probability about animals being moved

Question

I have this scenario:

1 animal with $30\%$ probability of be moved to Japan.
1 animal with $30\%$ probability of be moved to Japan.
1 animal with $30\%$ probability of be moved to Japan.
1 animal with $30\%$ probability of be moved to Japan.
1 animal with $30\%$ probability of be moved to Japan.
1 animal with $30\%$ probability of be moved to Japan.
1 animal with $30\%$ probability of be moved to China.
1 animal with 30% probability of be moved to Japan.
1 animal with $80\%$ probability of be moved to Brazil.
1 animal with $30\%$ probability of be moved to Japan.
1 animal with $20\%$ probability of be moved to Brazil.
1 animal with $30\%$ probability of be moved to Japan.
1 animal with $50\%$ probability of be moved to Mexico.
1 animal with $30\%$ probability of be moved to Japan.
(...)

Resuming, $10$ animals with $30\%$ of the probability of being moved to Japan.

Is that "right" to expect that $3$ animals gonna be moved to Japan?

The formula is: $30/100 \cdot 10 = 3$

Can I use Binomial Distribution for this scenario? If yes, how to elaborate the formula?

Thanks a lot!

Answer 1

The binomial distribution requires that all events have the same probability. The set-up here is new to me. I see that all the animals going to Japan have the same probability, so if the problem is restricted to them, it is possible. With $n=10$ animals, given the probability and the number $r=3$ that go, the rest is plug-and-chug.