Problem 5. The probability that an item has not passed the inspection, $p=0.2$. Find the probability that among 400 randomly selected items, between 70 and 100 items will not be inspected.
Solution. The answer to the question is given by formula (26) of the integral Laplace theorem (Theorem 10). We have: \( n=400 ; k_{1}=70 ; k_{2}=100 ; p=0.2 \); \( q=0.8 \). Therefore, \( x_{1}=-1.25 ; x_{2}=2.5 \) and \( p_{400}(70,100)=\Phi\left(x_{2}\right)-\Phi\left(x_{1}\right)= \Phi(2.5)+\Phi(1.25) \approx 0.4938+0.3944=0.8882 \). The property of the odd function \(\Phi(x)\) and the table of its values were used.

Answer: 0.8882

## Exercises

1. The probability of event \( A \) occurring in each of 243 independent trials is 0.25. Find: a) the probabilities that event \( A \) will occur: exactly 70 times; less than 70 times; b) the most probable number of occurrences of event \( A \) in these trials and the probability of this number.
2. The probability that a planted sapling will survive in a given area is 0.7. 2100 saplings were planted. Find the probability that from 1470 to 1500 saplings will survive.

## Answers

## 1. a) 0.0234; 0.915; b) 61; 0.4.2. 0.4236

## § 10. Probability of Deviation of Relative Frequency from Constant Probability of an Event

Let \( n \) independent trials be conducted, in each of which the probability of event \( A \) occurring is constant and equal to \( p(0<p<1) \). Then the probability that the deviation of the relative frequency \( \frac{m}{n} \) from the constant probability \( p \) does not exceed a given number \( \varepsilon \) in absolute value is calculated by the formula

\[
P\left(\left|\frac{m}{n}-p\right| \leqslant \varepsilon\right) \approx 2 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right)
\]

## Problems with Solutions