129. The probability of an event occurring in each of the independent trials is 0.8. How many trials need to be conducted to expect with a probability of 0.9 that the event will occur at least 75 times?
Solution. According to the condition, $p=0.8 ; q=0.2 ; k_{i}=75 ; k_{2}=n$; $P_{n}=(75, n)=0.9$.

Let's use the integral theorem of Laplace:

$$
P_{n}\left(k_{1} ; n\right)=\Phi\left(x^{\prime \prime}\right)-\Phi\left(x^{\prime}\right)=\Phi\left[\frac{k_{2}-n p}{\sqrt{n p q}}\right]-\Phi\left[\frac{k_{1}-n p}{\sqrt{n p q}}\right]
$$

Substituting the data from the problem, we get

$$
0.9=\Phi\left[\frac{n-0.8 n}{\sqrt{n \cdot 0.8 \cdot 0.2}}\right]-\Phi\left[\frac{75-0.8 n}{\sqrt{n \cdot 0.8 \cdot 0.2}}\right]
$$

or

$$
0.9=\Phi\left[\frac{\sqrt{n}}{2}\right]-\Phi\left[\frac{75-0.8 n}{0.4 \sqrt{n}}\right]
$$

Obviously, the number of trials $n>75$, so $\sqrt{n / 2}>\sqrt{75} / 2 \simeq$ $\simeq 4.33$. Since the Laplace function is increasing and $\Phi(4) \simeq 0.5$, we can set $\Phi(\sqrt{n} / 2)=0.5$. Therefore,

$$
0.9=0.5-\Phi\left[\frac{75-0.8 n}{0.4 \sqrt{n}}\right]
$$

Thus,

$$
\Phi\left[\frac{75-0.8 n}{0.4 \sqrt{n}}\right]=-0.4
$$

From the table in Appendix 2, we find $\Phi(1.28)=0.4$. Hence, from the relation (*), considering that the Laplace function is odd, we get

$$
(75-0.8 n) /(0.4 \sqrt{n})=-1.28
$$

Solving this equation as a quadratic equation in terms of $\sqrt{n}$, we get $\sqrt{n}=10$. Therefore, the required number of trials $n=100$.