Chapter 13 · Probability

Example 3: Card numbered 1 to 10: even, given it is more than 3

ExamplesConditional probability · Section 13.2

The question

Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Exam answer

Let $A$ ⁠: the number on the card drawn is even, and $B$ ⁠: the number on the card drawn is greater than 3. We have to find $P(A|B)$ ⁠.

The sample space is

S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}

with all 10 outcomes equally likely.

A = \{2, 4, 6, 8, 10\}

B = \{4, 5, 6, 7, 8, 9, 10\}

A\cap B = \{4, 6, 8, 10\}

P(B) = \frac{7}{10},\quad P(A\cap B) = \frac{4}{10}

By the definition of conditional probability,

\begin{aligned} P(A|B) &= \frac{P(A\cap B)}{P(B)} \\ &= \dfrac{\;\frac{4}{10}\;}{\frac{7}{10}} = \frac{4}{10}\times\frac{10}{7} \end{aligned}

$\therefore\ P(A|B) = \dfrac{4}{7}$ ⁠.

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

The one idea this problem is built on

The information "more than 3" removes the cards 1, 2 and 3. Seven equally likely cards are left, and four of them are even. So the probability is $\frac{4}{7}$ ⁠. The definition of conditional probability does exactly this, in fractions of the whole box:

P(A|B) = \frac{P(A \cap B)}{P(B)}.

Given, to find, and the plan

Given: Ten cards numbered 1 to 10, one drawn at random (each card equally likely). The number on the card is known to be more than 3.
To find: The probability that the number is even, given that it is more than 3.
Plan: Name the events $A$ (even) and $B$ (more than 3); list $S$ ⁠, $A$ ⁠, $B$ and $A \cap B$ ⁠; find $P(B)$ and $P(A \cap B)$ ⁠; divide.

Basics you need

1Events are sets of outcomes

"Even and more than 3" is the intersection $A \cap B$ ⁠: the numbers in both lists. Basics: events as sets

2Equally likely outcomes

$P(E) = \frac{n(E)}{n(S)}$ when each outcome is equally likely. Basics: the classical definition

3Conditional probability

$P(A|B) = \frac{P(A \cap B)}{P(B)}$ when $P(B) \ne 0$ (Definition 1, Section 13.2). Lesson: conditional probability

Solution

Name the events

Let $A$ ⁠: the number on the card drawn is even, and $B$ ⁠: the number on the card drawn is greater than 3. We have to find $P(A|B)$ ⁠.

"It is known that the number is more than 3" is the information we are given, so it is the event after the bar. "Even" is what we want the probability of, so it goes before the bar.

Lesson: conditional probability
List the sample space and the events
$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\},$ $n(S) = 10,$ $A = \{2, 4, 6, 8, 10\},$ $B = \{4, 5, 6, 7, 8, 9, 10\},$ $A \cap B = \{4, 6, 8, 10\}.$
The cards are mixed thoroughly and drawn at random, so each of the 10 numbers is equally likely. $A \cap B$ keeps the even numbers that are also more than 3; the even card 2 is left out because it is not more than 3.

Basics: events as sets
Find the probabilities
$P(A) = \frac{5}{10}, \quad P(B) = \frac{7}{10},$ $P(A \cap B) = \frac{4}{10}.$
Each is the number of cards in the event over 10. Only $P(B)$ and $P(A \cap B)$ are needed for $P(A|B)$ ⁠; $P(A)$ is used in the check.

Basics: the classical definition
Divide, using the definition
$\begin{aligned} P(A|B) &= \frac{P(A \cap B)}{P(B)} = \dfrac{\;\frac{4}{10}\;}{\frac{7}{10}} \\ &= \frac{4}{10} \times \frac{10}{7} = \frac{4}{7}. \end{aligned}$

The ten cards. Given $B$ ⁠, only the seven shaded cards (4 to 10) count; the four ringed ones are even. So $P(A|B) = \frac{4}{7}$ ⁠.

$P(B) = \frac{7}{10} \ne 0$ ⁠, so the definition applies. The tens cancel, leaving 4 even cards out of the 7 cards that are more than 3.

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

Result

$P(A|B) = \dfrac{4}{7}$ ⁠.

(Textbook answer: $\frac{4}{7}$ ⁠.)

Check it with numbers

Counting inside $B$ ⁠. $\frac{n(A \cap B)}{n(B)} = \frac{4}{7}$ ⁠, the same value.

The complement adds up. Given $B$ ⁠, the odd cards are 5, 7 and 9, so $P(A'|B) = \frac{3}{7}$ ⁠, and $\frac{4}{7} + \frac{3}{7} = 1$ ⁠, as Property 3 of Section 13.2.1 requires.

Decimals. $\frac{4}{7} \approx 0.5714$ ⁠, a little more than $P(A) = \frac{5}{10} = 0.5$ ⁠. That makes sense: the removed cards 1, 2, 3 contain two odd numbers and only one even number.

Common mistakes

Swapping the events. $P(B|A) = \frac{4}{5}$ is the probability of "more than 3" given "even", a different question.
Dividing by 10. $\frac{4}{10}$ is $P(A \cap B)$ ⁠, not the conditional probability; once we know the card is more than 3, only 7 cards remain.
Counting 3 as "more than 3". "More than 3" starts at 4.

What to take away

The given information goes after the bar and becomes the new sample space.
With equally likely outcomes, $P(A|B) = \frac{n(A \cap B)}{n(B)}$ ⁠.
A conditional probability can be larger or smaller than the plain probability; here it is larger.

Every fact used, and where it comes from

Fact	Used in	Source
Events as sets; "both" is the intersection	Step 2	Class XI, Probability (Basics: events as sets)
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Step 3	Class XI, Probability (Basics: the classical definition)
$P(A\|B) = \frac{P(A \cap B)}{P(B)}$ ⁠, $P(B) \ne 0$	Step 4	Definition 1, Section 13.2 (Lesson: conditional probability)
$P(A\|B) = \frac{n(A \cap B)}{n(B)}$	Check	Section 13.2, equation (1)
$P(E'\|F) = 1 - P(E\|F)$	Check	Property 3, Section 13.2.1

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Example 3: Card numbered 1 to 10: even, given it is more than 3

Basics you need

Solution

Name the events

List the sample space and the events

Find the probabilities

Divide, using the definition

Every fact used, and where it comes from