80. A coin is flipped 10 times. What is the probability that heads will appear exactly three times?
Solution. Let $A_{10.3}$ be the event that a coin lands heads up three times out of ten tosses. In this case, the probability of landing heads is $\frac{1}{2}$, i.e., $p=\frac{1}{2}$. Then $q=1-\frac{1}{2}=$ $=\frac{1}{2}$. Substituting these values into the Bernoulli formula, we get

$$
\begin{aligned}
P\left(A_{10.3}\right)= & C_{10}^{3} \cdot\left(\frac{1}{2}\right)^{3} \cdot\left(\frac{1}{2}\right)^{7}=\frac{10!}{3!\cdot 7!} \cdot\left(\frac{1}{2}\right)^{10}= \\
& =\frac{7!\cdot 8 \cdot 9 \cdot 10}{7!\cdot 1 \cdot 2 \cdot 3 \cdot 2^{10}}=\frac{8 \cdot 3 \cdot 5}{1024}=\frac{15}{128}
\end{aligned}
$$