Example. A youth radio station, to attract listeners, holds prize and super-prize draws among them. Prizes are drawn over sixteen hours (one prize per hour), and super-prizes are drawn during four evening hours (also one super-prize per hour). The probability that listeners will win a prize is 0.3, and a super-prize is 0.02.

Find the probability that over 30 days

a) listeners will win three super-prizes;

b) listeners will win between 130 and 160 prizes.
Solution. Super prize draws represent independent trials with two outcomes: if the listeners win a super prize - this is a success, if the super prize does not go to anyone - this is a failure. The probability of success in all trials is the same. A similar situation applies to prize draws. Therefore, in both cases (a) and (b), the Bernoulli formula applies:

$$
\mathrm{P}\left(\mu_{n}=k\right)=C_{n}^{k} p^{k} q^{n-k}
$$

where $\mu_{n}$ is the number of successes in $n$ trials.

In case (a) $n=30 \cdot 4=120, p=0.02$, and we are interested in the probability $\mathrm{P}\left(\mu_{n}=k\right)$ for $k=3:$

$$
\mathrm{P}\left(\mu_{n}=3\right)=C_{120}^{3} \cdot 0.02^{3} \cdot 0.98^{117}
$$

In case (b) $n=30 \cdot 16=480, p=0.3$, and we are interested in the probability

$$
\mathrm{P}\left(130 \leqslant \mu_{n} \leqslant 160\right)=\sum_{k=130}^{160} C_{480}^{k} \cdot 0.3^{k} \cdot 0.7^{480-k}
$$

Case (a).

1. Choose an approximate formula for calculating the value of $C_{120}^{3} \cdot 0.02^{3} \cdot 0.98^{117}$. For this, we determine whether conditions (5) or conditions (6) are met.

Since in case (a) $p=0.02$ and $k=3$ are small, we check conditions (6):

$$
n=120 \gg 9=k^{2}, \quad n p^{2}=0.048 \ll 1, \quad k p=0.06 \ll 1
$$

Therefore, formulas (3) and (4) are applicable.

2. Find the approximate value of $C_{120}^{3} \cdot 0.02^{3} \cdot 0.98^{117}$ using formula (3). We have $a=n p=2.4$ and using a calculator we get:

$$
\frac{2.4^{3}}{3!} e^{-2.4} \approx 0.209
$$

3. Calculate the relative error of formula (3):

$$
\max \left\{k p, \frac{k^{2}}{n-k}+\frac{n p^{2}}{q}\right\}=\max \left\{0.06, \frac{9}{117}+\frac{0.048}{0.98}\right\} \approx 0.126
$$

4. Round the number 0.209 considering the found error of formula (3). We get the approximate value of the desired probability:

$$
\mathrm{P}\left(\mu_{n}=3\right) \approx 0.2
$$

Case (b).

1. Choose an approximate formula for calculating the value of $C_{480}^{k} \cdot 0.3^{k} \cdot 0.7^{480-k}$ for $130 \leqslant k \leqslant 160$. For this, we determine whether conditions (5) or conditions (6) are met.

Since in case (b) the values of $k$ are not small, we check conditions (5):

$$
n p q=480 \cdot 0.3 \cdot 0.7=100.8 \gg 1, \quad n p=144 \approx k \in[130,160]
$$

Therefore, formulas (1) and (2) are applicable.

2. Find the approximate value of $\mathrm{P}\left(130 \leqslant \mu_{n} \leqslant 160\right)$ using formula (2):

$$
\begin{aligned}
\mathrm{P}\left(130 \leqslant \mu_{n} \leqslant 160\right) \approx \Phi\left(\frac{k_{2}-n p}{\sqrt{n p q}}\right)-\Phi\left(\frac{k_{1}-n p}{\sqrt{n p q}}\right)= \\
\quad=\Phi\left(\frac{160-144}{\sqrt{100.8}}\right)-\Phi\left(\frac{130-144}{\sqrt{100.8}}\right) \approx 0.8629
\end{aligned}
$$

3. Calculate the relative error of formula (2):

$$
\frac{2(k-n p)}{n p q}+\frac{1+p q}{10 n p q}=\frac{2 \cdot 16}{100.8}+\frac{1.21}{1008} \approx 0.32
$$

4. Round the number 0.8644 considering the found error of formula (2). We get the approximate value of the desired probability:

$$
\mathrm{P}\left(130 \leqslant \mu_{n} \leqslant 160\right) \approx 0.9
$$

Answer. a) $0.2 ;$ b) 0.9 .

Conditions of the problems.

1. In a lottery, large and small prizes are raffled. The probability that a large prize will fall on a lottery ticket is 0.001, and a small prize - 0.01. 1000 tickets were purchased. Find the probability that

a) there will be 2 large prizes;

b) there will be from 5 to 15 small prizes.

2. Students perform 15 typical calculations in mathematics over two years, each containing 20 problems. The probability of incorrectly solving a separate problem using the computer package SOLVER.VM is 0.01, without the package - 0.2. Find the probability that over 2 years

a) a student who constantly uses the SOLVER.VM package will solve no more than 5 problems incorrectly;

b) a student who does not use the SOLVER.VM package will solve from 50 to 70 problems incorrectly.

3. It is known that left-handers make up 1% on average, and people who are equally proficient with both hands - 10% (the rest are