Chapter 13 · Probability

Example 5: Die thrown three times: 4 on the third, given 6 then 5

ExamplesConditional probability · Section 13.2

The question

A die is thrown three times. Events A and B are defined as below:

A : 4 on the third throw

B : 6 on the first and 5 on the second throw

Find the probability of A given that B has already occurred.

Exam answer

The sample space has $6\times 6\times 6 = 216$ equally likely outcomes $(x, y, z)$ ⁠.

$B$ ⁠: 6 on the first and 5 on the second throw.

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

$A$ ⁠: 4 on the third throw. Then

A\cap B = \{(6,5,4)\}

P(B) = \frac{6}{216},\quad P(A\cap B) = \frac{1}{216}

By the definition of conditional probability,

\begin{aligned} P(A|B) &= \frac{P(A\cap B)}{P(B)} \\ &= \dfrac{\;\frac{1}{216}\;}{\frac{6}{216}} = \frac{1}{216}\times\frac{216}{6} \end{aligned}

$\therefore\ P(A|B) = \dfrac{1}{6}$ ⁠.

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

The one idea this problem is built on

Once the first two throws are known, only the third throw is still open. Knowing $B$ leaves six equally likely outcomes, $(6,5,1)$ to $(6,5,6)$ ⁠, and exactly one of them has a 4 on the third throw. The definition of conditional probability gives the same $\frac{1}{6}$ ⁠:

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

Given, to find, and the plan

Given: A fair die thrown three times; $A$ ⁠: 4 on the third throw; $B$ ⁠: 6 on the first and 5 on the second throw.
To find: $P(A|B)$ ⁠.
Plan: Count the sample space; list $B$ and $A \cap B$ (and count $A$ ⁠); find $P(B)$ and $P(A \cap B)$ ⁠; divide. Then compare the answer with $P(A)$ ⁠.

Basics you need

1Counting ordered outcomes

Three throws with 6 faces each give $6 \times 6 \times 6$ ordered triples. Basics: multiply the choices

2Equally likely outcomes

$P(E) = \frac{n(E)}{n(S)}$ ⁠. Basics: the classical definition

3Conditional probability

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

Solution

Count the sample space

Write an outcome as $(x, y, z)$ ⁠: the numbers on the first, second and third throws. Each can be any of 1 to 6, so
$n(S) = 6 \times 6 \times 6 = 216.$
A fair die makes each face equally likely on every throw, and the throws do not affect each other, so all 216 ordered triples are equally likely. Order matters: $(6,5,4)$ and $(4,5,6)$ are different results.

Basics: multiply the choices
List the events

$A$ fixes the third number as 4 and leaves the first two free:
$A = \{(1,1,4), (1,2,4), \ldots, (6,6,4)\},$ $n(A) = 6 \times 6 = 36.$
$B$ fixes the first two numbers and leaves the third free:
$\begin{aligned} B = \{&(6,5,1), (6,5,2), (6,5,3), \\ &(6,5,4), (6,5,5), (6,5,6)\}, \end{aligned}$ $A \cap B = \{(6,5,4)\}.$
The only triple that has 6 first, 5 second and 4 third is $(6,5,4)$ ⁠. We count $A$ instead of writing all 36 triples; $P(A)$ is used in the check.

Basics: events as sets
Find the probabilities
$P(B) = \frac{6}{216}, \quad P(A \cap B) = \frac{1}{216}.$
Equally likely outcomes: count over 216.

Basics: the classical definition
Divide, using the definition
$\begin{aligned} P(A|B) &= \frac{P(A \cap B)}{P(B)} = \dfrac{\;\frac{1}{216}\;}{\frac{6}{216}} \\ &= \frac{1}{216} \times \frac{216}{6} = \frac{1}{6}. \end{aligned}$

Given $B$ ⁠, only these six triples are possible, and they are equally likely. One of them, $(6,5,4)$ ⁠, has a 4 on the third throw.

$P(B) = \frac{6}{216} \ne 0$ ⁠. The 216s cancel, leaving 1 outcome out of the 6 in $B$ ⁠.

Definition 1, Section 13.2. Lesson: conditional probability

Result

$P(A|B) = \dfrac{1}{6}$ ⁠.

(Textbook answer: $\frac{1}{6}$ ⁠.)

Check it with numbers

Counting inside $B$ ⁠. $\frac{n(A \cap B)}{n(B)} = \frac{1}{6}$ ⁠.

Compare with $P(A)$ ⁠. $P(A) = \frac{36}{216} = \frac{1}{6}$ as well. Knowing the first two throws does not change the chance of a 4 on the third: $P(A|B) = P(A)$ ⁠, so $A$ and $B$ are independent events (Definition 2, Section 13.4). Indeed $P(A)\,P(B) = \frac{1}{6} \times \frac{6}{216} = \frac{1}{216}$ ⁠, which equals $P(A \cap B)$ ⁠.

Decimals. $\frac{1}{6} \approx 0.1667$ ⁠, while $P(A \cap B) = \frac{1}{216} \approx 0.0046$ is the much smaller chance of the exact sequence 6, 5, 4 before anything is known.

Common mistakes

Answering $\frac{1}{216}$ ⁠. That is $P(A \cap B)$ ⁠, the chance of 6, 5, 4 from the start; the first two throws are already known.
Thinking that the earlier throws make a 4 more or less likely. The die has no memory; the answer equals $P(A) = \frac{1}{6}$ ⁠.
Ignoring order. $B$ needs 6 on the first and 5 on the second throw, so $(5,6,4)$ is not in $B$ ⁠.

What to take away

With repeated throws, write outcomes as ordered triples; there are $6 \times 6 \times 6 = 216$ ⁠.
Fixing some throws leaves only the free throws to count.
When $P(A|B) = P(A)$ ⁠, the condition carries no information about $A$ ⁠: the events are independent.

Every fact used, and where it comes from

Fact	Used in	Source
$6 \times 6 \times 6 = 216$ ordered outcomes	Steps 1, 2	Counting principle, Class XI (Basics: multiply the choices)
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Step 3, Check	Class XI, Probability (Basics: the classical definition)
$P(A\|B) = \frac{P(A \cap B)}{P(B)}$ ⁠, $P(B) \ne 0$	Step 4	Definition 1, Section 13.2 (Lesson: conditional probability)
Independent: $P(A\|B) = P(A)$ ⁠; equivalently $P(A \cap B) = P(A)\,P(B)$	Check	Definitions 2 and 3, Section 13.4 (Lesson: independent events)

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Example 5: Die thrown three times: 4 on the third, given 6 then 5

Basics you need

Solution

Count the sample space

List the events

Find the probabilities

Divide, using the definition

Every fact used, and where it comes from