The probability of observing an eagle on a given day is 0.21, and the probability for a hawk is 0.17. 2% of the viewers observe both of these birds of prey on the same day. Are these events independent of each other?

Question

anonymous · Answer

independent Because hawks and eagles are free to roam around and there is no rule that if hawks come out eagles shouldn't or viceversa

anonymous · Answer

Needs to be explained using math

anonymous · Answer

probability of spotting and eagle and a hawk = probability of spotting an eagle * probability of spotting a hawk ===>P(H and E ) = P(H) * P(E) == > 0.21 * 0.17

anonymous · Answer

If they were independent events, then the probability of them both coming out is $$
(0.21)(0.17) = 0.0357
$$However the observed probability is $$
0.02
$$So these events must not be independent.
The probability that an eagle would come out given that a hawk came out is $$
(0.02)/(0.17) = 0.1176
$$Which is much less likely than normally $0.21$.
So maybe these birds know better than to compete.

anonymous · Answer

@Loulita hope this helps.