Chapter 13 · Probability

Exercise 13.1 · Q8: Die thrown three times: 4 on the third, given 6 then 5

Exercise 13.1Conditional probability · Section 13.2

The question

Determine $P(E|F)$ ⁠.

A die is thrown three times,

$E$ : 4 appears on the third toss, $F$ : 6 and 5 appears respectively on first two tosses

Exam answer

A die thrown three times has $6^3 = 216$ equally likely outcomes $(a, b, c)$ ⁠.

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

and $E\cap F = \{(6, 5, 4)\}$ ⁠.

P(F) = \frac{6}{216}

P(E\cap F) = \frac{1}{216}

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \frac{1}{216}\div\frac{6}{216} \end{aligned}

$\therefore\ P(E|F) = \dfrac{1}{6}$ ⁠.

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

The one idea this problem is built on

Knowing $F$ fixes the first two throws as 6 and 5, so only the third throw is still uncertain. The new sample space is just the 6 possible third throws, and $E$ needs one particular value, 4. So the answer is 1 out of 6.

Given, to find, and the plan

Given: A die thrown three times. Write an outcome as $(a, b, c)$ ⁠: the first, second and third throws. $E$ ⁠: the third throw is 4. $F$ ⁠: the first throw is 6 and the second is 5. We take the die to be fair, so all outcomes are equally likely.
To find: $P(E|F)$ ⁠.
Plan: (1) Count the sample space. (2) List $F$ and $E\cap F$ ⁠. (3) Find $P(F)$ ⁠, $P(E\cap F)$ and divide.

Basics you need

1Counting ordered outcomes

Three throws with 6 results each give $6\times 6\times 6 = 216$ outcomes. Basics: multiply the choices

2Classical probability

$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes. Basics: P(E) = n(E)/n(S)

3Conditional probability

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

Solution

Count the sample space
$n(S) = 6\times 6\times 6 = 6^3 = 216.$
Each throw can show any of 1 to 6, independently of the others, and the order matters: $(6, 5, 4)$ and $(4, 5, 6)$ are different outcomes. A fair die makes all 216 equally likely, so each has probability $\frac{1}{216}$ ⁠.

Fundamental principle of counting (Class XI). Basics: multiply the choices
List $F$ and $E\cap F$
$\begin{aligned} F = \{&(6, 5, 1), (6, 5, 2), (6, 5, 3), \\ &(6, 5, 4), (6, 5, 5), (6, 5, 6)\}, \end{aligned}$ $E\cap F = \{(6, 5, 4)\}.$
In $F$ the first two places are fixed at 6 and 5; the third can be any of 6 values, so $n(F) = 6$ ⁠. Of these, only $(6, 5, 4)$ has a 4 in the third place. (On its own, $E$ has $6\times 6 = 36$ outcomes, any first and second throw with a 4 third, but only one of them lies in $F$ ⁠.)

"And" is the intersection of the two sets. Basics: events as sets

The 6 outcomes of $F$ ⁠. Once $F$ is known, these are the only outcomes left, and just one of them, $(6, 5, 4)$ ⁠, is in $E$ ⁠.
Find the probabilities and divide
$P(F) = \frac{6}{216},\qquad P(E\cap F) = \frac{1}{216},$ $\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} = \frac{1/216}{6/216} \\ &= \frac{1}{216}\times\frac{216}{6} = \frac{1}{6}. \end{aligned}$
Each probability is a count over 216, and $P(F) = \frac{6}{216} = \frac{1}{36} \ne 0$ ⁠, so the definition applies. The 216s cancel, leaving $\frac{n(E\cap F)}{n(F)} = \frac16$ ⁠.

Definition 1, Section 13.2. Lesson: conditional probability

Result

$P(E|F) = \dfrac{1}{6}$ ⁠.

(Textbook answer: $\frac16$ ⁠.)

Check it with numbers

Without any condition, $P(E) = \frac{36}{216} = \frac16 \approx 0.1667$ ⁠, the same as $P(E|F) = \frac16 \approx 0.1667$ ⁠. That is what we expect: the first two throws have no effect on the third, so knowing them does not change the chance of a 4. (Section 13.4 calls such events independent.)

The same question, with the events called $A$ and $B$ ⁠, is worked in the book as Example 5 of Section 13.2, with the same answer $\frac16$ ⁠.

Common mistakes

Answering $\frac{1}{216}$ ⁠. That is $P(E\cap F)$ ⁠, the chance of the whole sequence 6, 5, 4. The question asks for $E$ given $F$ ⁠, so divide by $P(F)$ ⁠.
Taking $n(S) = 18$ or 36. Three throws give $6^3 = 216$ ordered outcomes.
Thinking the earlier throws make a 4 more or less likely. A die has no memory; the conditional probability is still $\frac16$ ⁠.

What to take away

When $F$ fixes some of the stages, the new sample space consists of the remaining stages only.
For separate throws of a die, conditioning on earlier throws leaves the chance for a later throw unchanged.

Every fact used, and where it comes from

Fact	Used in	Source
$n(S) = 6^3 = 216$ for three throws	Step 1	Class XI, Permutations and Combinations (Basics: multiply the choices)
$P(E) = \frac{n(E)}{n(S)}$	Step 3, Check	Class XI, Probability (Basics: classical probability)
$P(E\|F) = \frac{P(E\cap F)}{P(F)}$ ⁠, $P(F) \ne 0$	Step 3	Definition 1, Section 13.2 (Lesson: conditional probability)
The same problem worked as an example	Check	Example 5, Section 13.2 (Lesson: Examples 1 to 7)

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Exercise 13.1 · Q8: Die thrown three times: 4 on the third, given 6 then 5

Basics you need

Solution

Count the sample space

List FFF and E∩FE\cap FE∩F

Find the probabilities and divide

Every fact used, and where it comes from

List $F$ and $E\cap F$