Chapter 13 · Probability

Example 11: Die thrown twice: odd on the first and odd on the second

ExamplesIndependent events · Section 13.4

The question

An unbiased die is thrown twice. Let the event A be ‘odd number on the first throw’ and B the event ‘odd number on the second throw’. Check the independence of the events A and B.

Exam answer

The 36 outcomes $(x, y)$ are equally likely.

$A$ ⁠: odd number on the first throw, so $x$ is 1, 3 or 5 and $y$ is any of 6 numbers: $n(A) = 3\times 6 = 18$ ⁠.

Similarly $n(B) = 6\times 3 = 18$ ⁠.

$A\cap B$ ⁠: odd number on both throws, so $n(A\cap B) = 3\times 3 = 9$ ⁠.

P(A) = \frac{18}{36} = \frac{1}{2}

P(B) = \frac{18}{36} = \frac{1}{2}

P(A\cap B) = \frac{9}{36} = \frac{1}{4}

$A$ and $B$ are independent if $P(A\cap B) = P(A)\,P(B)$ ⁠.

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

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

Test independence with the product rule. $A$ and $B$ are independent exactly when

P(A \cap B) = P(A)\,P(B).

Counting the 36 equally likely outcomes gives all three numbers. (We expect independence, because one throw of a die cannot influence the other; the calculation confirms it.)

Given, to find, and the plan

Given: An unbiased die thrown twice; $A$ ⁠: odd number on the first throw; $B$ ⁠: odd number on the second throw.
To find: Whether $A$ and $B$ are independent.
Plan: Count $n(A)$ ⁠, $n(B)$ and $n(A \cap B)$ among the 36 outcomes; find the probabilities; compare $P(A \cap B)$ with $P(A)\,P(B)$ ⁠.

Basics you need

1Two throws as ordered pairs

$(x, y)$ ⁠: $x$ on the first throw, $y$ on the second; $6 \times 6 = 36$ pairs, all equally likely for an unbiased die. Basics: multiply the choices

2Equally likely outcomes

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

3Independent events

$A$ and $B$ are independent if $P(A \cap B) = P(A)\,P(B)$ (Definition 3, Section 13.4). Lesson: independent events

Solution

Count the outcomes in each event

For $A$ ⁠, the first number is 1, 3 or 5 and the second is any of 1 to 6:
$n(A) = 3 \times 6 = 18.$
For $B$ ⁠, the first number is anything and the second is odd:
$n(B) = 6 \times 3 = 18.$
For $A \cap B$ ⁠, both numbers are odd:
$n(A \cap B) = 3 \times 3 = 9.$
Each count multiplies the number of choices for the first throw by the number for the second. Listing is not needed, but the grid in Step 3 shows all of them.

Basics: multiply the choices
Find the three probabilities
$P(A) = \frac{18}{36} = \frac{1}{2}, \quad P(B) = \frac{18}{36} = \frac{1}{2},$ $P(A \cap B) = \frac{9}{36} = \frac{1}{4}.$
All 36 outcomes are equally likely ("unbiased die", as the book assumes), so each probability is a count over 36.

Basics: the classical definition
Apply the independence test
$\begin{aligned} P(A)\,P(B) &= \frac{1}{2} \times \frac{1}{2} \\ &= \frac{1}{4} = P(A \cap B). \end{aligned}$
So $A$ and $B$ are independent events.

The 36 outcomes. $A$ is the three odd columns, $B$ the three odd rows. They cross in $3 \times 3 = 9$ cells, which is $\frac{1}{4}$ of the grid, exactly $\frac{1}{2} \times \frac{1}{2}$ ⁠.

Definition 3 asks whether "both" has the probability of the product. Both sides equal $\frac{1}{4}$ ⁠, so the events are independent.

Definition 3, Section 13.4. Lesson: independent events

Result

$P(A \cap B) = \frac{1}{4} = P(A)\,P(B)$ ⁠, so $A$ and $B$ are independent events.

(Textbook answer: A and B are independent events.)

Check it with numbers

By conditional probability. $P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{9}{18} = \frac{1}{2} = P(A)$ ⁠: knowing that the second throw is odd leaves the chance of an odd first throw at one half.

Decimals. $P(A)\,P(B) = 0.5 \times 0.5 = 0.25$ and $P(A \cap B) = \frac{9}{36} = 0.25$ ⁠.

The two experiments are independent. This is an instance of Remarks (iii), Section 13.4: an event about the first throw and an event about the second throw always pass the product test.

Common mistakes

Stopping at "obviously independent". The question asks you to check; show $P(A \cap B) = P(A)\,P(B)$ with numbers.
Using 6 outcomes instead of 36. Two throws give ordered pairs; "odd on the first throw" is 18 pairs, not 3 faces (though the probability $\frac{1}{2}$ comes out the same).
Counting $A \cap B$ as $18 + 18$ ⁠. "Both odd" needs the first and the second odd: $3 \times 3 = 9$ ⁠.

What to take away

Separate throws of a die (or tosses of a coin) give independent events; the product test confirms it.
Counts of ordered pairs multiply: choices for the first times choices for the second.
Always write the conclusion in words. Two balls drawn with replacement are independent in the same way; Exercise 13.2, Question 13 uses this.

Every fact used, and where it comes from

Fact	Used in	Source
Counting ordered pairs: $3 \times 6$ ⁠, $6 \times 3$ ⁠, $3 \times 3$	Step 1	Counting principle, Class XI (Basics: multiply the choices)
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Step 2	Class XI, Probability (Basics: the classical definition)
Independent if $P(A \cap B) = P(A)\,P(B)$	Step 3	Definition 3, Section 13.4 (Lesson: independent events)
$P(A\|B) = \frac{n(A \cap B)}{n(B)}$ ⁠; independence as $P(A\|B) = P(A)$	Check	Section 13.2, equation (1); Definition 2, Section 13.4
Independent experiments	Check	Remarks (iii), Section 13.4

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Example 11: Die thrown twice: odd on the first and odd on the second

Basics you need

Solution

Count the outcomes in each event

Find the three probabilities

Apply the independence test

Every fact used, and where it comes from