Chapter 13 · Probability

Example 10: Die: are 'multiple of 3' and 'even' independent?

ExamplesIndependent events · Section 13.4

The question

A die is thrown. If E is the event ‘the number appearing is a multiple of 3’ and F be the event ‘the number appearing is even’ then find whether E and F are independent?

Exam answer

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

E = \{3, 6\},\quad F = \{2, 4, 6\}

E\cap F = \{6\}

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

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

P(E\cap F) = \frac{1}{6}

$E$ and $F$ are independent if $P(E\cap F) = P(E)\,P(F)$ ⁠.

P(E)\,P(F) = \frac{1}{3}\times\frac{1}{2} = \frac{1}{6}

Thus $P(E\cap F) = P(E)\,P(F)$ ⁠.

$\therefore\ E$ and $F$ are independent events.

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

The one idea this problem is built on

Independence is a test with numbers, not a feeling. Two events are independent exactly when the probability of both equals the product of their probabilities:

P(E \cap F) = P(E)\,P(F).

Find the three numbers and compare them.

Given, to find, and the plan

Given: One throw of a fair die; $E$ ⁠: the number is a multiple of 3; $F$ ⁠: the number is even.
To find: Whether $E$ and $F$ are independent.
Plan: List $S$ ⁠, $E$ ⁠, $F$ and $E \cap F$ ⁠; find $P(E)$ ⁠, $P(F)$ ⁠, $P(E \cap F)$ ⁠; compare $P(E \cap F)$ with $P(E)\,P(F)$ ⁠; confirm with $P(E|F)$ ⁠.

Basics you need

1Events as sets

"A multiple of 3" on a die is $\{3, 6\}$ ⁠; "even" is $\{2, 4, 6\}$ ⁠. Basics: events as sets

2Equally likely outcomes

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

3Independent events

$E$ and $F$ are independent if $P(E \cap F) = P(E)\,P(F)$ (Definition 3, Section 13.4); equivalently $P(E|F) = P(E)$ when $P(F) \ne 0$ (Definition 2). Lesson: independent events

Solution

List the sample space and the events
$S = \{1, 2, 3, 4, 5, 6\},$ $E = \{3, 6\}, \quad F = \{2, 4, 6\},$ $E \cap F = \{6\}.$
The multiples of 3 from 1 to 6 are 3 and 6; the even numbers are 2, 4 and 6. The only number that is both is 6.

Basics: events as sets
Find the three probabilities
$P(E) = \frac{2}{6} = \frac{1}{3}, \quad P(F) = \frac{3}{6} = \frac{1}{2},$ $P(E \cap F) = \frac{1}{6}.$
A fair die makes the six faces equally likely, so each probability is a count over 6.

Basics: the classical definition
Apply the independence test
$\begin{aligned} P(E)\,P(F) &= \frac{1}{3} \times \frac{1}{2} \\ &= \frac{1}{6} = P(E \cap F). \end{aligned}$
So $E$ and $F$ are independent events.

The six faces. $E = \{3, 6\}$ and $F = \{2, 4, 6\}$ share only 6. One face out of six is exactly $\frac{2}{6} \times \frac{3}{6}$ ⁠.

Definition 3 asks whether the probability of "both" equals the product. Here both sides are exactly $\frac{1}{6}$ ⁠, so the test is passed.

Definition 3, Section 13.4. Lesson: independent events
See what independence means here
$\begin{aligned} P(E|F) &= \frac{P(E \cap F)}{P(F)} = \dfrac{\;\frac{1}{6}\;}{\frac{1}{2}} \\ &= \frac{1}{3} = P(E). \end{aligned}$
Among the even faces 2, 4, 6, exactly one in three is a multiple of 3, the same share as among all six faces (two in six). Learning that the number is even does not change the chance that it is a multiple of 3. That is Definition 2 of independence.

Definition 2, Section 13.4. Lesson: independent events

Result

$P(E \cap F) = \frac{1}{6} = P(E)\,P(F)$ ⁠, so $E$ and $F$ are independent events.

(Textbook answer: E and F are independent events.)

Check it with numbers

Both ways round. $P(F|E) = \frac{1/6}{1/3} = \frac{1}{2} = P(F)$ ⁠: among the multiples of 3 (3 and 6), half are even, just as half of all faces are even.

Decimals. $P(E) = \frac{1}{3} \approx 0.3333$ and $P(F) = 0.5$ ⁠; their product and $P(E \cap F)$ are both $\frac{1}{6} \approx 0.1667$ ⁠. The exact fractions show that they are equal, not merely close.

Independent is not "disjoint". $E$ and $F$ share the face 6, so they are not mutually exclusive, yet they are independent.

Common mistakes

Concluding "dependent" because $E$ and $F$ overlap. Overlapping says nothing about independence; only the test $P(E \cap F) = P(E)\,P(F)$ decides.
Confusing independent with mutually exclusive. Mutually exclusive events with non-zero probabilities are never independent (Remarks (ii), Section 13.4).
Comparing with the sum $P(E) + P(F)$ ⁠. The test uses the product.

What to take away

To test independence: find $P(E)$ ⁠, $P(F)$ ⁠, $P(E \cap F)$ and check whether $P(E \cap F) = P(E)\,P(F)$ ⁠.
End with the conclusion in words: "independent" or "not independent (dependent)".
For the same idea with a coin and a die, see Exercise 13.2, Question 4; for a die with coloured faces, Question 5.

Every fact used, and where it comes from

Fact	Used in	Source
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Step 2	Class XI, Probability (Basics: the classical definition)
Independent if $P(E \cap F) = P(E)\,P(F)$	Step 3	Definition 3, Section 13.4 (Lesson: independent events)
$P(E\|F) = \frac{P(E \cap F)}{P(F)}$	Step 4, Check	Definition 1, Section 13.2 (Lesson: conditional probability)
Independent if $P(E\|F) = P(E)$ and $P(F\|E) = P(F)$	Step 4, Check	Definition 2, Section 13.4
Independent and mutually exclusive are different ideas	Check, Mistakes	Remarks (ii), Section 13.4

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Example 10: Die: are 'multiple of 3' and 'even' independent?

Basics you need

Solution

List the sample space and the events

Find the three probabilities

Apply the independence test

See what independence means here

Every fact used, and where it comes from