Chapter 13 · Probability

Exercise 13.3 · Q4: Knows or guesses: knew the answer, given it is correct, $P = \frac{12}{13}$

Exercise 13.3Bayes' theorem · Section 13.5

The question

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\frac34$ be the probability that he knows the answer and $\frac14$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac14$ ⁠. What is the probability that the student knows the answer given that he answered it correctly?

Exam answer

Let $E_1$ ⁠: the student knows the answer, $E_2$ ⁠: the student guesses the answer, and $A$ ⁠: the answer is correct.

$E_1$ and $E_2$ form a partition of the sample space (he either knows or guesses).

$P(E_1) = \frac34$ and $P(E_2) = \frac14$

A student who knows the answer answers correctly, so $P(A|E_1) = 1$ ⁠. Also $P(A|E_2) = \frac14$ ⁠.

By the theorem of total probability,

\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &= \frac34\cdot 1 + \frac14\cdot\frac14 \\ &= \frac{12}{16} + \frac{1}{16} = \frac{13}{16} \end{aligned}

By Bayes' theorem,

\begin{aligned} P(E_1|A) &= \frac{P(E_1)P(A|E_1)}{P(A)} \\ &= \frac34 \div \frac{13}{16} \\ &= \frac34 \times \frac{16}{13} \end{aligned}

$\therefore$ The probability that the student knows the answer, given that it is correct, is $\dfrac{12}{13}$ ⁠.

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

The one idea this problem is built on

A correct answer can come from knowing or from a lucky guess. Knowing always gives a correct answer, while guessing rarely does, so a correct answer is strong evidence that the student knew. Bayes' theorem measures how strong: the "knows" route to a correct answer, divided by both routes together.

Given, to find, and the plan

Given: The student knows the answer with probability $\frac34$ and guesses with probability $\frac14$ ⁠; a guess is correct with probability $\frac14$ ⁠. Not written, but meant: a student who knows the answer gives the correct answer.
To find: The probability that he knew the answer, given that his answer is correct: $P(E_1|A)$ ⁠.
Plan: (1) Name the events; check the partition. (2) List the four probabilities, including $P(A|E_1) = 1$ ⁠. (3) Total probability for $P(A)$ ⁠. (4) Bayes' theorem.

Basics you need

1A certain event has probability 1

If knowing the answer always leads to a correct answer, then $P(\text{correct} \mid \text{knows}) = 1$ ⁠. Basics: rules of probability

2Conditional probability (Section 13.2)

"A student who guesses will be correct with probability $\frac14$ " is a statement about guessers only: $P(A|E_2) = \frac14$ ⁠. Lesson: conditional probability

3Total probability and Bayes' theorem (Section 13.5)

For a partition $E_1, E_2$ ⁠: $P(A) = \sum P(E_i)P(A|E_i)$ and $P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(A)}$ ⁠. Lesson: Bayes' theorem

4Fractions with a common denominator

$\frac34 = \frac{12}{16}$ ⁠, so $\frac34 + \frac{1}{16} = \frac{13}{16}$ ⁠; dividing by $\frac{13}{16}$ is multiplying by $\frac{16}{13}$ ⁠. Basics: dividing by a fraction

Solution

Name the events

Let
- $E_1$ ⁠: the student knows the answer;
- $E_2$ ⁠: the student guesses the answer;
- $A$ ⁠: the student answers the question correctly.
$E_1$ and $E_2$ form a partition of the sample space: the question says he "either knows the answer or guesses", so exactly one happens, and $\frac34 + \frac14 = 1$ ⁠.

Knowing and guessing are the two possible causes; a correct answer is the result we observe. Bayes' theorem works backwards from the result to the cause.

Partition: pairwise disjoint, exhaustive, nonzero probabilities (Section 13.5.1). Lesson: Bayes' theorem
The four probabilities
$\begin{aligned} P(E_1) &= \frac34, \\ P(E_2) &= \frac14, \end{aligned}$ $\begin{aligned} P(A|E_1) &= 1, \\ P(A|E_2) &= \frac14. \end{aligned}$
The first three come straight from the question. $P(A|E_1) = 1$ is not printed, but it is what "knows the answer" means: a student who knows the answer marks it correctly. The book's answer $\frac{12}{13}$ uses this value too (Step 4).

A certain event has probability 1. Basics: rules of probability
The chance of a correct answer

Since $E_1$ and $E_2$ form a partition, by the theorem of total probability
$\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &= \frac34\cdot 1 + \frac14\cdot\frac14 \\ &= \frac{12}{16} + \frac{1}{16} = \frac{13}{16}. \end{aligned}$

The tree for Question 4. Knowing leads to a correct answer every time (the rose path, $\frac34$ ⁠); guessing leads to a correct answer only a quarter of the time (the green path, $\frac{1}{16}$ ⁠).

A correct answer comes either from knowing or from a lucky guess; these two routes cannot both happen, so their path probabilities add.

Multiply along a path, add the paths. Basics: tree diagrams
Knew it, given it is correct: Bayes' theorem

By Bayes' theorem,
$\begin{aligned} P(E_1|A) &= \frac{P(E_1)P(A|E_1)}{P(A)} \\ &= \frac34 \div \frac{13}{16} \\ &= \frac34 \times \frac{16}{13} = \frac{12}{13}. \end{aligned}$
The top is the "knows" path, $\frac34 \times 1 = \frac34$ ⁠; the bottom is every route to a correct answer, $\frac{13}{16}$ ⁠. Out of every 13 correct answers, on average 12 come from knowing and 1 from a lucky guess.

Dividing by a fraction is multiplying by its flip. Basics: dividing by a fraction
A quick sense check

The guess route gets the rest: $P(E_2|A) = \frac{1}{16} \div \frac{13}{16} = \frac{1}{13}$ ⁠, and $\frac{12}{13} + \frac{1}{13} = 1$ ⁠. The correct answer raised the chance that he knew from $\frac34 = 0.75$ to $\frac{12}{13} \approx 0.923$ ⁠.

A correct answer is more likely after knowing than after guessing (1 against $\frac14$ ⁠), so seeing it must make "knows" more likely, never less.

$P(E_1|A) + P(E_2|A) = 1$ for a partition. Lesson: Bayes' theorem

Result

The probability that the student knows the answer, given that he answered it correctly, is $\frac{12}{13}$ ⁠. (Textbook answer: $\frac{12}{13}$ ⁠.)

Check it with numbers

Picture 64 students answering the question. $\frac34$ of them, 48, know the answer and are all correct. The other 16 guess, and $\frac14$ of them, 4, are lucky. So $48 + 4 = 52$ answers are correct and 48 of these came from knowing: $\frac{48}{52} \approx 0.9231$ ⁠, the same as $\frac{12}{13} \approx 0.9231$ ⁠.

Common mistakes

Answering $\frac34$ ⁠. That is $P(E_1)$ ⁠, the chance before we know the answer was correct. The correct answer is extra information, and it changes the chance.
Leaving out $P(A|E_1)$ ⁠. It is not printed, but it is needed: it equals 1, because a student who knows marks the right answer.
Using $\frac14$ as $P(A)$ ⁠. $\frac14$ is the chance of a correct guess only; overall a correct answer has probability $\frac{13}{16}$ ⁠.

What to take away

Look for probabilities that the question leaves unsaid but means, such as "knows the answer" giving $P(\text{correct}) = 1$ ⁠. State them in your answer.
A result that is much more likely under one cause (here 1 against $\frac14$ ⁠) pushes the probability of that cause up.
Book Example 18 and Question 5 show the opposite effect: a rare cause stays fairly unlikely even after a positive test.

Every fact used, and where it comes from

Fact	Used in	Source
Partition of a sample space	Step 1	Section 13.5.1 (Lesson: Bayes' theorem)
A certain event has probability 1	Step 2	Class XI axioms (Basics: rules of probability)
Conditional probability $P(A\|E)$	Step 2	Section 13.2 (Lesson: conditional probability)
Theorem of total probability	Step 3	Section 13.5.2 (Lesson: Bayes' theorem)
Bayes' theorem	Steps 4, 5	Section 13.5 (Examples 15 to 21)
Dividing by a fraction	Step 4	Arithmetic (Basics: dividing by a fraction)

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Exercise 13.3 · Q4: Knows or guesses: knew the answer, given it is correct, P=1213P = \frac{12}{13}P=1312​

Basics you need

Solution

Name the events

The four probabilities

The chance of a correct answer

Knew it, given it is correct: Bayes' theorem

A quick sense check

Every fact used, and where it comes from

Exercise 13.3 · Q4: Knows or guesses: knew the answer, given it is correct, $P = \frac{12}{13}$