Probability that the owl is still in cage

Question

Can anybody help me in this I am not getting how to start it

Answer 1

There are $6$ birds and each has $\frac{1}{6}$ chance of flying.

There are $2$ flying events, so for a specific bird to fly, owl in this case, has $2\times\frac{1}{6}= \frac{1}{3}$ chance.

That also means that there is other $\frac{2}{3}$ chance for it to stay in the cage.

If it flew, there is $2\times\frac{1}{8}=\frac{1}{4}$ chance for it to fly back. ($8$ birds in the second cage now)

Chance for it to fly there and back thus is $\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$

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Now we have $2$ cases when the owl is in the first cage, if it stayed or if it returned, which yields:

$$\frac{2}{3}+\frac{1}{12}=\frac{9}{12}=\frac{3}{4}$$