Problem 2. The probability of hitting the ring with a ball in one throw is 0.7. Find the probability that in five throws there will be hits in the ring: a) no more than three; b) no less than four.
Solution. Given: $n=5 ; p=0.7 ; q=1-p=0.3$. We need to calculate $p_{5}(k \leqslant 3)$ and $p_{5}(k \geqslant 4)$.

First, let's find $p_{5}(k \geqslant 4)$. We have

$$
\begin{aligned}
& p_{5}(k \geqslant 4)=p_{5}(4)+p_{5}(5)=C_{5}^{4} \cdot 0.7^{4} \cdot 0.3+C_{5}^{5} \cdot 0.7^{5} \cdot 0.3^{0}= \\
&=5 \cdot 0.2401 \cdot 0.3+0.1681=0.5282
\end{aligned}
$$

The events "not more than 3 hits" and "not less than 4 hits" are mutually exclusive, so $p_{5}(k \leqslant 3)=1-p_{5}(k \geqslant 4)=1-0.5282=$ $=0.4718$.

Answer. a) 0.5282 ; b) 0.4718.