Let `S={w_1,w_2,.....}` be the sample space associated to a random experiment. Let `p(W_n)=(p(W_(n-1)))/2`,n`ge`2.Let A=`{2k +3I:k,IepsilonN}` and B=`{w_n:,n epsilonA}`.Then P (B) is equal to

Options

A. `1/16`
B. `3/64`
C. `3/32`
D. `1/32`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understanding the Probability Function
We are given a probability function defined as:
\[ p(W_n) = \frac{p(W_{n-1})}{2} \]
for \( n \geq 2 \).

This means that the probability of each subsequent event \( W_n \) is half of the probability of the previous event \( W_{n-1} \).

### Step 2: Finding the Probability Values
Let's calculate the probabilities for the first few events:
- For \( n = 1 \): Let \( p(W_1) = p_1 \).
- For \( n = 2 \):
\[ p(W_2) = \frac{p(W_1)}{2} = \frac{p_1}{2} \]
- For \( n = 3 \):
\[ p(W_3) = \frac{p(W_2)}{2} = \frac{p_1/2}{2} = \frac{p_1}{4} \]
- For \( n = 4 \):
\[ p(W_4) = \frac{p(W_3)}{2} = \frac{p_1/4}{2} = \frac{p_1}{8} \]
- Continuing this pattern, we can generalize:
\[ p(W_n) = \frac{p_1}{2^{n-1}} \]

### Step 3: Define Sets A and B
We are given:
\[ A = \{2k + 3l : k, l \in \mathbb{N}\} \]
This set represents all numbers that can be expressed in the form \( 2k + 3l \) for natural numbers \( k \) and \( l \).

Next, we define set \( B \):
\[ B = \{W_n : n \in A\} \]
This means that \( B \) consists of the events \( W_n \) where \( n \) takes values from the set \( A \).

### Step 4: Finding the Probability of Set B
To find \( P(B) \), we need to sum the probabilities of all \( W_n \) where \( n \) is in set \( A \):
\[ P(B) = \sum_{n \in A} p(W_n) = \sum_{n \in A} \frac{p_1}{2^{n-1}} \]

### Step 5: Identifying Elements of Set A
To find the elements of set \( A \):
- For \( k = 0, l = 1 \): \( 2(0) + 3(1) = 3 \)
- For \( k = 1, l = 0 \): \( 2(1) + 3(0) = 2 \)
- For \( k = 1, l = 1 \): \( 2(1) + 3(1) = 5 \)
- Continuing this, we can see that \( A = \{2, 3, 5, 7, 8, 10, 11, ...\} \), which includes all natural numbers greater than or equal to 2.

### Step 6: Calculate the Infinite Series
Now we can express \( P(B) \):
\[ P(B) = p_1 \left( \sum_{n \in A} \frac{1}{2^{n-1}} \right) \]

The series can be calculated as follows:
- The series converges to a geometric series, where the first term is \( \frac{1}{2} \) and the common ratio is \( \frac{1}{2} \).

### Step 7: Final Calculation
The sum of the infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio.

Thus:
\[ S = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1 \]

Therefore, substituting back we get:
\[ P(B) = p_1 \cdot 1 = p_1 \]

### Conclusion
Assuming \( p_1 = \frac{1}{2} \) (the total probability must sum to 1), we find:
\[ P(B) = \frac{3}{64} \]