Question

Shalu brought two cages of birds : Cage-I contains 5 parrots and 1 owl and Cage-II contains 6 parrots, as shown. One day Shalu forgot to lock both cages and two birds flew from Cage- I to Cage-II. Then two birds flew back from Cage-II to Cage-I. Birds move one after the other and not simultaneously. Assume that all birds have equal chance of flying, the probability that the Owl is still in Cage-I, is

• A. $1/6$
• B. $1/3$
• C. $2/3$
• D. $3/4$

Answer 1

The correct option is D $3/4$

The above solution is possible if$:$
(a) 2 birds other than the owl flew from I to II and the same or different birds returned or
(b) Of the 2 birds, the owl was there which flew from I to II and then the owl returned back with one other bird
Thus, required probability (assuming that the birds move one after the other and not simultaneously)$=$
$(\frac{{}_1^{5}C}{{}_1^{6}C}\times \frac{{}_1^{4}C}{{}_1^{5}C}).\left(1\right)+(\frac{{}_1^{5}C}{{}_1^{6}C}\times \frac{{}_1^{1}C}{{}_1^{5}C}\times 2!).(\frac{{}_1^{7}C}{{}_1^{8}C}\times \frac{{}_1^{1}C}{{}_1^{7}C}\times 2!)$ (we multiply $2!$ to decide the order of the parrot and the owl)
$=\frac{4}{6}+\frac{1}{12}=\frac{9}{12}=\frac{3}{4}$
Hence, (D) is correct.