Three coins are tossed into the air simultaneously and for each coin it is recorded whether an eagle or a sun appears. What is the probability that each of the following events will occur?
a) What is get two eagles.
b) That at least one eagle is obtained.
c) That there is no eagle.
a) There are $2^3=8$ possible outcomes when three coins are tossed. To get two eagles, we can have either EEE, EES, ESE, or SEE (where E represents an eagle and S represents a sun). So there are four outcomes that satisfy this condition. Thus, the probability is $\frac{4}{8}=\boxed{\frac{1}{2}}$.
b) We can find the probability of the complement event (that no eagles are obtained) and subtract from 1. To get no eagles, we must have all three coins showing suns, which has probability $\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}$. So the probability of getting at least one eagle is $1-\frac{1}{8}=\boxed{\frac{7}{8}}$.
c) To get no eagles, all three coins must show suns. This has probability $\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}$. Thus, the probability of getting no eagles is $\boxed{\frac{1}{8}}$.