Chapter 13 · Probability

Example 4: School of 1000 students: Class XII, given a girl

ExamplesConditional probability · Section 13.2

The question

In a school, there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in Class XII given that the chosen student is a girl?

Exam answer

Let $E$ ⁠: the student chosen studies in Class XII, and $F$ ⁠: the student chosen is a girl. We have to find $P(E|F)$ ⁠.

Girls in Class XII: $\frac{10}{100}\times 430 = 43$ ⁠.

P(F) = \frac{430}{1000} = \frac{43}{100}

P(E\cap F) = \frac{43}{1000}

By the definition of conditional probability,

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \dfrac{\;\frac{43}{1000}\;}{\frac{43}{100}} = \frac{43}{1000}\times\frac{100}{43} \end{aligned}

$\therefore\ P(E|F) = \dfrac{1}{10} = 0.1$ ⁠.

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

The one idea this problem is built on

"Given a girl" shrinks the school to the 430 girls. Among them, 10% study in Class XII, so the answer must be $\frac{1}{10}$ ⁠. The definition of conditional probability confirms this from probabilities measured over the whole school:

P(E|F) = \frac{P(E \cap F)}{P(F)}.

Given, to find, and the plan

Given: 1000 students, 430 of them girls; 10% of the 430 girls study in Class XII. One student is chosen at random (each student equally likely).
To find: The probability that the student studies in Class XII, given that the student is a girl.
Plan: Name the events; turn the percentage into a number of students; write $P(F)$ and $P(E \cap F)$ as exact fractions of 1000; divide. The book works in decimals (0.43, 0.043); we use fractions and convert at the end.

Basics you need

1A percentage is a fraction

$10\% = \frac{10}{100} = \frac{1}{10}$ ⁠, so 10% of 430 is $\frac{1}{10} \times 430 = 43$ ⁠.

2Equally likely outcomes

Each student is an outcome; $P(E) = \frac{n(E)}{n(S)}$ with $n(S) = 1000$ ⁠. Basics: the classical definition

3Conditional probability

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

4Dividing fractions

Dividing by $\frac{43}{100}$ is multiplying by $\frac{100}{43}$ ⁠. Basics: dividing by a fraction

Solution

Name the events

Let $E$ ⁠: the student chosen studies in Class XII, and $F$ ⁠: the student chosen is a girl. We have to find $P(E|F)$ ⁠.

"Given that the chosen student is a girl" is the information, so $F$ goes after the bar.

Lesson: conditional probability
Count the girls in Class XII
$\begin{aligned} n(E \cap F) &= 10\% \times 430 \\ &= \frac{10}{100} \times 430 = 43. \end{aligned}$
$E \cap F$ means "studies in Class XII and is a girl". The question says that 10% of the girls study in Class XII, which is exactly this group. (This answers the book's "(Why?)" next to $P(E \cap F)$ ⁠.) We do not know how many boys are in Class XII, so we cannot find $P(E)$ ⁠; luckily it is not needed.

Basics: events as sets
Write the two probabilities as exact fractions
$P(F) = \frac{430}{1000} = \frac{43}{100},$ $P(E \cap F) = \frac{43}{1000}.$
Each student is equally likely to be chosen, so each probability is a count over 1000. In decimals these are the book's $0.43$ and $0.043$ ⁠; fractions avoid any rounding.

Basics: the classical definition
Divide, using the definition
$\begin{aligned} P(E|F) &= \frac{P(E \cap F)}{P(F)} = \dfrac{\;\frac{43}{1000}\;}{\frac{43}{100}} \\ &= \frac{43}{1000} \times \frac{100}{43} = \frac{1}{10}. \end{aligned}$
As a decimal, $\frac{1}{10} = 0.1$ ⁠, the book's answer.

Drawn to scale. Top: the whole school, 430 of 1000 are girls. Bottom: given a girl, only the 430 girls count, and 43 of them are in Class XII: $\frac{43}{430} = \frac{1}{10}$ ⁠.

$P(F) = \frac{43}{100} \ne 0$ ⁠, so we may divide. The 43s cancel and $\frac{100}{1000} = \frac{1}{10}$ ⁠.

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

Result

$P(E|F) = \dfrac{1}{10} = 0.1$ ⁠.

(Textbook answer: $0.1$ ⁠.)

Check it with numbers

Counting among the girls. $\frac{n(E \cap F)}{n(F)} = \frac{43}{430} = \frac{1}{10}$ ⁠, the same value.

The book's decimals. $P(F) = 0.43$ and $P(E \cap F) = 0.043$ ⁠; their quotient is 0.1, which matches our $\frac{1}{10}$ ⁠.

Common sense. The question already says that 10% of the girls are in Class XII. "10% of the girls" is a conditional probability in words, so the answer had to be $\frac{1}{10}$ ⁠; the calculation confirms it.

Common mistakes

Answering 0.043. That is $P(E \cap F)$ ⁠, the chance of picking a Class XII girl from the whole school. Given a girl, we divide by $P(F)$ ⁠.
Using 10% as $P(E \cap F)$ ⁠. The 10% is a share of the girls, not of all 1000 students.
Trying to find $P(E)$ ⁠. The number of boys in Class XII is not given, and the definition does not need it.

What to take away

Phrases like "10% of the girls" are conditional probabilities: the group named after "of" is the condition.
Turn percentages and decimals into exact fractions before dividing.
Only $P(E \cap F)$ and $P(F)$ are needed for $P(E|F)$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$10\% = \frac{10}{100}$	Step 2	Arithmetic of percentages
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Step 3	Class XI, Probability (Basics: the classical definition)
$P(E\|F) = \frac{P(E \cap F)}{P(F)}$ ⁠, $P(F) \ne 0$	Step 4	Definition 1, Section 13.2 (Lesson: conditional probability)
$P(E\|F) = \frac{n(E \cap F)}{n(F)}$	Check	Section 13.2, equation (1)

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Example 4: School of 1000 students: Class XII, given a girl

Basics you need

Solution

Name the events

Count the girls in Class XII

Write the two probabilities as exact fractions

Divide, using the definition

Every fact used, and where it comes from