Chapter 13 · Probability

Exercise 13.2 · Q4: Coin and die: are 'head' and '3' independent?

Exercise 13.2Independent events · Section 13.4

The question

A fair coin and an unbiased die are tossed. Let $A$ be the event ‘head appears on the coin’ and $B$ be the event ‘3 on the die’. Check whether $A$ and $B$ are independent events or not.

Exam answer

The sample space is

\begin{aligned} S = \{&H1, H2, H3, H4, H5, H6, \\ &T1, T2, T3, T4, T5, T6\} \end{aligned}

so $n(S) = 12$ ⁠, all outcomes equally likely.

A = \{H1, H2, H3, H4, H5, H6\}

B = \{H3, T3\},\quad A\cap B = \{H3\}

P(A) = \frac{6}{12} = \frac{1}{2}

P(B) = \frac{2}{12} = \frac{1}{6}

P(A\cap B) = \frac{1}{12}

P(A)\,P(B) = \frac{1}{2}\times\frac{1}{6} = \frac{1}{12}

Thus $P(A\cap B) = P(A)\,P(B)$ ⁠.

$\therefore\ A$ and $B$ are independent events.

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

The one idea this problem is built on

To check independence, compute both sides of the test $P(A\cap B) = P(A)\,P(B)$ and compare. If they are equal, the events are independent; if not, they are not. Here the coin cannot influence the die, and the numbers confirm it.

Given, to find, and the plan

Given: A fair coin and an unbiased die are tossed together. $A$ ⁠: head on the coin. $B$ ⁠: 3 on the die.
To find: Whether $A$ and $B$ are independent.
Plan: (1) List the 12 equally likely outcomes. (2) Write $A$ ⁠, $B$ and $A\cap B$ as sets. (3) Find $P(A)$ ⁠, $P(B)$ ⁠, $P(A\cap B)$ ⁠. (4) Compare $P(A\cap B)$ with $P(A)\,P(B)$ and conclude in words.

Basics you need

1Sample space of a combined experiment

Two choices for the coin and six for the die give $2\times 6 = 12$ outcomes. A fair coin and an unbiased die make all 12 equally likely. Basics: multiply the choices

2Events are sets of outcomes

"Head appears" is the set of all outcomes starting with H, whatever the die shows. Basics: events as sets

3The independence test

$A$ and $B$ are independent if $P(A\cap B) = P(A)\,P(B)$ (Definition 3, Section 13.4), and dependent if the two sides differ (Remarks (i)). Lesson: independent events

Solution

List the sample space

Write H or T for the coin followed by the number on the die:
$\begin{aligned} S = \{&H1, H2, H3, H4, H5, H6, \\ &T1, T2, T3, T4, T5, T6\}. \end{aligned}$
So $n(S) = 12$ ⁠, and the 12 outcomes are equally likely.

The coin has 2 equally likely faces and the die 6, so each pair (coin face, die number) has the same chance: there are $2\times 6 = 12$ of them. Listing them lets us count events exactly, as the book does in Examples 10 to 12.

Fundamental principle of counting (Class XI). Basics: multiply the choices
Write the events as sets
$\begin{aligned} A &= \{H1, H2, H3, H4, H5, H6\}, \\ B &= \{H3, T3\}, \\ A\cap B &= \{H3\}. \end{aligned}$
$A$ contains every outcome with a head, whatever the die shows (6 outcomes). $B$ contains every outcome with a 3, whatever the coin shows (2 outcomes). Both happen only in H3, the one cell where the row of heads meets the column of threes (see the figure).

$A\cap B$ is the set of outcomes in both $A$ and $B$ ⁠. Basics: events as sets
Find the three probabilities
$\begin{aligned} P(A) &= \frac{6}{12} = \frac{1}{2}, \\ P(B) &= \frac{2}{12} = \frac{1}{6}, \\ P(A\cap B) &= \frac{1}{12}. \end{aligned}$

The 12 outcomes as a grid. $A$ is the blue row of heads (6 cells), $B$ the gold column of threes (2 cells), and $A\cap B$ is the single cell H3 (framed).

With equally likely outcomes each probability is a count over 12: 6 cells for $A$ ⁠, 2 for $B$ ⁠, 1 for $A\cap B$ ⁠. Reducing $\frac{6}{12}$ and $\frac{2}{12}$ gives the familiar values for a coin and a die.

$P(E) = \frac{n(E)}{n(S)}$ ⁠. Basics: classical probability
Apply the independence test
$\begin{aligned} P(A)\,P(B) &= \frac{1}{2}\times\frac{1}{6} \\ &= \frac{1}{12} = P(A\cap B). \end{aligned}$
The two sides are equal, so $A$ and $B$ are independent events.

Definition 3 says exactly this: equal means independent. In words, knowing that a head appeared does not change the chance of a 3: among the 6 head outcomes, 1 has a 3, so $P(B|A) = \frac{1/12}{1/2} = \frac16 = P(B)$ ⁠. That matches common sense, since the coin and the die do not affect each other (Remarks (iii) calls such experiments independent).

Definition 3, Section 13.4: $P(A\cap B) = P(A)\,P(B)$ means independent. Lesson: Examples 10 to 14

Result

$P(A\cap B) = \frac{1}{12} = \frac12\times\frac16 = P(A)\,P(B)$ ⁠, so $A$ and $B$ are independent events.

(Textbook answer: A and B are independent.)

Check it with numbers

In decimals, $P(A\cap B) = \frac{1}{12} \approx 0.0833$ ⁠, and $P(A)\,P(B) = \frac12\times\frac16 \approx 0.0833$ (with $\frac12 = 0.5$ and $\frac16 \approx 0.1667$ ⁠). Another view: of the 2 outcomes with a 3, exactly 1 has a head, so $P(A|B) = \frac12 = P(A)$ ⁠.

Common mistakes

Concluding from intuition alone. "The coin and die have nothing to do with each other" is the right instinct, but the question says "check", so the product test must be shown with numbers.
Writing $B = \{3\}$ ⁠. In the combined experiment $B$ is $\{H3, T3\}$ ⁠, and $P(B) = \frac{2}{12}$ ⁠, not $\frac{1}{12}$ ⁠.
Mixing up independent and mutually exclusive. $A$ and $B$ do overlap (H3), so they are not mutually exclusive; they are independent.

What to take away

To check independence: find $P(A)$ ⁠, $P(B)$ ⁠, $P(A\cap B)$ ⁠, compare $P(A\cap B)$ with $P(A)\,P(B)$ ⁠, and conclude in words.
For two separate experiments (a coin and a die), any event about one is independent of any event about the other.
A grid of the combined sample space makes $A$ ⁠, $B$ and $A\cap B$ easy to count.

Every fact used, and where it comes from

Fact	Used in	Source
Fundamental principle of counting: $2\times 6 = 12$ outcomes	Step 1	Class XI, Permutations and Combinations
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Step 3	Class XI, Probability (recalled in Section 13.1)
Independent $\iff P(A\cap B) = P(A)\,P(B)$ ⁠; dependent otherwise	Step 4	Definition 3 and Remarks (i), Section 13.4 (Lesson: independent events)
$P(B\|A) = \frac{P(A\cap B)}{P(A)}$	Step 4, Check	Definition 1, Section 13.2
Independent experiments	Step 4	Remarks (iii), Section 13.4

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Exercise 13.2 · Q4: Coin and die: are 'head' and '3' independent?

Basics you need

Solution

List the sample space

Write the events as sets

Find the three probabilities

Apply the independence test

Every fact used, and where it comes from