Let \(E_1\) be the event that one parrot and one owl flew from Cage-I
\(E_2\) 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.
\(=\left(5_{C_1} \times 1_{C_1}\right)\left(7_{C_1} \times 1_{C_1}\right)+\left(5_{C_1} \times 1_{C_1}\right)\left(7_{C_2}\right)+\left(5_{C_2}\right)\left(8_{C_2}\right)\)
Probability that the owl is still in Cage-I \(=\mathrm{P}\left(E_1 \cap A\right)+\mathrm{P}\left(E_2 \cap A\right)\)
\(\frac{\left(5_{C_1} \times 1_{C_1}\right)\left(7_{C_1} \times 1_{C_1}\right)+\left(5_{C_2}\right)\left(8_{C_2}\right)}{\left(5_{C_1} \times 1_{C_1}\right)\left(7_{C_1} \times 1_{C_1}\right)+\left(5_{C_1} \times 1_{C_1}\right)\left(7_{C_2}\right)+\left(5_{C_2}\right)\left(8_{C_2}\right)}\)
\(=\frac{35+280}{35+105+280}\)
\(=\frac{315}{420}\)
\(=\frac{3}{4}\)
(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 \(P\left(E_1 / A\right)\)
\(P\left(E_1 / A\right)=\frac{\mathrm{P}\left(E_1 \cap A\right)}{\mathrm{P}\left(E_1 \cap A\right)+\mathrm{P}\left(E_2 \cap A\right)}\quad(\text {by Baye's Theorem})\)
\(=\cfrac{\frac{35}{420}}{\frac{315}{420}}\)
\(=\frac{1}{9}\)
