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:

1. 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)
2. 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)

Answer 1

Let E₁ be the event that one parrot and one owl flew from cage-I.

E₂ be the event that two parrots flew from Cage-I.

A be the event that the owl is still in cage-I.

i. Total ways for A to happen:

From cage I, 1 parrot and 1 owl flew and then from cage II, 1 parrot and 1 owl.

Flew back + From cage-I, 1 parrot and 1 owl flew and then from Cage-II, 2 parrots.

Flew back + From cage-I, 2 parrots flew and then from Cage-II, 2 parrots came back.

= `(5_(c_1) xx 1_(c_1))(7_(c_1) × 1_(c_1)) + (5_(c_1) xx 1_(c_1)) + (5_(c_1) xx 1_(c_1))(7_(c_2)) + (5_(c_2))(8_(c_2))`

Probability that the owl is still in cage-I = P(E₁ ∩ A) + P(E₂ ∩ A)

= `((5_(c_1) xx 1_(c_1))(7_(c_1) xx 1_(c_1)) + (5_(c_2))(8_(c_2)))/((5_(c_1) xx 1_(c_1))(7_(c_1) xx 1_(c_1)) + (5_(c_1) xx 1_(c_1))(7_(c_2)) + (5_(c_2))(8_(c_2))`

= `(35 + 280)/(35 + 105 + 280)`

= `315/420`

= `3/4`

ii. The probability that one parrot and the owl flew from Cage-I to Cage-II given that the owl is still in cage-I is `P(E_1/A)`.

`P(E_1/A) = (P(E_1 cap A))/(P(E_1 cap A) + P(E_2 cap A))`   ...(By Baye’s Theorem)

= `(35/420)/(315/420)`

= `1/9`