Chapter 13 · Probability

Exercise 13.1 · Q13: Question bank: easy, given multiple choice

Exercise 13.1Conditional probability · Section 13.2

The question

An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Exam answer

Total number of questions $= 300 + 200 + 500 + 400$ $= 1400$ ⁠, each equally likely to be selected.

Let $E$ ⁠: the question is easy, and $F$ ⁠: the question is multiple choice.

n(F) = 500 + 400 = 900

n(E\cap F) = 500

By the definition of conditional probability,

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \frac{500}{1400}\div\frac{900}{1400} \end{aligned}

$\therefore\ P(E|F) = \dfrac{5}{9}$ ⁠.

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

The one idea this problem is built on

Being told the question is multiple choice throws away the 500 True / False questions. The new sample space is the 900 multiple choice questions, and 500 of them are easy. So the answer is $\frac{500}{900} = \frac59$ ⁠: the conditional probability is a share of the given group, not of the whole bank.

Given, to find, and the plan

Given: A bank of questions: 300 easy True / False, 200 difficult True / False, 500 easy multiple choice, 400 difficult multiple choice. One question is selected at random, so every question is equally likely.
To find: $P(E|F)$ ⁠, where $E$ ⁠: the question is easy, and $F$ ⁠: the question is multiple choice.
Plan: (1) Put the numbers in a two-way table and find the total. (2) Read off $n(F)$ and $n(E\cap F)$ ⁠. (3) Divide $P(E\cap F)$ by $P(F)$ ⁠.

Basics you need

1"At random" means equally likely

Each of the questions in the bank has the same chance of being selected, so $P(A) = \frac{n(A)}{n(S)}$ ⁠. Basics: P(E) = n(E)/n(S)

2Two-way classification

Each question is either easy or difficult, and either True / False or multiple choice. "Easy and multiple choice" is one cell of the table. Basics: two events cut S into four regions

3Conditional probability

$P(E|F) = \dfrac{P(E\cap F)}{P(F)}$ ⁠, provided $P(F) \ne 0$ (Definition 1, Section 13.2); for equally likely outcomes this is $\dfrac{n(E\cap F)}{n(F)}$ ⁠. Lesson: conditional probability

Solution

Arrange the bank in a table and count it
$n(S) = 300 + 200 + 500 + 400 = 1400.$
The four groups do not overlap (each question is in exactly one), so the total is their sum. Each question has probability $\frac{1}{1400}$ of being selected.

The four cells are the four regions cut out by "easy" and "multiple choice". Basics: two events cut S into four regions

The question bank as a two-way table. The given event $F$ (multiple choice) is the right column, 900 questions; its easy cell, 500 questions, is $E\cap F$ ⁠. The True / False column drops out once we know the question is multiple choice.
Name the events and count them

Let $E$ ⁠: the question is easy, and $F$ ⁠: it is multiple choice. Then
$\begin{aligned} n(F) &= 500 + 400 = 900, \\ n(E\cap F) &= 500. \end{aligned}$
So $P(F) = \frac{900}{1400}$ and $P(E\cap F) = \frac{500}{1400}$ ⁠.

$F$ is the whole multiple choice column, easy and difficult together. $E\cap F$ is the questions that are both easy and multiple choice: the single cell 500. The 300 easy True / False questions are in $E$ but not in $F$ ⁠.

"And" is the intersection. Basics: events as sets
Divide
$\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} = \frac{500/1400}{900/1400} \\ &= \frac{500}{1400}\times\frac{1400}{900} = \frac{5}{9}. \end{aligned}$
$P(F) \ne 0$ ⁠, so the definition applies. The 1400s cancel, leaving $\frac{500}{900}$ ⁠, and dividing top and bottom by 100 gives $\frac{5}{9}$ ⁠.

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

Result

The probability that the question is easy, given that it is multiple choice, is $P(E|F) = \dfrac{5}{9}$ ⁠.

(Textbook answer: $\frac59$ ⁠.)

Check it with numbers

Of the 900 multiple choice questions, 500 are easy: $\frac{500}{900} = \frac59 \approx 0.5556$ ⁠. Among the whole bank, 800 of the 1400 questions are easy: $P(E) = \frac{800}{1400} = \frac47 \approx 0.5714$ ⁠. Easy questions are a slightly smaller share of the multiple choice part than of the whole bank, because the True / False part is $\frac{300}{500} = 0.6$ easy.

Common mistakes

Dividing by the whole bank. $\frac{500}{1400} = \frac{5}{14}$ is $P(E\cap F)$ ⁠, the chance of "easy and multiple choice", not "easy given multiple choice".
Reversing the condition. $\frac{500}{800} = \frac58$ is $P(F|E)$ ⁠: the share of easy questions that are multiple choice.
Counting all easy questions on top. The 300 easy True / False questions are outside $F$ ⁠; only the 500 easy multiple choice questions count.

What to take away

With a two-way table, $P(E|F)$ is a cell divided by the total of its row or column: the one named after "given that".
The grand total cancels, so $P(E|F) = \dfrac{n(E\cap F)}{n(F)}$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Selection at random: $P(A) = \frac{n(A)}{n(S)}$	Steps 1, 2	Class XI, Probability (Basics: classical probability)
Disjoint groups: the total is the sum of the counts	Step 1	Class XI, Sets (Basics: two events cut S into four regions)
$P(E\|F) = \frac{P(E\cap F)}{P(F)}$ ⁠, $P(F) \ne 0$	Step 3	Section 13.2, Definition 1 (Lesson: conditional probability)

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Exercise 13.1 · Q13: Question bank: easy, given multiple choice

Basics you need

Solution

Arrange the bank in a table and count it

Name the events and count them

Divide

Every fact used, and where it comes from