Question

A hunter's chance of shooting an animal at a distance r is (r²/a²). He fires when r = 2a & if he misses he reloads & fires when r = 3a, 4a,..... If he misses at a distance '50a', the animal escapes. Find the probability that the anmial escapes

Answer 1

51/100

Answer 2

Assume the probability of hitting an animal at distance $r$ is $P(hit|r) = \frac{a^2}{r^2}$ for $r > a$, based on the context of similar problems, as the stated formula $P(hit|r) = \frac{r^2}{a^2}$ yields probabilities greater than 1, which is invalid.

The hunter fires at distances $r = 2a, 3a, 4a, \dots, 50a$. The probability of missing a shot at distance $r$ is $P(miss|r) = 1 - P(hit|r) = 1 - \frac{a^2}{r^2}$. For a shot fired at distance $ka$, the probability of missing is $P(miss|ka) = 1 - \frac{a^2}{(ka)^2} = 1 - \frac{a^2}{k^2 a^2} = 1 - \frac{1}{k^2}$.

The animal escapes if the hunter misses every shot fired at distances 2a, 3a, ..., 50a. Since the shots are independent events given the distances, the probability of the animal escaping is the product of the probabilities of missing at each of these distances. Let $E$ be the event that the animal escapes. $P(E) = P(\text{miss at 2a}) \times P(\text{miss at 3a}) \times \dots \times P(\text{miss at 50a})$ The distances are $ka$ for $k$ ranging from 2 to 50. $P(E) = \prod_{k=2}^{50} P(miss|ka) = \prod_{k=2}^{50} \left(1 - \frac{1}{k^2}\right)$

We can simplify the term inside the product: $1 - \frac{1}{k^2} = \frac{k^2 - 1}{k^2} = \frac{(k-1)(k+1)}{k^2}$. So, the product becomes: $P(E) = \prod_{k=2}^{50} \frac{(k-1)(k+1)}{k^2}$ Expanding the product: $P(E) = \frac{(2-1)(2+1)}{2^2} \times \frac{(3-1)(3+1)}{3^2} \times \frac{(4-1)(4+1)}{4^2} \times \dots \times \frac{(50-1)(50+1)}{50^2}$ $P(E) = \frac{1 \cdot 3}{2 \cdot 2} \times \frac{2 \cdot 4}{3 \cdot 3} \times \frac{3 \cdot 5}{4 \cdot 4} \times \dots \times \frac{49 \cdot 51}{50 \cdot 50}$

We can rearrange the terms in the product: $P(E) = \left(\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \dots \times \frac{49}{50}\right) \times \left(\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \dots \times \frac{51}{50}\right)$

The first part of the product is a telescoping product: $\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \dots \times \frac{49}{50} = \frac{1}{50}$.

The second part of the product is also a telescoping product: $\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \dots \times \frac{51}{50} = \frac{51}{2}$.

Multiplying these two results gives the probability of the animal escaping: $P(E) = \frac{1}{50} \times \frac{51}{2} = \frac{51}{100}$.

The final answer is $\boxed{51/100}$.