Question

A random variable $X$ has the following probability distribution:
$X = x$: 0, 1, 2, 3, 4, 5
$P(X=x)$: $k$, $3k$, $5k$, $7k$, $9k$, $11k$
Then the value of $k$ is:

• A. $\frac{1}{36}$
• B. $\frac{1}{18}$
• C. $\frac{1}{12}$
• D. $\frac{1}{6}$

Hint:

The coefficients $1, 3, 5, 7, 9, 11$ are the first 6 odd natural numbers. The sum of the first $n$ odd natural numbers is always $n^2$. Here, $6^2 = 36$, so $36k = 1 \implies k = 1/36$.

Answer 1

The Correct Option is A

Step 1: Rule for a probability table.
All the probabilities in a distribution must add up to exactly 1.

Step 2: List the probabilities.
They are $k, 3k, 5k, 7k, 9k, 11k$.

Step 3: Add them.
\[ k + 3k + 5k + 7k + 9k + 11k = 36k \]

Step 4: Set equal to 1.
\[ 36k = 1 \]

Step 5: Solve for $k$.
\[ k = \frac{1}{36} \]

Step 6: Conclusion.
\[ \boxed{ k = \frac{1}{36} } \]