125. The probability of an event occurring in each of 100 independent trials is constant and equal to \( p = 0.8 \). Find the probability that the event will occur: a) at least 75 times and no more than 90 times; b) at least 75 times; c) no more than 74 times.
Solution. We will use the integral Laplace theorem:

$$
P_{n}\left(k_{1}, k_{2}\right)=\Phi\left(x^{\prime \prime}\right)-\Phi\left(x^{\prime}\right),
$$

where $\Phi(x)$ is the Laplace function,

$$
x^{\prime}=\left(k_{1}-n p\right) / \sqrt{\overline{n p q}}, \quad x^{\prime \prime}=\left(k_{2}-n p\right) / \sqrt{n p q} .
$$

a) According to the condition, $n=100 ; p=0.8 ; q=0.2 ; k_{1}=75 ; k_{2}=90$. Let's calculate $x^{\prime}$ and $x^{\prime \prime}$ :

$$
\begin{gathered}
x^{\prime}=\frac{k_{1}-n p}{\sqrt{n p q}}=\frac{75-100 \cdot 0.8}{\sqrt{100 \cdot 0.8 \cdot 0.2}}=-1.25 ; \\
x^{\prime \prime}=\frac{k_{2}-n p}{\sqrt{n p q}}=\frac{90-100 \cdot 0.8}{\sqrt{100 \cdot 0.8 \cdot 0.2}}=2.5 .
\end{gathered}
$$

Considering that the Laplace function is odd, i.e., $\Phi(-x)=-\Phi(x)$, we get

$$
P_{100}(75 ; 90)=\Phi(2.5)-\Phi(-1.25)=\Phi(2.5)+\Phi(1.25) .
$$

From the table in Appendix 2, we find:

$$
\Phi(2.5)=0.4938 ; \quad \Phi(1.25)=0.3944 .
$$

The required probability

$$
P_{100}(75 ; 90)=0.4938+0.3944=0.8882 \text {. }
$$

b) The requirement that the event occurs at least 75 times means that the number of occurrences of the event can be 75 or $76, \ldots$, or 100. Thus, in this case, we should take $k_{1}=75, k_{2}=100$. Then

$$
\begin{gathered}
x^{\prime}=\frac{k_{1}-n p}{\sqrt{n p q}}=\frac{75-100 \cdot 0.8}{\sqrt{100 \cdot 0.8 \cdot 0.2}}=-1.25 ; \\
x^{\prime \prime}=\frac{k_{2}-n p}{\sqrt{n p q}}=\frac{100-100 \cdot 0.8}{\sqrt{100 \cdot 0.8 \cdot 0.2}}=5 .
\end{gathered}
$$

From the table in Appendix 2, we find $\Phi(1.25)=0.3944 ; \Phi(5)=0.5$. The required probability

$$
\begin{aligned}
P_{100}(75 ; 100) & =\Phi(5)-\Phi(-1.25)=\Phi(5)+\Phi(1.25)= \\
& =0.5+0.3944=0.8944 .
\end{aligned}
$$

c) The events "A occurs at least 75 times" and "A occurs no more than 74 times" are opposite, so the sum of the probabilities of these events is one. Therefore, the required probability

$$
P_{100}(0 ; 74)=1-P_{100}(75 ; 100)=1-0.8944=0.1056
$$