Chapter 13 · Probability

Exercise 13.1 · Q10: Black and red dice: sum above 9 given black 5; sum 8 given red below 4

Exercise 13.1Conditional probability · Section 13.2

The question

A black and a red dice are rolled.

(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Exam answer

Write an outcome as $(b, r)$ ⁠: $b$ on the black die, $r$ on the red die. There are $6\times 6 = 36$ equally likely outcomes.

(a) Let $E$ ⁠: the sum is greater than 9, and $F$ ⁠: the black die shows 5.

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

so $E\cap F = \{(5, 5), (5, 6)\}$ ⁠.

By the definition of conditional probability,

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \frac{2}{36}\div\frac{6}{36} \end{aligned}

$P(E|F) = \frac{1}{3}$ ⁠.

(b) Let $E$ ⁠: the sum is 8, and $F$ ⁠: the red die shows 1, 2 or 3.

n(F) = 6\times 3 = 18

E\cap F = \{(6, 2), (5, 3)\}

By the definition of conditional probability,

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \frac{2}{36}\div\frac{18}{36} \end{aligned}

$\therefore\ P(E|F) = \dfrac{1}{9}$ ⁠.

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

The one idea this problem is built on

Two dice give 36 equally likely ordered outcomes. "Given that" shrinks the world to the outcomes of the given event, and we count, inside that smaller world, how many outcomes also have the required sum. So in each part $P(E|F) = \dfrac{n(E\cap F)}{n(F)}$ ⁠.

Given, to find, and the plan

Given: A black die and a red die, rolled together. Write an outcome as $(b, r)$ ⁠, with $b$ the number on the black die and $r$ the number on the red die. The dice are fair, so the $6\times 6 = 36$ outcomes are equally likely, each with probability $\frac{1}{36}$ ⁠.
To find: (a) $P(\text{sum} > 9 \mid b = 5)$ ⁠. (b) $P(\text{sum} = 8 \mid r < 4)$ ⁠.
Plan: In each part: (1) name the events $E$ (about the sum) and $F$ (the given information); (2) list $F$ and $E\cap F$ ⁠; (3) find $P(F)$ ⁠, $P(E\cap F)$ and divide.

Basics you need

1The 36 outcomes of two dice

The dice have different colours, so $(5, 6)$ (black 5, red 6) and $(6, 5)$ (black 6, red 5) are different outcomes: $6\times 6 = 36$ in all. Basics: multiply the choices

2Classical probability

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

3Conditional probability

$P(E|F) = \dfrac{P(E\cap F)}{P(F)}$ ⁠, provided $P(F) \ne 0$ (Definition 1, Section 13.2). For equally likely outcomes this is $\dfrac{n(E\cap F)}{n(F)}$ ⁠. Lesson: conditional probability

4"Greater than" and "less than"

"Greater than 9" means 10, 11 or 12, not 9. "Less than 4" means 1, 2 or 3, not 4. Basics: events as sets

Part (a): sum greater than 9, given the black die shows 5

Name the events and list $F$

Let $E$ ⁠: the sum is greater than 9, and $F$ ⁠: the black die shows 5. Then
$\begin{aligned} F = \{&(5, 1), (5, 2), (5, 3), \\ &(5, 4), (5, 5), (5, 6)\}, \end{aligned}$
so $n(F) = 6$ and $P(F) = \frac{6}{36}$ ⁠.

The black die is fixed at 5 and the red die can still show any of its 6 numbers. These 6 outcomes are the new, smaller sample space.

An event is the set of outcomes that make it happen. Basics: events as sets

All 36 outcomes, black die across and red die down; each cell shows the sum $b + r$ ⁠. Gold and blue cells form the given event $F$ ⁠; the blue ones are also in $E$ ⁠. (a) 2 of the 6 cells in the column "black 5" have a sum above 9. (b) 2 of the 18 cells in the rows "red 1, 2, 3" have sum 8.
Find $E\cap F$

With $b = 5$ ⁠, the sum is $5 + r$ ⁠, and $5 + r > 9$ means $r > 4$ ⁠, so $r = 5$ or $6$ ⁠:
$E\cap F = \{(5, 5), (5, 6)\},$ $P(E\cap F) = \frac{2}{36}.$
The sums in $F$ are 6, 7, 8, 9, 10, 11. Only 10 and 11 are greater than 9. A sum of exactly 9, $(5, 4)$ ⁠, does not count, because "greater than 9" excludes 9.

"And" is the intersection of the two sets. Basics: events as sets
Divide
$\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} = \frac{2/36}{6/36} \\ &= \frac{2}{36}\times\frac{36}{6} = \frac{1}{3}. \end{aligned}$
$P(F) = \frac{6}{36} \ne 0$ ⁠, so the definition applies. The 36s cancel, leaving $\frac{n(E\cap F)}{n(F)} = \frac26 = \frac13$ ⁠.

Definition 1, Section 13.2. Lesson: conditional probability

Part (b): sum 8, given the red die shows less than 4

Name the events and count $F$

Let $E$ ⁠: the sum is 8, and $F$ ⁠: the red die shows 1, 2 or 3. The black die is free, so
$n(F) = 6\times 3 = 18,\qquad P(F) = \frac{18}{36}.$
For each of the 3 allowed red numbers, the black die can show any of 6 numbers. "Less than 4" does not include 4, so the red die has 3 choices, not 4.

Multiply the choices. Basics: multiply the choices
Find $E\cap F$

The outcomes with sum 8 are $(2, 6)$ ⁠, $(3, 5)$ ⁠, $(4, 4)$ ⁠, $(5, 3)$ ⁠, $(6, 2)$ ⁠. Keep only those with $r < 4$ ⁠:
$E\cap F = \{(6, 2), (5, 3)\},$ $P(E\cap F) = \frac{2}{36}.$
Going the other way gives the same list: with $r = 1, 2, 3$ the black die must show $b = 8 - r = 7, 6, 5$ ⁠. A die cannot show 7, so only $r = 2$ ( $b = 6$ ⁠) and $r = 3$ ( $b = 5$ ⁠) work.

"And" is the intersection of the two sets. Basics: events as sets
Divide
$\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} = \frac{2/36}{18/36} \\ &= \frac{2}{36}\times\frac{36}{18} = \frac{1}{9}. \end{aligned}$
$P(F) = \frac{18}{36} \ne 0$ ⁠. The 36s cancel: 2 favourable outcomes among the 18 outcomes of $F$ ⁠, and $\frac{2}{18} = \frac19$ ⁠.

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

Result

(a) $P(E|F) = \frac{1}{3}$ (b) $P(E|F) = \frac{1}{9}$ ⁠.

(Textbook answer: (a) $\frac13$ ⁠, (b) $\frac19$ ⁠.)

Check it with numbers

(a) Think of the red die alone: once the black die shows 5, the sum exceeds 9 exactly when the red die shows 5 or 6, a chance of $\frac26 \approx 0.3333$ ⁠. Without the condition, a sum above 9 happens in 6 of the 36 outcomes, $\frac{6}{36} \approx 0.1667$ ⁠: knowing the black die is 5 doubles the chance.

(b) Without the condition, a sum of 8 happens in 5 of the 36 outcomes, $\frac{5}{36} \approx 0.1389$ ⁠. Given $r < 4$ it falls to $\frac{2}{18} \approx 0.1111$ ⁠: a small red number makes a sum of 8 harder to reach.

Common mistakes

Dividing by 36. $\frac{2}{36}$ is $P(E\cap F)$ ⁠. The question gives information, so divide by $P(F)$ as well.
Counting a sum of 9 in (a). "Greater than 9" means 10, 11, 12; the outcome $(5, 4)$ has sum 9 and is not in $E$ ⁠.
Counting $(4, 4)$ in (b). Its red die shows 4, which is not less than 4, so it is outside $F$ ⁠.
Mixing up the dice. In (a) the black die is fixed; in (b) the red die is restricted. Writing every outcome as $(b, r)$ keeps them apart.

What to take away

With two dice, list outcomes as ordered pairs; the condition picks out a column, a row or a band of the 6 by 6 grid.
For equally likely outcomes, $P(E|F) = \dfrac{n(E\cap F)}{n(F)}$ ⁠: count only inside $F$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Two dice give $6\times 6 = 36$ equally likely ordered outcomes	Given	Class XI, Permutations and Combinations (Basics: multiply the choices)
$P(E) = \frac{n(E)}{n(S)}$	Both parts	Class XI, Probability (Basics: classical probability)
$P(E\|F) = \frac{P(E\cap F)}{P(F)}$ ⁠, $P(F) \ne 0$	Both parts, last step	Section 13.2, Definition 1 (Lesson: conditional probability)

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Exercise 13.1 · Q10: Black and red dice: sum above 9 given black 5; sum 8 given red below 4

Basics you need

Part (a): sum greater than 9, given the black die shows 5

Name the events and list FFF

Find E∩FE\cap FE∩F

Divide

Part (b): sum 8, given the red die shows less than 4

Name the events and count FFF

Find E∩FE\cap FE∩F

Divide

Every fact used, and where it comes from

Name the events and list $F$

Find $E\cap F$

Name the events and count $F$

Find $E\cap F$