	Class XI, Probability (Basics: classical probability)
Union and intersection of sets	Part (iii)	Class XI, Sets (Basics: events as sets)
$P(A\|B) = \frac{P(A\cap B)}{P(B)}$ ⁠, $P(B) \ne 0$	All parts	Section 13.2, Definition 1 (Lesson: conditional probability)
$P((A\cup B)\|F)$ $= P(A\|F) + P(B\|F)$ $-\, P((A\cap B)\|F)$	Part (iii), check	Section 13.2.1, Property 2

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Exercise 13.1 · Q11: Die with E={1,3,5}E = \{1,3,5\}E={1,3,5}⁠, F={2,3}F = \{2,3\}F={2,3}⁠, G={2,3,4,5}G = \{2,3,4,5\}G={2,3,4,5}⁠: six conditional probabilities

Basics you need

Part (i): P(E∣F)P(E|F)P(E∣F) and P(F∣E)P(F|E)P(F∣E)

The probabilities of EEE⁠, FFF⁠, GGG

Intersect and divide both ways

Part (ii): P(E∣G)P(E|G)P(E∣G) and P(G∣E)P(G|E)P(G∣E)

Intersect and divide both ways

Part (iii): P((E∪F)∣G)P((E\cup F)|G)P((E∪F)∣G) and P((E∩F)∣G)P((E\cap F)|G)P((E∩F)∣G)

The union and the intersection, cut down to GGG

Divide by P(G)P(G)P(G)

Check the first answer with Property 2

Every fact used, and where it comes from

Exercise 13.1 · Q11: Die with $E = \{1,3,5\}$ ⁠, $F = \{2,3\}$ ⁠, $G = \{2,3,4,5\}$ ⁠: six conditional probabilities

Part (i): $P(E|F)$ and $P(F|E)$

The probabilities of $E$ ⁠, $F$ ⁠, $G$

Part (ii): $P(E|G)$ and $P(G|E)$

Part (iii): $P((E\cup F)|G)$ and $P((E\cap F)|G)$

The union and the intersection, cut down to $G$

Divide by $P(G)$