A hunter’s chance of shooting an animal at a distance r is `(a^(2))/(r^(2)) ( r gt a)` . He fires when r =2a if he misses, he reloads and fires again where r = 3a . Further if he misses at r = 3a then he tries again at r = 4a . This process continuous till r= na . If he misses at a distance na, the animal escape. Find odd against the event that animal is shot.

Options

A. `n + 1 : 2n `
B. `n + 1 : n -1 `
C. `n - 1 : 2n `
D. `n - 1 : n + 1`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the probability of hitting the animal at various distances.

The probability of hitting the animal at a distance \( r \) is given by:

\[ P(\text{hit at } r) = \frac{a^2}{r^2} \]

### Step 2: Calculate the probabilities for each distance.

1. When \( r = 2a \):
\[ P(\text{hit at } 2a) = \frac{a^2}{(2a)^2} = \frac{a^2}{4a^2} = \frac{1}{4} \]

2. When \( r = 3a \):
\[ P(\text{hit at } 3a) = \frac{a^2}{(3a)^2} = \frac{a^2}{9a^2} = \frac{1}{9} \]

3. When \( r = 4a \):
\[ P(\text{hit at } 4a) = \frac{a^2}{(4a)^2} = \frac{a^2}{16a^2} = \frac{1}{16} \]

4. Continuing this pattern, for \( r = na \):
\[ P(\text{hit at } na) = \frac{a^2}{(na)^2} = \frac{a^2}{n^2a^2} = \frac{1}{n^2} \]

### Step 3: Calculate the probability of missing at each distance.

The probability of missing at each distance is:

1. At \( r = 2a \):
\[ P(\text{miss at } 2a) = 1 - P(\text{hit at } 2a) = 1 - \frac{1}{4} = \frac{3}{4} \]

2. At \( r = 3a \):
\[ P(\text{miss at } 3a) = 1 - P(\text{hit at } 3a) = 1 - \frac{1}{9} = \frac{8}{9} \]

3. At \( r = 4a \):
\[ P(\text{miss at } 4a) = 1 - P(\text{hit at } 4a) = 1 - \frac{1}{16} = \frac{15}{16} \]

4. At \( r = na \):
\[ P(\text{miss at } na) = 1 - P(\text{hit at } na) = 1 - \frac{1}{n^2} = \frac{n^2 - 1}{n^2} \]

### Step 4: Calculate the overall probability of missing all shots.

The overall probability of missing all shots until \( r = na \) is the product of the probabilities of missing at each distance:

\[ P(\text{miss all}) = P(\text{miss at } 2a) \cdot P(\text{miss at } 3a) \cdot P(\text{miss at } 4a) \cdots P(\text{miss at } na) \]

This can be expressed as:

\[ P(\text{miss all}) = \left( \frac{3}{4} \right) \cdot \left( \frac{8}{9} \right) \cdot \left( \frac{15}{16} \right) \cdots \left( \frac{n^2 - 1}{n^2} \right) \]

### Step 5: Calculate the probability of hitting the animal.

The probability of hitting the animal \( P(\text{hit}) \) is given by:

\[ P(\text{hit}) = 1 - P(\text{miss all}) \]

### Step 6: Calculate the odds against the event that the animal is shot.

The odds against the event that the animal is shot can be calculated as:

\[ \text{Odds against hitting} = \frac{P(\text{miss})}{P(\text{hit})} \]

Substituting the values we have:

\[ \text{Odds against hitting} = \frac{P(\text{miss all})}{1 - P(\text{miss all})} \]

### Step 7: Final expression for odds against hitting.

After calculating the probabilities and substituting them back into the odds formula, we find:

\[ \text{Odds against hitting} = \frac{n + 1}{n - 1} \]