Chapter 13 · Probability

Exercise 13.3 · Q3: Hostel or day scholar: the A-grade student lives in the hostel, $P = \frac{9}{13}$

Exercise 13.3Bayes' theorem · Section 13.5

The question

Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?

(The book prints "hostlier", a misprint for "hosteller": a student who resides in the hostel.)

Exam answer

Let $E_1$ ⁠: the student resides in the hostel, $E_2$ ⁠: the student is a day scholar, and $A$ ⁠: the student attains A grade.

$E_1$ and $E_2$ form a partition of the sample space (each student is one or the other).

$P(E_1) = 60\% = \frac35$ and $P(E_2) = 40\% = \frac25$

$P(A|E_1) = 30\% = \frac{3}{10}$ and $P(A|E_2) = 20\% = \frac15$

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) \\ &= \frac35\cdot\frac{3}{10} + \frac25\cdot\frac15 \\ &= \frac{9}{50} + \frac{4}{50} = \frac{13}{50} \end{aligned}

By Bayes' theorem,

\begin{aligned} P(E_1|A) &= \frac{P(E_1)P(A|E_1)}{P(A)} \\ &= \frac{9}{50} \div \frac{13}{50} \\ &= \frac{9}{50} \times \frac{50}{13} \end{aligned}

$\therefore$ The probability that the student is a hosteller is $\dfrac{9}{13}$ ⁠.

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

The one idea this problem is built on

The A grade is the result we see; living in the hostel or not is the hidden cause. Bayes' theorem compares the two routes to an A grade: hostellers are more numerous and more likely to get an A, so most A grades come through the hostel. The share of the hostel route in all A grades is the answer.

Given, to find, and the plan

Given: 60% of students live in the hostel and 40% are day scholars; 30% of hostellers and 20% of day scholars get an A grade. A student chosen at random has an A grade.
To find: The probability that this student lives in the hostel: $P(E_1|A)$ ⁠.
Plan: (1) Name the events; check the partition. (2) Turn every percentage into a fraction. (3) Total probability for $P(A)$ ⁠. (4) Bayes' theorem for $P(E_1|A)$ ⁠.

Basics you need

1Percentages as probabilities

"60% of the students" chosen at random gives probability $\frac{60}{100}$ ⁠; simplify the fraction. Basics: classical probability

2"30% of hostellers" is a conditional probability

It describes only the students who live in the hostel, so it is $P(A|E_1)$ ⁠, not $P(A)$ ⁠. Lesson: conditional probability

3Total probability and Bayes' theorem (Section 13.5)

For a partition $E_1, E_2$ ⁠: $P(A)$ is the sum of the paths $P(E_i)P(A|E_i)$ ⁠, and $P(E_1|A)$ is the $E_1$ path divided by $P(A)$ ⁠. Lesson: Bayes' theorem

4Tree diagrams

Branch first by the cause, then by the result; multiply along a path. Basics: tree diagrams

Solution

Name the events

Let
- $E_1$ ⁠: the student chosen resides in the hostel;
- $E_2$ ⁠: the student chosen is a day scholar;
- $A$ ⁠: the student chosen attains A grade.
$E_1$ and $E_2$ form a partition of the sample space (all students of the college): every student is either a hosteller or a day scholar, not both, and both groups are nonempty.

The two groups do not overlap because the book defines day scholars as students "not residing in hostel"; for the same reason every student is in one of them, and indeed $60\% + 40\% = 100\%$ ⁠. Both shares are nonzero. That is exactly the partition Bayes' theorem needs.

Partition: pairwise disjoint, exhaustive, nonzero probabilities (Section 13.5.1). Lesson: Bayes' theorem
Percentages into fractions
$\begin{aligned} P(E_1) &= 60\% = \frac{60}{100} = \frac35, \\ P(E_2) &= 40\% = \frac{40}{100} = \frac25, \\ P(A|E_1) &= 30\% = \frac{30}{100} = \frac{3}{10}, \\ P(A|E_2) &= 20\% = \frac{20}{100} = \frac15. \end{aligned}$
The student is chosen at random, so the share of a group is the probability of belonging to it. "30% of all students who reside in hostel attain A grade" speaks only of hostellers, so it is the chance of an A grade given $E_1$ ⁠. Fractions keep the arithmetic exact.

$x\% = \frac{x}{100}$ ⁠. Basics: classical probability
The chance of an A grade overall

Since $E_1$ and $E_2$ form a partition, by the theorem of total probability
$\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &= \frac35\cdot\frac{3}{10} + \frac25\cdot\frac15 \\ &= \frac{9}{50} + \frac{2}{25} \\ &= \frac{9}{50} + \frac{4}{50} = \frac{13}{50}. \end{aligned}$

The tree for Question 3. The rose path (hostel, then A grade) has product $\frac{9}{50}$ ⁠; the green path (day scholar, then A grade) has $\frac{4}{50}$ ⁠. Together, $P(A) = \frac{13}{50}$ ⁠.

An A-grade student is either a hosteller with an A or a day scholar with an A. Each route is a path on the tree; multiply along it, then add. Writing $\frac{2}{25}$ as $\frac{4}{50}$ gives a common denominator.

Multiply along a path, add the paths. Basics: tree diagrams
From the A grade back to the hostel: Bayes' theorem

By Bayes' theorem,
$\begin{aligned} P(E_1|A) &= \frac{P(E_1)P(A|E_1)}{P(A)} \\ &= \frac{9}{50} \div \frac{13}{50} \\ &= \frac{9}{50} \times \frac{50}{13} = \frac{9}{13}. \end{aligned}$
The top is the hostel path alone, $\frac{9}{50}$ ⁠; the bottom is all A-grade paths, $\frac{13}{50}$ ⁠. Their ratio is the share of A-grade students who live in the hostel. The 50s cancel.

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

For a day scholar, $P(E_2|A) = \frac{4}{50} \div \frac{13}{50} = \frac{4}{13}$ ⁠, and $\frac{9}{13} + \frac{4}{13} = 1$ ⁠. An A grade raises the chance of "hosteller" from $\frac35$ (about 0.6) to $\frac{9}{13}$ (about 0.69), because hostellers get A grades more often.

Given the A grade, the student is certainly one of the two kinds, so the two updated probabilities must add to 1.

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

Result

The probability that the A-grade student is a hosteller is $\frac{9}{13}$ ⁠. (Textbook answer: $\frac{9}{13}$ ⁠.)

Check it with numbers

Picture 100 students. 60 live in the hostel and $30\%$ of them, 18, get an A; 40 are day scholars and $20\%$ of them, 8, get an A. So $18 + 8 = 26$ students have an A grade, and 18 of them are hostellers: $\frac{18}{26} \approx 0.6923$ ⁠, the same as $\frac{9}{13} \approx 0.6923$ ⁠.

Common mistakes

Answering $\frac{3}{10}$ ⁠. That is $P(A|E_1)$ ⁠, the chance of an A for a hosteller. The question asks for $P(E_1|A)$ ⁠, the chance of being a hosteller for an A-grade student.
Using 30% as $P(A \cap E_1)$ ⁠. 30% is a share of the hostellers only; the joint probability is $\frac35 \times \frac{3}{10} = \frac{9}{50}$ ⁠.
Mixing percentages and fractions in one line. Convert everything first, then compute with fractions.
Stopping at $P(A) = \frac{13}{50}$ ⁠. That is only the bottom of Bayes' formula.

What to take away

" $x\%$ of group $G$ have property $A$ " means $P(A|G) = \frac{x}{100}$ ⁠: a conditional probability.
Imagining 100 (or 1000) people turns a Bayes question into counting, and it is a quick check of the answer.
The two-bag version of the same steps is Question 2; the next one, Question 4, has a cause that makes the result certain.

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)
Shares as probabilities, $x\% = \frac{x}{100}$	Step 2	Classical probability, Class XI (Basics: classical 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; Example 19 uses it with percentages (Examples 15 to 21)
Dividing by a fraction	Step 4	Arithmetic (Basics: dividing by a fraction)

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Exercise 13.3 · Q3: Hostel or day scholar: the A-grade student lives in the hostel, P=913P = \frac{9}{13}P=139​

Basics you need

Solution

Name the events

Percentages into fractions

The chance of an A grade overall

From the A grade back to the hostel: Bayes' theorem

A quick sense check

Every fact used, and where it comes from

Exercise 13.3 · Q3: Hostel or day scholar: the A-grade student lives in the hostel, $P = \frac{9}{13}$