73. a) Two hunters simultaneously saw a fox and shot at it simultaneously. Suppose that each of these hunters at such a distance usually hits and kills the fox in one out of three cases. What is the probability that the fox will be killed?

b) Solve the same problem for the case of three hunters, assuming that the accuracy of all these hunters is the same as in problem a).

c) Solve the same problem for the case of $n$ hunters.
73. a) The experiment described in the problem involves two hunters shooting at a fox simultaneously. The frequency of hits and misses of one hunter, of course, does not depend on the results of the simultaneous shot of the second: if they shoot at the fox many times simultaneously, the first will hit the target one-third of the time, and the second will hit the target both during the successful shots of the first hunter and during the unsuccessful ones. When calculating probabilities, these two hunters can be replaced by two people, each of whom independently draws a ticket from a box containing two tickets marked "M" (i.e., "miss") and one ticket marked "H" ("hit"). The number of tickets drawn with the mark "H" indicates the number of hits on the fox.

Combining the three outcomes of the first person drawing a ticket with the three such outcomes of the second person drawing, we get the following nine equally likely outcomes of our experiment:

![](https://cdn.mathpix.com/cropped/2024_05_21_68334347b64402497ba5g-238.jpg?height=124&width=660&top_left_y=767&top_left_x=220)

(The first result listed is the result of the first ticket draw). Out of these nine outcomes, five are favorable, i.e., those in which at least one person draws a ticket marked "H" (in our table, these are the outcomes in the third column or the third row). Thus, the probability that the hunters will kill the fox is \(\frac{5}{9}\).

b) Combining each of the nine outcomes of the previous problem with the two independent misses and one hit of the third hunter, we get 27 equally likely results of the simultaneous "shots" of the three hunters. Out of these 27 outcomes, at least one of the first two hunters will hit the fox in \(5 \cdot 3 = 15\) cases (recall that in the previous problem, hits occurred in five out of nine cases); in the remaining 12 cases, both the first and the second hunter will miss, but in one-third of these cases (i.e., in four cases), the third hunter will hit the fox. Thus, at least one hit will occur in \(15 + 4 = 19\) out of 27 equally likely outcomes, and the desired probability is \(\frac{19}{27}\).

c) In the case of simultaneous shots by \(n\) hunters, we similarly to cases a) and b) get \(3^n\) equally likely outcomes of the experiment. However, counting the number of outcomes corresponding to at least one hunter hitting the fox is significantly more complex than in the simplest cases considered above. Nevertheless, such a calculation can be performed (it is similar to the calculation carried out in the first solution of problem 78a)). However, we will not present it here, as the problem can be solved more simply using the following reasoning. We will count the number of outcomes corresponding to all \(n\) hunters missing. The number of outcomes corresponding to at least one of them hitting will be the difference between the total number of equally likely outcomes and this number. But the number of outcomes corresponding to all hunters missing is already simple to count: the first hunter, obviously, out of \(3^n\) shots will miss in \(\frac{2}{3} \cdot 3^n = 2 \cdot 3^{n-1}\) cases. Out of these \(2 \cdot 3^{n-1}\) unsuccessful shots for the first hunter, the second hunter will miss in \(\frac{2}{3} \cdot 2 \cdot 3^{n-1} = 2^2 \cdot 3^{n-2}\) cases. Out of these \(2^2 \cdot 3^{n-2}\) unsuccessful shots for the first two hunters, the third hunter will miss in \(\frac{2}{3} \cdot 2^2 \cdot 3^{n-2} = 2^3 \cdot 3^{n-3}\) cases. Continuing this reasoning, we conclude that all \(n\) hunters will miss simultaneously in \(2^n\) cases out of \(3^n\). Thus, the number of outcomes corresponding to at least one hunter hitting is \(3^n - 2^n\), and the desired probability in this problem is \(\frac{3^n - 2^n}{3^n} = 1 - \left(\frac{2}{3}\right)^n\).