MathsCBSE Class 12 · Board exam 2025, set 65/7/1 · 4 marks · Case-based

CBSE Class 12 Mathematics 2025: question 38

Set in the CBSE Class 12 board exam of 2025, paper 65/7/1, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedProbability

Case Study – 3

Based upon the results of regular medical check-ups in a hospital, it was found that out of 1000 people, 700 were very healthy, 200 maintained average health and 100 had a poor health record.

Let $A_{1}$ : People with good health,
$A_{2}$ : People with average health,
and $A_{3}$ : People with poor health.

During a pandemic, the data expressed that the chances of people contracting the disease from category $A_{1}$ , $A_{2}$ and $A_{3}$ are 25%, 35% and 50%, respectively.

Based upon the above information, answer the following questions :

(i)
A person was tested randomly. What is the probability that he/she has contracted the disease ?
2 marks
(ii)
Given that the person has not contracted the disease, what is the probability that the person is from category $A_{2}$ ?
2 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 65/7/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Convert 700, 200, 100 out of 1000 into probabilities of each health category.

Hint 2

For (i) use the theorem of total probability over $A_{1}, A_{2}, A_{3}$ .

Hint 3

For (ii) the chances of NOT contracting are $1 - 0.25$ , $1 - 0.35$ , $1 - 0.5$ ; use Bayes' theorem.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Let $A$ : the person has contracted the disease. $P (A_{1}) = \frac{7}{10}$ , $P (A_{2}) = \frac{2}{10}$ , $P (A_{3}) = \frac{1}{10}$ ; $P (A ∣ A_{1}) = \frac{25}{100}$ , $P (A ∣ A_{2}) = \frac{35}{100}$ , $P (A ∣ A_{3}) = \frac{50}{100}$ . By the theorem of total probability:
$P (A) = \frac{7}{10} \cdot \frac{25}{100} + \frac{2}{10} \cdot \frac{35}{100} + \frac{1}{10} \cdot \frac{50}{100}$ .
1½ marks
(i) $P (A) = \frac{175 + 70 + 50}{1000} = \frac{295}{1000} = 0.295$ (or $\frac{59}{200}$ ).
½ mark
(ii) $P (\overset{ˉ}{A} ∣ A_{1}) = \frac{75}{100}$ , $P (\overset{ˉ}{A} ∣ A_{2}) = \frac{65}{100}$ , $P (\overset{ˉ}{A} ∣ A_{3}) = \frac{50}{100}$ . By Bayes' theorem:
$P (A_{2} ∣ \overset{ˉ}{A}) = \frac{P ( A _{2} ) P ( A ˉ ∣ A _{2} )}{P ( A _{1} ) P ( A ˉ ∣ A _{1} ) + P ( A _{2} ) P ( A ˉ ∣ A _{2} ) + P ( A _{3} ) P ( A ˉ ∣ A _{3} )} = \frac{\frac{2}{10} \cdot \frac{65}{100}}{\frac{7}{10} \cdot \frac{75}{100} + \frac{2}{10} \cdot \frac{65}{100} + \frac{1}{10} \cdot \frac{50}{100}}$ .
1½ marks
(ii) $= \frac{130}{525 + 130 + 50} = \frac{130}{705} = \frac{26}{141}$ .
½ mark

Answer: (i) $0.295$ (= $\frac{59}{200}$ ); (ii) $\frac{26}{141}$

Where the marks are

(i) Applies total probability with correct priors and conditional probabilities (1½ marks)
(i) Obtains $P (A) = 0.295$ (½ mark)
(ii) Applies Bayes' theorem with complementary probabilities 75%, 65%, 50% (1½ marks)
(ii) Obtains $\frac{26}{141}$ (½ mark)

Revise the chapter

Chapter: Probability (notes, formulas, where marks are lost and every past question)
Topic: Bayes' Theorem

Practise this chapterBoard-style questions from Probability, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 65/7/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Mathematics 2025: question 38

Hints

Step-by-step solution

Revise the chapter

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