Chapter 13 · Probability

Exercise 13.2 · Q5: Die with red and green numbers: are 'even' and 'red' independent?

Exercise 13.2Independent events · Section 13.4

The question

A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let $A$ be the event, ‘the number is even,’ and $B$ be the event, ‘the number is red’. Are $A$ and $B$ independent?

Exam answer

$S = \{1, 2, 3, 4, 5, 6\}$ ⁠, all outcomes equally likely.

A = \{2, 4, 6\},\quad B = \{1, 2, 3\}

A\cap B = \{2\}

P(A) = \frac{3}{6} = \frac{1}{2}

P(B) = \frac{3}{6} = \frac{1}{2}

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

P(A)\,P(B) = \frac{1}{2}\times\frac{1}{2} = \frac{1}{4}

Since $\frac{1}{6} \ne \frac{1}{4}$ ⁠, $P(A\cap B) \ne P(A)\,P(B)$ ⁠.

$\therefore\ A$ and $B$ are not independent.

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

The one idea this problem is built on

Independence is decided by numbers, not by how the events sound. Compute $P(A\cap B)$ and $P(A)\,P(B)$ ⁠: equal means independent, unequal means dependent. Here only one of the three red numbers is even, so learning "red" makes "even" less likely, and the test fails.

Given, to find, and the plan

Given: A die with 1, 2, 3 marked in red and 4, 5, 6 in green is tossed. $A$ ⁠: the number is even. $B$ ⁠: the number is red.
To find: Whether $A$ and $B$ are independent.
Plan: (1) Write the sample space and the events as sets. (2) Find $P(A)$ ⁠, $P(B)$ ⁠, $P(A\cap B)$ ⁠. (3) Compare $P(A\cap B)$ with $P(A)\,P(B)$ ⁠. (4) Explain the result with $P(A|B)$ ⁠.

Basics you need

1Events are sets of outcomes

"Even" is $\{2, 4, 6\}$ ⁠; "red" is the set of numbers printed in red. Basics: events as sets

2Classical probability

A die is assumed fair unless stated otherwise (the book takes equally likely outcomes throughout the chapter), so $P(E) = \frac{n(E)}{6}$ ⁠. Basics: P(E) = n(E)/n(S)

3The independence test

Independent if $P(A\cap B) = P(A)\,P(B)$ (Definition 3, Section 13.4); dependent, that is not independent, if the two sides differ (Remarks (i)). Lesson: independent events

Solution

Sample space and events
$\begin{aligned} S &= \{1, 2, 3, 4, 5, 6\}, \\ A &= \{2, 4, 6\}, \\ B &= \{1, 2, 3\}, \\ A\cap B &= \{2\}. \end{aligned}$

The six faces. Red faces form $B = \{1, 2, 3\}$ ⁠; the ringed even faces form $A = \{2, 4, 6\}$ ⁠. Only face 2 is both, so $A\cap B = \{2\}$ ⁠.

The colours tell us which numbers are red: exactly 1, 2 and 3. The even numbers are 2, 4 and 6. Only 2 is in both lists.

An event is the set of outcomes for which it happens. Basics: events as sets
Find the three probabilities
$\begin{aligned} P(A) &= \frac{3}{6} = \frac{1}{2}, \\ P(B) &= \frac{3}{6} = \frac{1}{2}, \\ P(A\cap B) &= \frac{1}{6}. \end{aligned}$
Each of the six faces is equally likely, so every probability is a count out of 6: three even faces, three red faces, one face that is both.

$P(E) = \frac{n(E)}{n(S)}$ ⁠. Basics: classical probability
Apply the independence test
$\begin{aligned} P(A)\,P(B) &= \frac{1}{2}\times\frac{1}{2} \\ &= \frac{1}{4}. \end{aligned}$
But $P(A\cap B) = \frac16$ ⁠, and $\frac16 \ne \frac14$ ⁠. So $P(A\cap B) \ne P(A)\,P(B)$ ⁠, and $A$ and $B$ are not independent.

Definition 3 needs the two sides to be equal. They differ ( $\frac16 = \frac{2}{12}$ while $\frac14 = \frac{3}{12}$ ⁠), so by Remarks (i) the events are dependent.

Definition 3 and Remarks (i), Section 13.4. Lesson: independent events
See why: knowing "red" changes the chance of "even"

By the definition of conditional probability,
$\begin{aligned} P(A|B) &= \frac{P(A\cap B)}{P(B)} \\ &= \frac{1}{6}\div\frac{1}{2} = \frac{1}{3}. \end{aligned}$
Without any information, $P(A) = \frac12$ ⁠. Once we know the number is red, the chance that it is even drops to $\frac13$ ⁠: among the red numbers 1, 2, 3 only one is even.

This is Definition 2 at work: independence would need $P(A|B) = P(A)$ ⁠. Here $\frac13 \ne \frac12$ ⁠, so the occurrence of $B$ does affect $A$ ⁠, which is what "not independent" means.

Definition 1, Section 13.2, and Definition 2, Section 13.4. Lesson: conditional probability

Result

$P(A\cap B) = \frac16$ but $P(A)\,P(B) = \frac14$ ⁠. Since $P(A\cap B) \ne P(A)\,P(B)$ ⁠, $A$ and $B$ are not independent.

(Textbook answer: A and B are not independent.)

Check it with numbers

In decimals, $P(A\cap B) = \frac16 \approx 0.1667$ and $P(A)\,P(B) = 0.5\times 0.5 = 0.25$ ⁠: different. Also $P(B|A) = \frac{1/6}{1/2} = \frac13$ ⁠: among the even numbers 2, 4, 6 only 2 is red, again different from $P(B) = \frac12$ ⁠.

Common mistakes

Deciding from the wording. "Even" and "red" sound unrelated, but the colouring links them. Only the product test settles the question.
Thinking "not independent" means "mutually exclusive". $A$ and $B$ share the outcome 2, so they are not mutually exclusive either. These are two different properties.
Stopping at the numbers. End with the conclusion in words: " $A$ and $B$ are not independent".

What to take away

The test $P(A\cap B) = P(A)\,P(B)$ decides independence; any difference means dependent.
Two events on the same throw of one die can be dependent even if they sound unrelated.
$P(A|B)$ shows the reason: here knowing "red" lowers the chance of "even" from $\frac12$ to $\frac13$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Outcomes equally likely unless stated otherwise	Step 2	Section 13.1 (last sentence)
$P(E) = \frac{n(E)}{n(S)}$	Step 2	Class XI, Probability (recalled in Section 13.1)
Independent $\iff P(A\cap B) = P(A)\,P(B)$ ⁠; dependent otherwise	Step 3	Definition 3 and Remarks (i), Section 13.4 (Lesson: independent events)
$P(A\|B) = \frac{P(A\cap B)}{P(B)}$ ⁠; independence as $P(A\|B) = P(A)$	Step 4	Definition 1, Section 13.2; Definition 2, Section 13.4
Same kind of test on one die	Step 3	Example 10, Section 13.4 (Lesson: Examples 10 to 14)

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Exercise 13.2 · Q5: Die with red and green numbers: are 'even' and 'red' independent?

Basics you need

Solution

Sample space and events

Find the three probabilities

Apply the independence test

See why: knowing "red" changes the chance of "even"

Every fact used, and where it comes from