74. A hunter shoots at a fox running away from him for the first time from a distance of $100 \mu$; let the probability of hitting the fox from this distance be ${ }^{1} / 2$ (i.e., from a distance of $100 m$ the hunter hits the running fox as often as he misses). In the event of a miss, the hunter reloads his gun and shoots again, but by this time the fox has moved away another 50 m. In the case of a second miss, he reloads the gun and shoots a third (and final) time; during this time, the fox moves away another 50 m. Assuming the probability of hitting is inversely proportional to the square of the distance, determine the probability that the hunter will succeed in hitting the fox.
74. First, let's calculate the probability of hitting the target on the second and third shots. The second shot is taken from a distance of $150 \mu$, and the third shot from a distance of $200 \mu$. Since the probability of hitting the target is inversely proportional to the square of the distance, and the probability of hitting the target from a distance of 100 m is $\frac{1}{2}$, the probability of hitting the target on the second shot will be $\frac{1}{2}\left(\frac{100}{150}\right)^{2}=\frac{1}{2} \cdot \frac{4}{9}=\frac{2}{9}, \quad$ and on the third shot - $\frac{1}{2}\left(\frac{100}{200}\right)^{2}=\frac{1}{8}$.

From this, it follows that our problem is equivalent to the following. A person randomly draws a ball from a box containing one white and one black ball. If the drawn ball is black, he draws a ball from another box containing two white and $9-2=7$ black balls. Finally, if this ball is also black, he draws a ball from a third box containing one white and $8-1=7$ black balls. What is the probability that this person will draw a white ball at least once? This probability will be exactly equal to the probability that the hunter will shoot the fox.

We will assume that regardless of the result of the first and second draws, the person will still draw a ball from all three boxes (this assumption is similar to those made in solving problems 65 and 71, and serves the same purpose). Drawing a ball from the first box can have two equally likely outcomes, one of which results in a white ball and the other in a black ball. Drawing from the second box can have nine equally likely outcomes, two of which result in a white ball and seven in a black ball. Drawing from the third box can have eight equally likely outcomes, only one of which results in a white ball. By combining the two outcomes of the first draw with the nine outcomes of the second and the eight outcomes of the third, we get a total of $2 \cdot 9 \cdot 8=144$ equally likely outcomes of the experiment. The unfavorable outcomes will be the $1 \cdot 7 \cdot 7=49$ outcomes where a black ball is drawn from all three boxes. The remaining $144-49=95$ outcomes will be favorable. Therefore, the probability sought in the problem is

$$
\frac{95^{\circ}}{144} \approx 0.66
$$