Chapter 13 · Probability

Example 6: Two dice with sum 6: a 4 appears at least once

ExamplesConditional probability · Section 13.2

The question

A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Exam answer

The sample space has $6\times 6 = 36$ equally likely outcomes.

Let $E$ ⁠: number 4 appears at least once, and $F$ ⁠: the sum of the numbers appearing is 6.

\begin{aligned} E = \{&(4,1), (4,2), (4,3), \\ &(4,4), (4,5), (4,6), \\ &(1,4), (2,4), (3,4), \\ &(5,4), (6,4)\} \end{aligned}

\begin{aligned} F = \{&(1,5), (2,4), (3,3), \\ &(4,2), (5,1)\} \end{aligned}

E\cap F = \{(2,4), (4,2)\}

P(F) = \frac{5}{36},\quad P(E\cap F) = \frac{2}{36}

By the definition of conditional probability,

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \dfrac{\;\frac{2}{36}\;}{\frac{5}{36}} = \frac{2}{36}\times\frac{36}{5} \end{aligned}

$\therefore\ P(E|F) = \dfrac{2}{5}$ ⁠.

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

The one idea this problem is built on

Knowing the sum is 6 leaves only five equally likely outcomes. They are $(1,5), (2,4), (3,3), (4,2), (5,1)$ ⁠, and a 4 appears in two of them. So the answer is $\frac{2}{5}$ ⁠. The definition of conditional probability gives the same:

P(E|F) = \frac{P(E \cap F)}{P(F)}.

Given, to find, and the plan

Given: A fair die thrown twice; the sum of the two numbers is 6.
To find: The probability that the number 4 has appeared at least once, given the sum is 6.
Plan: Name $E$ (a 4 at least once) and $F$ (sum 6); list them as ordered pairs; find $P(F)$ and $P(E \cap F)$ ⁠; divide.

Basics you need

1Two throws as ordered pairs

$(a, b)$ means $a$ on the first throw and $b$ on the second; there are $6 \times 6 = 36$ pairs, all equally likely. Basics: multiply the choices

2"At least once" means one or both

A 4 on the first throw, on the second throw, or on both. Basics: do not count the overlap twice

3Conditional probability

$P(E|F) = \frac{P(E \cap F)}{P(F)}$ when $P(F) \ne 0$ (Definition 1, Section 13.2). Lesson: conditional probability

Solution

Name the events

Let $E$ ⁠: number 4 appears at least once, and $F$ ⁠: the sum of the numbers appearing is 6. We need $P(E|F)$ ⁠.

The sum is what we are told ("observed to be 6"), so $F$ goes after the bar.

Lesson: conditional probability
List the outcomes of each event
$\begin{aligned} E = \{&(4,1), (4,2), (4,3), (4,4), \\ &(4,5), (4,6), (1,4), (2,4), \\ &(3,4), (5,4), (6,4)\}, \end{aligned}$ $F = \{(1,5), (2,4), (3,3), (4,2), (5,1)\},$ $E \cap F = \{(2,4), (4,2)\}.$
$E$ has 6 pairs with a 4 first and 6 pairs with a 4 second, but $(4,4)$ is in both groups, so $n(E) = 6 + 6 - 1 = 11$ ⁠. For $F$ ⁠, the first number can be 1 to 5 and the second is then 6 minus it. Of these five pairs, only $(2,4)$ and $(4,2)$ contain a 4.

Basics: do not count the overlap twice
Find the probabilities
$P(E) = \frac{11}{36}, \quad P(F) = \frac{5}{36},$ $P(E \cap F) = \frac{2}{36}.$
All 36 pairs are equally likely, so each probability is a count over 36. $P(E)$ is not needed for $P(E|F)$ ⁠; it is used in the check.

Basics: the classical definition
Divide, using the definition
$\begin{aligned} P(E|F) &= \frac{P(E \cap F)}{P(F)} = \dfrac{\;\frac{2}{36}\;}{\frac{5}{36}} \\ &= \frac{2}{36} \times \frac{36}{5} = \frac{2}{5}. \end{aligned}$

All 36 outcomes (first throw across, second throw down). The shaded diagonal is $F$ ⁠, sum 6; the dots mark $E$ ⁠, a 4 at least once. Two shaded cells carry a dot, so $P(E|F) = \frac{2}{5}$ ⁠.

$P(F) = \frac{5}{36} \ne 0$ ⁠. The 36s cancel: 2 favourable pairs among the 5 pairs with sum 6.

Definition 1, Section 13.2. Basics: dividing by a fraction

Result

$P(E|F) = \dfrac{2}{5}$ ⁠.

(Textbook answer: $\frac{2}{5}$ ⁠.)

Check it with numbers

Counting inside $F$ ⁠. $\frac{n(E \cap F)}{n(F)} = \frac{2}{5}$ ⁠, the same value.

Decimals. $\frac{2}{5} = 0.4$ ⁠, while $P(E) = \frac{11}{36} \approx 0.3056$ ⁠. Knowing the sum is 6 rules out 9 of the 11 pairs with a 4, but it rules out even more of the pairs without a 4 (22 of 25), so the share of pairs containing a 4 rises.

The other way round. $P(F|E) = \frac{n(E \cap F)}{n(E)} = \frac{2}{11}$ ⁠, a different number: the order of the events matters.

Common mistakes

Counting $(4,4)$ twice and getting 12 pairs in $E$ instead of 11.
Ignoring the order of the throws. Lumping $(2,4)$ with $(4,2)$ and $(1,5)$ with $(5,1)$ leaves three "outcomes" with sum 6 and the wrong answer $\frac{1}{3}$ ⁠; those three are not equally likely, because $(3,3)$ happens in only one way.
Dividing by 36. $\frac{2}{36}$ is $P(E \cap F)$ ⁠; given the sum is 6, only 5 outcomes remain.

What to take away

For two dice, list ordered pairs; a 6 by 6 grid shows every event at once.
With equally likely outcomes, $P(E|F) = \frac{n(E \cap F)}{n(F)}$ ⁠: count inside the condition.
$P(E|F)$ and $P(F|E)$ are usually different.

Every fact used, and where it comes from

Fact	Used in	Source
$6 \times 6 = 36$ ordered pairs	Step 2	Counting principle, Class XI (Basics: multiply the choices)
$n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$	Step 2	Class XI, Sets (Basics: do not count the overlap twice)
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Step 3	Class XI, Probability (Basics: the classical definition)
$P(E\|F) = \frac{P(E \cap F)}{P(F)}$ ⁠, $P(F) \ne 0$	Step 4	Definition 1, Section 13.2 (Lesson: conditional probability)
$P(E\|F) = \frac{n(E \cap F)}{n(F)}$	Check	Section 13.2, equation (1)

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Example 6: Two dice with sum 6: a 4 appears at least once

Basics you need

Solution

Name the events

List the outcomes of each event

Find the probabilities

Divide, using the definition

Every fact used, and where it comes from