MathsCBSE Class 12 · Sample paper, session 2026-27 · 5 marks · Long answer

CBSE Class 12 Mathematics sample paper 2026-27: question 35

From the official CBSE sample question paper for 2026-27, section D, question 35, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35 · 5 marks · Long answerCompetency-basedProbability

A fitness app syncs with four types of smartwatches: Alpha, Beta, Gamma and Delta. 45% of the users use Alpha, 30% use Beta, 15% use Gamma and 10% use Delta.
Due to device errors, the probabilities of getting a false heart-rate alert from the Alpha, Beta, Gamma and Delta smartwatches are respectively 0.01, 0.02, 0.05, and 0.01.
Based on the given information, answer the following questions:

(i)
Find the probability that a randomly selected user gets a false alert.
0 marks
(ii)
If a false alert occurs, what is the probability that it came from a Beta smartwatch?
0 marks

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

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

Hint 1

Let $E_{1}$ to $E_{4}$ be the events of using each watch type. These form a partition of the sample space.

Hint 2

For (i), add $P (E_{i}) \times P (false alert ∣ E_{i})$ over all four brands.

Hint 3

For (ii), divide the Beta term by the total you found in (i).

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.

Define the events: $E_{1}, E_{2}, E_{3}, E_{4}$ are 'the user uses Alpha, Beta, Gamma, Delta' respectively, and $A$ is 'the user gets a false alert'. The events $E_{1}, \dots, E_{4}$ form a partition of the sample space, with $P (E_{1}) = 0.45, P (E_{2}) = 0.30, P (E_{3}) = 0.15, P (E_{4}) = 0.10$ .
1 mark
The conditional probabilities of a false alert are $P (A ∣ E_{1}) = 0.01, P (A ∣ E_{2}) = 0.02, P (A ∣ E_{3}) = 0.05, P (A ∣ E_{4}) = 0.01$ .
1 mark
(i) By the theorem of total probability, $P (A) = \sum P (E_{i}) P (A ∣ E_{i}) = 0.45 (0.01) + 0.30 (0.02) + 0.15 (0.05) + 0.10 (0.01)$
$= 0.0045 + 0.006 + 0.0075 + 0.001 = 0.019$ .
1½ marks
(ii) By Bayes' theorem, $P (E_{2} ∣ A) = \frac{P ( E _{2} ) P ( A ∣ E _{2} )}{P ( A )} = \frac{0.30 \times 0.02}{0.019} = \frac{0.006}{0.019} = \frac{6}{19}$ .
1½ marks

Answer: (i) 0.019; (ii) $\frac{6}{19}$

Where the marks are

Defines events and writes the prior probabilities (1 mark)
Writes the conditional probabilities of a false alert (1 mark)
Applies total probability to get $P (A) = 0.019$ (1½ marks)
Applies Bayes' theorem to get $6/19$ (1½ marks)

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 paperSample paper 2026-27: every question, by section.

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CBSE Class 12 Mathematics sample paper 2026-27: question 35

Hints

Step-by-step solution

Revise the chapter

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