Task 1. The germination rate of the given variety is $90 \%$. Determine the probability that out of 400 planted seeds, 350 will germinate.
Solution. Given: $n=400, p=0.9, q=0.1, k=350$. We need to calculate $p_{400}(350)$. Note that it is not easy to numerically implement the formulas of Bernoulli and Poisson. We will use the Laplace formula. First, we calculate $\sqrt{n p q}=\sqrt{400 \cdot 0.9 \cdot 0.1}=6$, then $k-n p=350-400 \cdot 0.9=-10$, next $x=-10: 6=-1.667$, and the value $\varphi(-1.667)=\varphi(1.667)=0.1$ is found from Table 2 in the appendix (interpolation is required using neighboring values). Finally, we have: $p_{400}(350)=0.1: 6=0.0167$.

Answer. 0.0167.

Task 2. The probability of event $A$ in each of 400 independent trials is 0.3. Find the probability of the following events:

1) event $A$ will occur 100 to 130 times inclusive;
2) event $A$ will occur no more than 20 times;
3) event $A$ will occur no less than 200 times.

Solution Given: $n=400, p=0.3, q=0.7$. We find $\sqrt{n p q}=\sqrt{400 \cdot 0.3 \cdot 0.7}=9.17$.

1) We take $k_{1}=100, k_{2}=130$.

We have $x_{1}=\frac{100-0.3 \cdot 400}{9.17}=-2.18 ; x_{2}=1.09 ; p_{400}(100 \leqslant k \leqslant 130)=$ $=\Phi(1.09)-\Phi(-2.18)=0.3621+0.4854=0.8475$. Here we used the oddness of the function $\Phi(x)$.

2) We take $k_{1}=1, k_{2}=20$. We calculate $x_{1}=-13, x_{2}=-10.87$.

We have $p_{400}(1 \leqslant k \leqslant 20) \approx \Phi(-10.87)-\Phi(-13)=0.5+0.5=0$.

3) We take $k_{1}=200, k_{2}=400$. We calculate $x_{1}=8.7 ; x_{2} \approx 30$.

We have $p_{400}(200 \leqslant k \leqslant 400) \approx \Phi(30)-\Phi(8.7)=0$.

Answer. 1) 0.8475; 2) 0; 3) 0.

Task 3. The probability of manufacturing a part of the highest grade on a given machine is 0.4. Find the probability that among 26 randomly selected parts, half will be of the highest grade.

Solution. We need to find the probability $p_{26}(13)$. The answer is given by the local Laplace theorem (Theorem 9). We will provide the necessary calculations:

$p=0.4 ; q=1-0.4=0.6 ; n=26 ; m=13, n p=26 \cdot 0.4=10.4 ;$ $n p q=6.24 ; \sqrt{n p q}=2.5 ; m-n p=13-10.4=2.6 ; x=\frac{m-n p}{\sqrt{n p q}}=$ $=\frac{2.6}{2.5}=1.04 ; \varphi(x)=\varphi(1.04)=1.04$ (from Table 2 in the appendix); $p_{26}(13) \approx$ $\approx \frac{\varphi(x)}{\sqrt{n p q}}=\frac{0.2323}{2.5}=0.093$

Answer. 0.093.

Task 4. The probability of winning on one lottery ticket is 0.2. What are the probabilities that out of 100 tickets, wins will fall on: a) two tickets, b) at least one?

Solution. a) Using formula (25), we find $p_{100}(2)$.

Considering that $n=100, k=2, p=0.02, q=0.98$, we have: $\sqrt{n p q}=\sqrt{100 \cdot 0.02 \cdot 0.98}=1.4 ; x=\frac{k-n p}{\sqrt{n p q}}=\frac{2-100 \cdot 0.02}{1.4}=0$; $p_{100}(2)=\frac{1}{1.4} \varphi(0) \approx 0.2848$

b) The events "a win falls on at least one ticket" $(k \geqslant 1)$ and "a win does not fall on any ticket" ( $k=0$ ) are opposite, so the sum of their probabilities is one. Therefore, $p_{100}(k \geqslant 1)=1-p_{100}(0)$.

The probability $p_{100}(0)$ is found using formula (25) for $k=0, n=100$, $p=0.02 ; x=\frac{k-n p}{\sqrt{n p q}}=\frac{0-100 \cdot 0.02}{1.4} \approx-1.43$. The function $\varphi(x)$ is even,
so $\varphi(-1.43)=\varphi(1.43)=0.1435$. We get $p_{100}(k \geqslant 1)=1-$ $-\frac{1}{1.4} \cdot 0.1435 \approx 0.8072$

Answer. a) 0.2848; b) 0.1435; c) 0.8975