188. Let's denote by $p$ the probability of hitting the target with a single shot. Calculate the probability that the target will be hit, i.e., at least one hit will occur, in $n$ shots.

At the beginning of the section, it was noted that unlikely events can be considered practically impossible, and there are different degrees of such impossibility. Let's clarify this.

If methods of probability theory are applied to study some real phenomenon, each time it is necessary to choose some small number $\varepsilon$ as the allowable probability of error. This means that, when predicting the course of a real phenomenon using methods of probability theory, we are forced to admit some possibility of error in our forecast and only require that the probability of error does not exceed $\varepsilon$. With such an assumption, we have grounds to consider any event with a probability less than $\varepsilon$ as practically impossible, and any event with a probability greater than $1-\varepsilon$ as practically certain. The number $\varepsilon$ can also be called the allowable degree (or level) of uncertainty, and the number $1-\varepsilon$ the required degree (or level) of certainty. It is self-evident that the value of $\varepsilon$ is chosen in each specific problem with regard to practical requirements for the certainty of conclusions. Often used values are 0.01, 0.005, 0.001, 0.0001.

Let's illustrate this with an example of the problem of the number of shots needed to hit the target (see problem 188). Suppose, for definiteness, the probability of hitting the target with each shot is 0.2. How many shots are needed to hit the target? Clearly, no absolutely certain number of shots can be specified: it may happen that the target is hit with the first shot, and at the same time, it is not excluded that, for example, 100 or 200 shots will not result in a single hit. Therefore, let's not strive for absolute certainty and introduce some allowable degree of uncertainty $\varepsilon$. A practically acceptable value, for example, is $\varepsilon$ equal to 0.001. Asserting that the target will be hit with $n$ shots, we will make an error with a probability of $(1-0.2)^n = (0.8)^n$ (see the solution to problem 188). Now let's find $n$ such that

$$
(0.8)^n < 0.001.
$$

The smallest value of $n$ satisfying this inequality is 31 (it is easily found using logarithm tables). Thus, asserting that 31 shots are sufficient to hit the target, we can make an error with a probability not exceeding the allowable limit of 0.001. Therefore, under the accepted requirements for the level of certainty, hitting the target with 31 shots can be considered practically certain.

The number of shots found will change if the allowable degree of uncertainty is changed. In the table below, the values of $n$ corresponding to various allowable levels of uncertainty are given.

| $\boldsymbol{n}$ | $\boldsymbol{n}$ |
| :---: | :---: |
| 0.01 | 21 |
| 0.005 | 24 |
| 0.001 | 31 |
| 0.0001 | 42 |

Problem. Calculate the probability that a 6 will never appear in an unlimited number of rolls of a die.

Solution. First, let's calculate the probability of the event $B_n$ "a 6 does not appear in $n$ rolls." In solving problem 187, we found that $\mathbf{P}(B_6) = \left(\frac{5}{6}\right)^6$. Similarly, for any $n$ (by formula (6))

$$
\mathbf{P}(B_n) = \left(\frac{5}{6}\right)^n
$$

Let's denote the event "a 6 never appears in an unlimited number of rolls of the die" by $B$. From the event $B$, each of the events $B_1, B_2, \ldots, B_n, \ldots$ follows. Therefore, by the first property of probability,

$$
\begin{aligned}
& \mathbf{P}(B) \leqslant \mathbf{P}(B_1) = \frac{5}{6} \\
& \mathbf{P}(B) \leqslant \mathbf{P}(B_2) = \left(\frac{5}{6}\right)^2 \\
& \cdots \cdot \cdot \cdot \cdot \\
& \mathbf{P}(B) \leqslant \mathbf{P}(B_n) = \left(\frac{5}{6}\right)^n
\end{aligned}
$$

But the numbers $\frac{5}{6}, \left(\frac{5}{6}\right)^2, \ldots, \left(\frac{5}{6}\right)^n, \ldots$ as terms of a decreasing geometric progression become smaller than any preassigned positive number as $n$ increases. Therefore, $\mathbf{P}(B)$ is also smaller than any positive number, i.e., $\mathbf{P}(B) = 0$.

Thus, the probability that a 6 will never appear in an unlimited number of rolls of the die is zero.

Events we have encountered so far were either impossible or, if possible, had a probability greater than zero. Here we encounter an event with a probability of zero and yet it seems logically possible. We could not have obtained such a result if the probability of our event could be calculated by the rule given at the beginning of the section, namely as the ratio of the number of favorable outcomes of the experiment to the total number of all possible equally likely outcomes.

1) A. P. Kiselev, Algebra, part 2, 23rd edition, p. 78.

The result we obtained in solving our problem can also be interpreted as follows: no matter how high a degree of certainty we desire, we can specify a number of rolls such that the appearance of a 6 at least once will be practically certain. This is the real meaning of the assertion that a 6 will appear with probability 1 in an unlimited number of rolls of the die.

The formula for total probability. A system of events $A_1, A_2, A_3, \ldots, A_n$ is called complete if at least one of them will definitely occur (in other words, if the event $A_1 + A_2 + A_3 + \ldots + A_n$ is certain). Note that if the events $A_1, A_2, \ldots, A_n$ form a complete system and are pairwise mutually exclusive, then

$$
\mathbf{P}(A_1) + \mathbf{P}(A_2) + \ldots + \mathbf{P}(A_n) = 1
$$

(follows from formula (5) and property 4).

Property 7 (formula for total probability). Let a complete system of pairwise mutually exclusive events $A_1, A_2, A_3, \ldots, A_n$ be given. Then the probability of any event $B$ can be found by the formula

$$
\begin{array}{r}
\mathbf{P}(B) = \mathbf{P}(A_1) \mathbf{P}(B / A_1) + \mathbf{P}(A_2) \mathbf{P}(B / A_2) + \ldots \\
\ldots + \mathbf{P}(A_n) \mathbf{P}(B / A_n)
\end{array}
$$

To prove this, note that since one of the events $A_1, A_2, \ldots, A_n$ will definitely occur, the occurrence of event $B$ is equivalent to the occurrence of one of the events $B A_1, B A_2, \ldots, B A_n$; therefore,

$$
\mathbf{P}(B) = \mathbf{P}((A_1 B) + (A_2 B) + \ldots + (A_n B))
$$

Since the events $A_1 B, A_2 B, \ldots, A_n B$ are pairwise mutually exclusive (since $A_1, A_2, \ldots, A_n$ are pairwise mutually exclusive), by formula (5)

$$
\mathbf{P}(B) = \mathbf{P}(A_1 B) + \mathbf{P}(A_2 B) + \ldots + \mathbf{P}(A_n B)
$$

Using property 6, we get

$$
\mathbf{P}(B) = \mathbf{P}(A_1) \math
188. Let the desired probability be denoted by $x$. Then the probability of no hits in $n$ shots is $1-x$, and for one shot it is 1 - p (property 3). As in the previous problem, by formula (6)

$$
1-x=(1-p)^{n}
$$

from which

$$
x=1-(1-p)^{n}
$$