Example 1. The probability of event $A$ occurring in each of 900 independent trials is $p=0.8$. Find the probability that event $A$ will occur: a) 750 times; b) 710 times.
Solution. From the condition, it follows that $n=900, p=0.8$, therefore $q=0.2$; in the first case $k=750$, in the second case $k=710$. Since $n p q=900 \cdot 0.8 \cdot 0.2=144>10$, we can use formula (4.3.1) or (4.3.2)-(4.3.4).

Consider the first case. Using formula (4.3.3), we determine $x$:

$$
x=\frac{750-900 \cdot 0.8}{\sqrt{900 \cdot 0.8 \cdot 0.2}}=\frac{750-720}{\sqrt{144}}=\frac{30}{12}=2.5
$$

From the table of values of the function (4.3.4), we find

$$
\varphi(2.5) \approx 0.0175
$$

According to formula (4.3.2), we obtain the required probability

$$
P_{900}(750) \approx \frac{1}{12} \cdot \varphi(2.5)=\frac{1}{12} \cdot 0.0175 \approx 0.00146
$$

Similarly, we find the probability in the second case, taking into account the evenness of the function (4.3.4):

$$
\begin{gathered}
x=\frac{710-720}{12} \approx-0.83, \varphi(-0.83)=\varphi(0.83) \approx 0.2827 \\
P_{900}(710) \approx \frac{1}{12} \cdot 0.2827 \approx 0.0236
\end{gathered}
$$