Key Concepts and Formulas

• Axiom of Total Probability for Discrete Sample Spaces: For a discrete sample space $S = \{w_1, w_2, w_3, \dots\}$, the sum of probabilities of all elementary outcomes must be equal to 1. That is, $\sum_{n=1}^{\infty} P(w_n) = 1$.
• Sum of an Infinite Geometric Series: The sum of an infinite geometric series $a + ar + ar^2 + \dots$ is given by the formula $\frac{a}{1-r}$, provided that the common ratio $|r|<1$.
• Linear Combinations of Integers (Frobenius Coin Problem variant): Determining which integers can be expressed in the form $ax+by$ for given positive integers $a, b$ and non-negative (or positive) integers $x, y$. The smallest integer that cannot be expressed in this form (with $x,y \ge 0$) is called the Frobenius number. Here, the constraint $k,l \in \mathbb{N}$ means $k,l \ge 1$.

Step-by-Step Solution

Step 1: Determine the Probability Distribution $P(w_n)$

• What we are doing: We need to find an explicit formula for $P(w_n)$ for any $n \ge 1$.
• Why we are doing it: This is the foundational step to calculate $P(B)$, as $P(B)$ is a sum of these individual probabilities.

1. Establish the relationship between consecutive probabilities: We are given the recurrence relation $P(w_n) = \frac{P(w_{n-1})}{2}$ for $n \ge 2$. This implies that the probability of each outcome is half the probability of the preceding outcome, forming a geometric progression.

2. Express $P(w_n)$ in terms of $P(w_1)$: Using the recurrence relation repeatedly: $P(w_2) = \frac{P(w_1)}{2}$ $P(w_3) = \frac{P(w_2)}{2} = \frac{P(w_1)/2}{2} = \frac{P(w_1)}{2^2}$ Following this pattern, the general formula for $P(w_n)$ is: $P(w_n) = \frac{P(w_1)}{2^{n-1}}$ This expression allows us to find $P(w_1)$ using the total probability axiom.

3. Use the sum of probabilities to find $P(w_1)$: According to the axiom of total probability, the sum of probabilities of all elementary outcomes must be 1: $\sum_{n=1}^{\infty} P(w_n) = 1$ Substitute the expression for $P(w_n)$: $\sum_{n=1}^{\infty} \frac{P(w_1)}{2^{n-1}} = 1$ Factor out $P(w_1)$: $P(w_1) \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n-1} = 1$ The sum $\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n-1}$ is an infinite geometric series with the first term $a = \left(\frac{1}{2}\right)^{1-1} = 1$ (for $n=1$) and common ratio $r = \frac{1}{2}$. The sum of this series is $\frac{a}{1-r} = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$. Substituting this sum back into the equation: $P(w_1) \cdot 2 = 1 \implies P(w_1) = \frac{1}{2}$

4. Write the general formula for $P(w_n)$: Substitute $P(w_1) = \frac{1}{2}$ back into the general expression for $P(w_n)$: $P(w_n) = \frac{1/2}{2^{n-1}} = \frac{1}{2 \cdot 2^{n-1}} = \frac{1}{2^n}$ This is the explicit probability for any elementary outcome $w_n$.

Step 2: Characterize the Set of Indices A

• What we are doing: We need to precisely define which natural numbers $n$ belong to the set $A = \{2k + 3l : k, l \in \mathbb{N}\}$, where $\mathbb{N} = \{1, 2, 3, \dots\}$.
• Why we are doing it: The event $B$ is defined by these indices, so correctly identifying them is crucial for summing the probabilities.

1. Understand the definition and constraints of set A: The set $A$ contains integers $n$ that can be expressed as $2k+3l$, where $k$ and $l$ must be positive integers (i.e., $k \ge 1, l \ge 1$).

2. Find the smallest element in A: The smallest possible value for $2k+3l$ occurs when $k=1$ and $l=1$: $2(1) + 3(1) = 5$. Thus, any integer $n \in A$ must be greater than or equal to 5.

3. Test small integers to identify elements of A:

   • For $n=5$: $5 = 2(1)+3(1)$. So $5 \in A$.
   • For $n=6$: Can $6 = 2k+3l$ for $k,l \ge 1$?
     • If $l=1$, then $6 = 2k+3 \implies 2k=3 \implies k=3/2$, which is not an integer.
     • If $l \ge 2$, then $3l \ge 6$. This implies $2k = 6-3l \le 0$, which means $k \le 0$. But $k$ must be $\ge 1$. Therefore, $6$ cannot be expressed in the form $2k+3l$ with $k,l \ge 1$. So $6 \notin A$.
   • For $n=7$: $7 = 2(2)+3(1)$. So $7 \in A$. (Here $k=2, l=1$)
   • For $n=8$: $8 = 2(1)+3(2)$. So $8 \in A$. (Here $k=1, l=2$)
   • For $n=9$: $9 = 2(3)+3(1)$. So $9 \in A$. (Here $k=3, l=1$)
   • For $n=10$: $10 = 2(2)+3(2)$. So $10 \in A$. (Here $k=2, l=2$)

4. Characterize the set A: From the above observations, it appears that all integers $n \ge 5$, except for $n=6$, can be expressed in the form $2k+3l$ with $k,l \ge 1$.

   • For odd integers $n \ge 5$: Any odd integer $n \ge 5$ can be written as $n = 2m+1$ for some integer $m \ge 2$. We can then write $n = 2(m-1)+3(1)$. Since $m \ge 2$, $m-1 \ge 1$. So, we have $k=m-1 \ge 1$ and $l=1 \ge 1$. Thus, all odd integers $n \ge 5$ are in $A$.
   • For even integers $n \ge 8$: Any even integer $n \ge 8$ can be written as $n = 2m$ for some integer $m \ge 4$. We can then write $n = 2(m-3)+3(2)$. Since $m \ge 4$, $m-3 \ge 1$. So, we have $k=m-3 \ge 1$ and $l=2 \ge 1$. Thus, all even integers $n \ge 8$ are in $A$. Combining these, all integers $n \ge 5$ are in $A$, with the single exception of $n=6$. Therefore, $A = \{n \in \mathbb{N} : n \ge 5 \text{ and } n \ne 6\}$.

Step 3: Calculate P(B)

• What we are doing: We will sum the probabilities $P(w_n)$ for all $n$ belonging to the set $A$.
• Why we are doing it: This is the direct application of the definition of the probability of an event in a discrete sample space.

1. Express $P(B)$ as a sum: The event $B = \{w_n : n \in A\}$. Therefore, $P(B)$ is the sum of probabilities $P(w_n)$ for all $n \in A$: $P(B) = \sum_{n \in A} P(w_n)$

2. Substitute the explicit forms of $P(w_n)$ and set A: We found $P(w_n) = \frac{1}{2^n}$ and $A = \{n \in \mathbb{N} : n \ge 5 \text{ and } n \ne 6\}$. So, $P(B)$ is the sum of $\frac{1}{2^n}$ for $n=5, 7, 8, 9, \dots$. This sum can be conveniently calculated by summing all terms from $n=5$ to infinity and then subtracting the term for $n=6$ (since $6 \notin A$): $P(B) = \left( \sum_{n=5}^{\infty} \frac{1}{2^n} \right) - \frac{1}{2^6}$

3. Calculate the infinite geometric series: The series $\sum_{n=5}^{\infty} \frac{1}{2^n}$ is an infinite geometric series with:

   • First term $a = \frac{1}{2^5} = \frac{1}{32}$ (when $n=5$).
   • Common ratio $r = \frac{1}{2}$. The sum of this series is $\frac{a}{1-r} = \frac{1/32}{1 - 1/2} = \frac{1/32}{1/2} = \frac{1}{32} \times 2 = \frac{1}{16}$.

4. Complete the calculation for P(B): Now, substitute the sum back into the expression for $P(B)$: $P(B) = \frac{1}{16} - \frac{1}{2^6}$ $P(B) = \frac{1}{16} - \frac{1}{64}$ To subtract these fractions, we find a common denominator, which is 64: $P(B) = \frac{4}{64} - \frac{1}{64}$ $P(B) = \frac{3}{64}$

Common Mistakes & Tips

• Misinterpreting $\mathbb{N}$: In JEE, $\mathbb{N}$ usually denotes the set of natural numbers $\{1, 2, 3, \dots\}$. A common mistake is to assume $k,l \ge 0$, which would change the set $A$. Always confirm the definition of $\mathbb{N}$ if not explicitly stated.
• Incorrect Geometric Series Sum: Ensure the first term 'a' of the geometric series is correctly identified based on the starting index of the sum. For $\sum_{n=5}^{\infty} \frac{1}{2^n}$, the first term is $1/2^5$, not $1/2^1$.
• Arithmetic Errors: Be careful with fraction addition/subtraction. Double-check calculations, especially when dealing with powers of 2.

Summary

The problem required us to first establish the probability distribution for an infinite sample space using a given recurrence relation and the axiom of total probability, which led to $P(w_n) = \frac{1}{2^n}$. Next, we meticulously characterized the set $A$ of indices by analyzing the linear combination $2k+3l$ with the constraint $k,l \in \mathbb{N}$, finding that $A = \{n \in \mathbb{N} : n \ge 5 \text{ and } n \ne 6\}$. Finally, we calculated $P(B)$ by summing the probabilities $P(w_n)$ for all $n \in A$, using the formula for an infinite geometric series and adjusting for the excluded term. The final calculated probability is $\frac{3}{64}$.

The final answer is $\boxed{\frac{3}{64}}$, which corresponds to option (D).