1571. A regular coin was tossed 4 times. What is the probability of the coat of arms landing twice?
Solution. In this case, $A$ is the occurrence of heads in a separate trial; $p=P(A)=0.5$ and, consequently, $q=1-p=0.5$. Using the Bernoulli formula, we find the probability

$$
P_{2,4}=C_{4}^{2} p^{2} q^{4-2}=\frac{4 \cdot 3}{1 \cdot 2} \cdot(0.5)^{2} \cdot(0.5)^{2}=0.375
$$