MathsCBSE Class 12 · Board exam 2024, set 65/1/1 · 4 marks · Case-based

CBSE Class 12 Mathematics 2024: question 37

Set in the CBSE Class 12 board exam of 2024, paper 65/1/1, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q37 · 4 marks · Case-basedCompetency-basedProbability

According to recent research, air turbulence has increased in various regions around the world due to climate change. Turbulence makes flights bumpy and often delays the flights.

Assume that, an airplane observes severe turbulence, moderate turbulence or light turbulence with equal probabilities. Further, the chance of an airplane reaching late to the destination are 55%, 37% and 17% due to severe, moderate and light turbulence respectively.

On the basis of the above information, answer the following questions :

(i)
Find the probability that an airplane reached its destination late.
2 marks
(ii)
If the airplane reached its destination late, find the probability that it was due to moderate turbulence.
2 marks

Figure in the paper: Picture titled 'Turbulence intensity' showing three airplanes: Light (±1 meter), Moderate (±5 meters), Severe (±30 meters) vertical displacement with arrows. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 65/1/1.

Hints

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

Hint 1

Name the three turbulence events and the event 'reaches late'; each turbulence type has probability $\frac{1}{3}$ .

Hint 2

For (i), use the theorem of total probability.

Hint 3

For (ii), the question asks for a 'reverse' probability given the plane was late, so 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 $E_{1}, E_{2}, E_{3}$ be the events of severe, moderate and light turbulence, and $A$ the event that the airplane reaches late.
$P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}$ ; $P (A ∣ E_{1}) = \frac{55}{100}, P (A ∣ E_{2}) = \frac{37}{100}, P (A ∣ E_{3}) = \frac{17}{100}$
½ mark
(i) By the theorem of total probability:
$P (A) = \frac{1}{3} \cdot \frac{55}{100} + \frac{1}{3} \cdot \frac{37}{100} + \frac{1}{3} \cdot \frac{17}{100}$
1 mark
(i) $= \frac{1}{3} \cdot \frac{109}{100} = \frac{109}{300}$
½ mark
(ii) By Bayes' theorem:
$P (E_{2} ∣ A) = \frac{P ( E _{2} ) P ( A ∣ E _{2} )}{P ( A )} = \frac{\frac{1}{3} \cdot \frac{37}{100}}{\frac{109}{300}}$
1½ marks
(ii) $= \frac{37}{109}$
½ mark

Answer: (i) $\frac{109}{300}$ ; (ii) $\frac{37}{109}$

Where the marks are

(i) Defines the events and writes their probabilities (½ mark)
(i) Applies total probability correctly (1 mark)
(i) Gets $P (A) = \frac{109}{300}$ (½ mark)
(ii) Applies Bayes' theorem correctly (1½ marks)
(ii) Gets $\frac{37}{109}$ (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 paper2024 board paper, set 65/1/1: 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 2024: question 37

Hints

Step-by-step solution

Revise the chapter

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