CBSE Sample Paper 2024-25 · Section E · 4 marks

Q38: Case study: birds flying between two cages

Sample 2024-25Ch 13 · Probability

The question

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)

Exam answer

Let $E_1$ be the event that one parrot and the owl flew from Cage-I, $E_2$ the event that two parrots flew from Cage-I, and A the event that the owl is still in Cage-I.

Why?

Cage-I has 5 parrots and 1 owl, so the two birds that fly out are either one parrot and the owl, or two parrots. $E_1$ and $E_2$ cannot both happen, and one of them must. This split is what total probability and Bayes' theorem need.

(i)

P(E_1) = \frac{{}^5C_1 \times {}^1C_1}{{}^6C_2} = \frac{5}{15}

Why?

All birds are equally likely to fly, so every choice of 2 birds from the 6 is equally likely. There are ${}^6C_2 = \frac{6 \times 5}{2} = 15$ such choices.

For $E_1$ ⁠, choose 1 of the 5 parrots and the 1 owl: ${}^5C_1 \times {}^1C_1 = 5$ ways.

P(E_2) = \frac{{}^5C_2}{{}^6C_2} = \frac{10}{15}

Why?

For $E_2$ ⁠, choose 2 of the 5 parrots: ${}^5C_2 = \frac{5 \times 4}{2} = 10$ ways. Check: $\frac{5}{15} + \frac{10}{15} = 1$ ⁠.

After $E_1$ ⁠, Cage-II has 7 parrots and the owl; the owl is back in Cage-I only if 1 parrot and the owl fly back:

P(A \mid E_1) = \frac{{}^7C_1 \times {}^1C_1}{{}^8C_2} = \frac{7}{28}

Why?

Cage-II now holds its 6 parrots plus the parrot and the owl that arrived: 8 birds. Two of them fly back, in ${}^8C_2 = \frac{8 \times 7}{2} = 28$ equally likely ways.

The owl returns only if it is one of the two, with any one of the 7 parrots: ${}^7C_1 \times {}^1C_1 = 7$ ways.

After $E_2$ ⁠, the owl never left Cage-I:

P(A \mid E_2) = \frac{{}^8C_2}{{}^8C_2} = 1

Why?

Cage-II then has only parrots, so whichever 2 birds fly back, the owl stays in Cage-I. Every one of the 28 outcomes is favourable.

Probability that the owl is still in Cage-I

\begin{aligned} P(A) &= P(E_1 \cap A) + P(E_2 \cap A) \\ &= P(E_1)P(A \mid E_1) \\ &\quad + P(E_2)P(A \mid E_2) \\ &= \frac{5}{15}\cdot\frac{7}{28} + \frac{10}{15}\cdot 1 \\ &= \frac{35}{420} + \frac{280}{420} = \frac{315}{420} \end{aligned}

Why?

Theorem of total probability: A happens either with $E_1$ or with $E_2$ ⁠, and

P(E_i \cap A) = P(E_i)P(A \mid E_i)

$15 \times 28 = 420$ ⁠, and $\frac{10}{15} = \frac{280}{420}$ ⁠. Finally $\frac{315}{420} = \frac34$ ⁠, dividing top and bottom by 105.

\begin{aligned} \frac{5}{15}\cdot\frac{7}{28} &= \frac{5 \times 7}{15 \times 28} = \frac{35}{420} \\[4pt] \frac{10}{15}\cdot 1 &= \frac{10 \times 28}{15 \times 28} = \frac{280}{420} \\[4pt] \frac{35}{420} + \frac{280}{420} &= \frac{35 + 280}{420} = \frac{315}{420} \\[4pt] \frac{315}{420} &= \frac{315 \div 105}{420 \div 105} = \frac34 \end{aligned}

$\therefore$ The probability that the owl is still in Cage-I is $\dfrac34$ ⁠.

(ii) The required probability is $P(E_1 \mid A)$ ⁠. By Bayes' theorem,

Why?

Now we know the result (the owl is in Cage-I) and ask about the cause (which birds flew out first). Reversing a conditional probability like this is the job of Bayes' theorem:

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

with $P(A)$ from part (i).

\begin{aligned} &P(E_1 \mid A) \\ &= \frac{P(E_1 \cap A)}{P(E_1 \cap A) + P(E_2 \cap A)} \\ &= \frac{\dfrac{35}{420}}{\dfrac{315}{420}} = \frac{35}{315} \end{aligned}

Why?

The 420s cancel, and $\frac{35}{315} = \frac19$ ⁠, dividing top and bottom by 35.

\begin{aligned} \frac{35}{420} \div \frac{315}{420} &= \frac{35}{420} \times \frac{420}{315} \\ &= \frac{35}{315} \\ &= \frac{35 \div 35}{315 \div 35} = \frac19 \end{aligned}

The answer is small: most of the time the owl is in Cage-I because it never left (280 of the 315 parts).

$\therefore$ The probability that one parrot and the owl flew from Cage-I to Cage-II is $\dfrac19$ ⁠.

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: The owl can end up in Cage-I by two routes: the owl and a parrot fly out ( $\frac{5}{15}$ ⁠) and the owl flies back ( $\frac{7}{28}$ ⁠), or two parrots fly out ( $\frac{10}{15}$ ⁠) and the owl never leaves. Add the two routes: $\frac34$ ⁠.

(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 owl is already seen in Cage-I, and you want the chance the first flight was the owl and a parrot. Divide that route, $\frac{35}{420}$ ⁠, by the total $\frac{315}{420}$ from (i).

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