Let S = {1, 2, … 100}. The probability of choosing an integer k, `1lekle100` is proportional to log k. The conditional probability of choosing the integer 2, given that an even integer is chosen is

Options

A. `(log2)/(50log2+log(50)!)`
B. `(log2)/(100log2+log(50)!)`
C. `(log2)/(50log2+log(100)!)`
D. `(log2)/(100log2+(100)!)`

Answer

To solve the problem, we need to find the conditional probability of choosing the integer 2 given that an even integer is chosen. Let's break down the solution step by step.

### Step 1: Define the probability of choosing an integer
The probability of choosing an integer \( k \) (where \( 1 \leq k \leq 100 \)) is proportional to \( \log k \). We can express this as:
\[ P(k) = c \cdot \log k \]
where \( c \) is a constant of proportionality.

### Step 2: Identify the events
Let \( A \) be the event of choosing the integer 2, and let \( B \) be the event of choosing an even integer. We want to find \( P(A | B) \), which is the conditional probability of \( A \) given \( B \).

### Step 3: Calculate \( P(A \cap B) \)
The intersection \( A \cap B \) represents the event of choosing the integer 2, which is the only element in both events. Thus:
\[ P(A \cap B) = P(2) = c \cdot \log 2 \]

### Step 4: Calculate \( P(B) \)
The event \( B \) consists of all even integers from 1 to 100. The even integers in this range are:
\[ 2, 4, 6, \ldots, 100 \]
This is an arithmetic sequence where the first term \( a = 2 \) and the last term \( l = 100 \) with a common difference \( d = 2 \).

The number of even integers from 1 to 100 can be calculated as:
\[ n = \frac{100 - 2}{2} + 1 = 50 \]

Now, we can find \( P(B) \):
\[ P(B) = P(2) + P(4) + P(6) + \ldots + P(100) \]
Substituting the probabilities:
\[ P(B) = c \cdot \log 2 + c \cdot \log 4 + c \cdot \log 6 + \ldots + c \cdot \log 100 \]
Factoring out \( c \):
\[ P(B) = c \left( \log 2 + \log 4 + \log 6 + \ldots + \log 100 \right) \]

### Step 5: Simplify \( P(B) \)
Using the property of logarithms \( \log a + \log b = \log(ab) \), we can combine the logs:
\[ P(B) = c \cdot \log(2 \cdot 4 \cdot 6 \cdots 100) \]

The product \( 2 \cdot 4 \cdot 6 \cdots 100 \) can be expressed as:
\[ 2^{50} \cdot (1 \cdot 2 \cdot 3 \cdots 50) = 2^{50} \cdot 50! \]
Thus:
\[ P(B) = c \cdot \log(2^{50} \cdot 50!) = c \left( 50 \log 2 + \log(50!) \right) \]

### Step 6: Calculate \( P(A | B) \)
Now we can find the conditional probability:
\[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{c \cdot \log 2}{c \left( 50 \log 2 + \log(50!) \right)} \]
The \( c \) cancels out:
\[ P(A | B) = \frac{\log 2}{50 \log 2 + \log(50!)} \]

### Final Answer
Thus, the conditional probability of choosing the integer 2 given that an even integer is chosen is:
\[ \boxed{\frac{\log 2}{50 \log 2 + \log(50!)}} \]