Question

A hunter's chance of shooting an animal at a distance r is $\frac{a^2}{r^2}$ (r > a). 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

Let $r_k$ be the distance at which the k-th shot is fired. According to the problem, the distances are $r_1 = 2a$, $r_2 = 3a$, ..., $r_n = (n+1)a$. The hunter fires shots at distances 2a, 3a, ..., up to 50a. Let the shots be indexed by the distance. The distances are 2a, 3a, ..., 50a. Let $P(hit|r)$ be the probability of hitting the animal at distance r, which is given as $\frac{a^2}{r^2}$ for $r > a$. The probability of missing the animal at distance r is $P(miss|r) = 1 - P(hit|r) = 1 - \frac{a^2}{r^2}$.

The hunter fires at distance $r_1 = 2a$. If he misses, he fires at $r_2 = 3a$. If he misses again, he fires at $r_3 = 4a$, and so on. He continues this process. The last distance mentioned is 50a. The problem states that if he misses at a distance '50a', the animal escapes. This implies that the sequence of shots is at distances 2a, 3a, ..., 50a.

Let $M_i$ be the event that the hunter misses the shot fired at distance $(i+1)a$. The distances are $(1+1)a, (2+1)a, \dots, (49+1)a$, i.e., 2a, 3a, ..., 50a. The animal escapes if the hunter misses all the shots fired at these distances. Since the hunter only fires the next shot if the previous one is missed, the event that the animal escapes is the event that the hunter misses the shot at 2a AND misses the shot at 3a AND ... AND misses the shot at 50a. Let $E$ be the event that the animal escapes. $E = M_1 \cap M_2 \cap \dots \cap M_{49}$, where $M_i$ is the event of missing the shot at distance $(i+1)a$. The shots are independent events given the distances. Therefore, the probability of the animal escaping is the product of the probabilities of missing each shot in the sequence: $P(E) = P(\text{miss at 2a}) \times P(\text{miss at 3a}) \times \dots \times P(\text{miss at 50a})$.

The probability of missing at distance r is $1 - \frac{a^2}{r^2}$. For the shot at distance 2a, the probability of missing is $1 - \frac{a^2}{(2a)^2} = 1 - \frac{a^2}{4a^2} = 1 - \frac{1}{4}$. For the shot at distance 3a, the probability of missing is $1 - \frac{a^2}{(3a)^2} = 1 - \frac{a^2}{9a^2} = 1 - \frac{1}{9}$. For the shot at distance $(i+1)a$, the probability of missing is $1 - \frac{a^2}{((i+1)a)^2} = 1 - \frac{a^2}{(i+1)^2 a^2} = 1 - \frac{1}{(i+1)^2}$.

The distances are 2a, 3a, ..., 50a. These correspond to values of $(i+1)$ from 2 to 50. So, the probability of the animal escaping is the product: $P(E) = \left(1 - \frac{1}{2^2}\right) \times \left(1 - \frac{1}{3^2}\right) \times \left(1 - \frac{1}{4^2}\right) \times \dots \times \left(1 - \frac{1}{50^2}\right)$.

This is a product of the form $\prod_{n=2}^{N} \left(1 - \frac{1}{n^2}\right)$ with $N=50$. We can simplify the term $1 - \frac{1}{n^2} = \frac{n^2 - 1}{n^2} = \frac{(n-1)(n+1)}{n^2}$. So the product becomes: $P(E) = \prod_{n=2}^{50} \frac{(n-1)(n+1)}{n^2}$ $P(E) = \frac{(2-1)(2+1)}{2^2} \times \frac{(3-1)(3+1)}{3^2} \times \frac{(4-1)(4+1)}{4^2} \times \dots \times \frac{(50-1)(50+1)}{50^2}$ $P(E) = \frac{1 \times 3}{2 \times 2} \times \frac{2 \times 4}{3 \times 3} \times \frac{3 \times 5}{4 \times 4} \times \dots \times \frac{49 \times 51}{50 \times 50}$

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 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 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, we get the probability of the animal escaping: $P(E) = \frac{1}{50} \times \frac{51}{2} = \frac{51}{100}$.

The probability that the animal escapes is $\frac{51}{100}$.