Arka bought two cages of birds: Cage-I contains 5 parrots and 1 owl and Cage-II contains 6 parrots. One day Arka forgot to lock both cages and two birds flew from Cage-I to Cage-II (simultaneously). Then two birds flew back from cage-II to cage-I (simultaneously). Assume that all the birds have equal chances of flying. On the basis of the above information, answer the following questions:- (i) When two birds flew from Cage-I to Cage-II and two birds flew back from Cage-II to Cage-I then find the probability that the owl is still in Cage-I. (ii) When two birds flew from Cage-I to Cage-II and two birds flew back from Cage-II to Cage-I, the owl is still seen in Cage-I, what is the probability that one parrot and the owl flew from Cage-I to Cage-II?

MathsCBSE Class 12 · Sample paper, session 2024-25 · 4 marks · Case-based

CBSE Class 12 Mathematics sample paper 2024-25: question 38

From the official CBSE sample question paper for 2024-25, 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

Arka bought two cages of birds: Cage-I contains 5 parrots and 1 owl and Cage-II contains 6 parrots. One day Arka forgot to lock both cages and two birds flew from Cage-I to Cage-II (simultaneously). Then two birds flew back from cage-II to cage-I (simultaneously).
Assume that all the birds have equal chances of flying.
On the basis of the above information, answer the following questions:-

(i)
When two birds flew from Cage-I to Cage-II and two birds flew back from Cage-II to Cage-I then find the probability that the owl is still in Cage-I.
2 marks
(ii)
When two birds flew from Cage-I to Cage-II and two birds flew back from Cage-II to Cage-I, the owl is still seen in Cage-I, what is the probability that one parrot and the owl flew from Cage-I to Cage-II?
2 marks

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

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

Hint 1

Split into cases by which two birds flew out of Cage-I: either the owl went with a parrot, or two parrots went.

Hint 2

If the owl left, it must come back for it to be in Cage-I. Find that probability from the 8 birds now in Cage-II.

Hint 3

Part (ii) is a 'reverse' conditional probability, so use Bayes' theorem with the result of part (i) as the denominator.

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) Define the events: $E_{1}$ = one parrot and the owl fly from Cage-I, $E_{2}$ = two parrots fly from Cage-I, $A$ = the owl is still in Cage-I at the end. Two birds are chosen from 6 in $^{6} C_{2} = 15$ ways, so $P (E_{1}) = \frac{5 \times 1}{15} = \frac{1}{3}$ and $P (E_{2}) = \frac{^{5} C _{2}}{15} = \frac{10}{15} = \frac{2}{3}$ .
½ mark
(i) Under $E_{1}$ , Cage-II has 7 parrots and 1 owl. The owl is back in Cage-I only if it returns with one parrot, so $P (A ∣ E_{1}) = \frac{7 \times 1}{^{8} C _{2}} = \frac{7}{28} = \frac{1}{4}$ . Under $E_{2}$ , the owl never left, so $P (A ∣ E_{2}) = 1$ . By the theorem of total probability, $P (A) = P (E_{1}) P (A ∣ E_{1}) + P (E_{2}) P (A ∣ E_{2}) = \frac{1}{3} \cdot \frac{1}{4} + \frac{2}{3} \cdot 1$ .
1 mark
(i) Simplify: $P (A) = \frac{1}{12} + \frac{8}{12} = \frac{9}{12} = \frac{3}{4}$ . As a count check, $\frac{35 + 280}{420} = \frac{3}{4}$ .
½ mark
(ii) The required probability is $P (E_{1} ∣ A)$ .
½ mark
(ii) By Bayes' theorem, $P (E_{1} ∣ A) = \frac{P ( E _{1} ) P ( A ∣ E _{1} )}{P ( E _{1} ) P ( A ∣ E _{1} ) + P ( E _{2} ) P ( A ∣ E _{2} )}$ .
½ mark
(ii) Substitute: $P (E_{1} ∣ A) = \frac{1/12}{3/4} = \frac{1}{9}$ .
1 mark

Answer: (i) $\frac{3}{4}$ ; (ii) $\frac{1}{9}$

Where the marks are

(i) Defines the events and finds the probabilities of the two ways the birds can leave Cage-I (1/2 mark)
(i) Applies total probability with the correct conditional probabilities (1 mark)
(i) Gets $P (A) = 3/4$ (1/2 mark)
(ii) Identifies the required probability as $P (E_{1} ∣ A)$ (1/2 mark)
(ii) Writes Bayes' theorem (1/2 mark)
(ii) Gets $1/9$ (1 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 paperSample paper 2024-25: every question, by section.

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CBSE Class 12 Mathematics sample paper 2024-25: question 38

Hints

Step-by-step solution

Revise the chapter

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