`P(0),P(J)=(1/3)^j , j=1,2,3` then find the value of `P(J)` when J is even and positive integer

Options

A. `1/8`
B. `1/9`
C. `1/64`
D. `1/2`

Answer

To find the value of \( P(J) \) when \( J \) is an even and positive integer, we start with the given probability function:

\[ P(J) = \left( \frac{1}{3} \right)^j \]

where \( j = 1, 2, 3 \). Since we need \( J \) to be an even positive integer, we can consider the values \( J = 2, 4, 6, 8, \ldots \).

### Step 1: Calculate \( P(2) \)
First, we calculate \( P(2) \):

\[ P(2) = \left( \frac{1}{3} \right)^2 = \frac{1}{9} \]

### Step 2: Calculate \( P(4) \)
Next, we calculate \( P(4) \):

\[ P(4) = \left( \frac{1}{3} \right)^4 = \frac{1}{81} \]

### Step 3: Calculate \( P(6) \)
Now, we calculate \( P(6) \):

\[ P(6) = \left( \frac{1}{3} \right)^6 = \frac{1}{729} \]

### Step 4: Calculate \( P(8) \)
Next, we calculate \( P(8) \):

\[ P(8) = \left( \frac{1}{3} \right)^8 = \frac{1}{6561} \]

### Step 5: Sum the probabilities
Now we need to sum these probabilities for all even positive integers:

\[ P(J) = P(2) + P(4) + P(6) + P(8) + \ldots \]

This forms an infinite geometric series where the first term \( a = \frac{1}{9} \) and the common ratio \( r = \left( \frac{1}{3} \right)^2 = \frac{1}{9} \).

### Step 6: Use the formula for the sum of an infinite geometric series
The formula for the sum \( S \) of an infinite geometric series is given by:

\[ S = \frac{a}{1 - r} \]

Substituting the values we have:

\[ S = \frac{\frac{1}{9}}{1 - \frac{1}{9}} = \frac{\frac{1}{9}}{\frac{8}{9}} = \frac{1}{8} \]

### Final Answer
Thus, the value of \( P(J) \) when \( J \) is an even and positive integer is:

\[ \boxed{\frac{1}{8}} \]