Sample 2025-26 Class 12 Maths Q38 Solution: Case study: screen time, anxiety and low retention

The question

Case Study -3

Excessive use of screens can result in vision problems, obesity, sleep disorders, anxiety, low retention problems and can impede social and emotional comprehension and expression. It is essential to be mindful of the amount of time we spend on screens and to reduce our screen-time by taking regular breaks, setting time limits, and engaging in non-screen-based activities.

In a class of students of the age group 14 to 17, the students were categorised into three groups according to a feedback form filled by them. The first group constituted of the students who spent more than 4 hours per day on the mobile screen or the gaming screens, while the second group spent 2 to 4 hours /day on the same activities. The third group spent less than 2 hours /day on the same. The first group with the high screen time is 60% of all the students, whereas the second group with moderate screen time is 30% and the third group with low screen time is only 10% of the total number of students. It was observed that 80% students of first group faced severe anxiety and low retention issues, with 70% of second group, and 30% of third group having the same symptoms.

I. What is the total percentage of students who suffer from anxiety and low retention issues in the class? (2 marks)

II. A student is selected at random, and he is found to suffer from anxiety and low retention issues. What is the probability that he/she spends screen time more than 4 hours per day? (2 marks)

Exam answer

Let the events be

$E_1$ ⁠: the student is in the first group (more than 4 hours a day on screens),

$E_2$ ⁠: the student is in the second group (2 to 4 hours),

$E_3$ ⁠: the student is in the third group (less than 2 hours),

A: the student shows symptoms of anxiety and low retention.

\begin{aligned} P(E_1) &= \frac{60}{100},\ P(E_2) = \frac{30}{100}, \\ P(E_3) &= \frac{10}{100} \end{aligned}

Why?

Every student is in exactly one of the three groups, so $E_1$ ⁠, $E_2$ ⁠, $E_3$ do not overlap and together cover the whole class. Check: $60\% + 30\% + 10\% = 100\%$ ⁠.

\begin{aligned} P(A|E_1) &= \frac{80}{100},\ P(A|E_2) = \frac{70}{100}, \\ P(A|E_3) &= \frac{30}{100} \end{aligned}

Why?

"80% of the first group have the symptoms" is a probability within that group only, so it is the conditional probability $P(A|E_1)$ ⁠.

Watch out: it is not $P(A \cap E_1)$ ⁠, which would be a fraction of the whole class.

I. By the theorem of total probability,

Why?

A student with the symptoms comes from one of the three groups. Adding the three ways gives the theorem of total probability:

P(A) = \sum_{i = 1}^{3}P(E_i)P(A|E_i)

Each term is $P(E_i \cap A)$ ⁠, by the multiplication rule.

\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &\quad + P(E_3)P(A|E_3) \\ &= \frac{60}{100}\cdot\frac{80}{100} + \frac{30}{100}\cdot\frac{70}{100} \\ &\quad + \frac{10}{100}\cdot\frac{30}{100} \\ &= \frac{4800 + 2100 + 300}{10000} = \frac{72}{100} \end{aligned}

Why?

Each product has denominator $100 \times 100 = 10000$ ⁠. The numerators are $60 \times 80 = 4800$ ⁠, $30 \times 70 = 2100$ and $10 \times 30 = 300$ ⁠.

Their sum is 7200, and $\frac{7200}{10000} = \frac{72}{100}$ ⁠, that is 72%.

\begin{aligned} 4800 + 2100 &= 6900 \\ 6900 + 300 &= 7200 \\[4pt] \frac{7200}{10000} &= \frac{7200 \div 100}{10000 \div 100} = \frac{72}{100} \end{aligned}

$\therefore$ 72% of the students suffer from anxiety and low retention issues.

II. By Bayes' theorem,

Why?

Now we know the result (the student has the symptoms) and ask for the cause (the group). Reversing a conditional probability like this is Bayes' theorem:

P(E_1|A) = \frac{P(E_1 \cap A)}{P(A)}

The top is the first term of the total probability sum, and the bottom is the whole sum from part I.

\begin{aligned} P(E_1|A) &= \frac{P(E_1)P(A|E_1)}{P(A)} \\ &= \frac{\frac{60}{100}\cdot\frac{80}{100}}{\frac{72}{100}} = \frac{48}{72} \end{aligned}

Why?

The top is $\frac{4800}{10000} = \frac{48}{100}$ ⁠. Dividing by $\frac{72}{100}$ ⁠, the hundreds cancel, leaving $\frac{48}{72}$ ⁠.

\begin{aligned} \frac{48}{100} \div \frac{72}{100} &= \frac{48}{100} \times \frac{100}{72} \\ &= \frac{48}{72} \end{aligned}

Divide the top and the bottom by 24: $\frac{48}{72} = \frac23$ ⁠.

\frac{48}{72} = \frac{48 \div 24}{72 \div 24} = \frac{2}{3}

$\therefore$ Required probability $= \dfrac23$

Where this comes from

Chapter 13, Probability: Total probability and Bayes' theorem.

Question type: I. Two routes to one result; II. Bayes' theorem

How to spot it: click to see

I. Type 7. Two routes to one result: total probability

Spot it: the result can happen in two (or more) ways, such as a ball from bag I or from bag II, and you want its overall chance.

Here: Anxiety can come from any of the three screen-time groups (60%, 30%, 10%). Multiply each by its affected share (80%, 70%, 30%) and add: $\frac{72}{100}$ ⁠, so 72%.

II. Type 8. Bayes' theorem: from the result back to the cause

Spot it: the result has already happened (a red ball, a faulty item, an A grade), and you are asked for the chance that it came from one cause. It often comes as a case study or a 5-mark question.

Here: The student already has the symptoms, and you want the chance the cause is more than 4 hours of screen time. Divide that route, $\frac{48}{100}$ ⁠, by the total $\frac{72}{100}$ from I.

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Q38: Case study: screen time, anxiety and low retention