Question

Let \( X \) be a random variable having discrete uniform distribution on \( \{1, 3, 5, 7, \dots, 99\} \). Then \( E(X \mid X \text{ is not a multiple of 15}) \) equals

• A. \( \frac{2365}{47} \)
• B. \( \frac{2365}{50} \)
• C. 50
• D. 47

Hint:

For a uniform distribution, the expected value is the sum of the values divided by the number of values.

Answer 1

The Correct Option is A

Step 1: Describe the random variable.
The random variable $X$ is uniformly distributed over the odd integers from $1$ to $99$.
This set can be written as:

\[ \{1,3,5,7,\ldots,99\} \] 

There are exactly $50$ such values, since the odd numbers follow the pattern $2n+1$ for $n=0$ to $49$.

Step 2: Compute the total sum of all possible values.
The sum of the first $n$ odd numbers is known to be $n^2$.
Here, $n=50$, so the sum of all odd numbers from $1$ to $99$ is:

\[ 1+3+5+\cdots+99 = 50^2 = 2500 \]

Step 3: Identify values that are multiples of 15.
Among these odd numbers, the values that are divisible by $15$ are:

\[ 15,\;45,\;75 \]

There are $3$ such numbers, and their total is:

\[ 15+45+75 = 135 \]

Step 4: Exclude multiples of 15 from the sum.
Removing these values from the overall total gives:

\[ 2500 - 135 = 2365 \]

This is the sum of all odd numbers between $1$ and $99$ that are not divisible by $15$.

Step 5: Compute the conditional expectation.
After excluding the three multiples of $15$, the remaining number of possible values is:

\[ 50 - 3 = 47 \]

Since the distribution is uniform, the conditional expectation is simply the average of these remaining values:

\[ E(X \mid X \text{ is not a multiple of } 15) = \frac{2365}{47} \]