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 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:
$X = x$: 0, 1, 2, 3, 4, 5
$P(X=x)$: $k$, $3k$, $5k$, $7k$, $9k$, $11k$
Then the value of $k$ is:
Show 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 $36k = 1 \implies k = 1/36$.
Updated On: May 31, 2026
- $\frac{1}{36}$
- $\frac{1}{18}$
- $\frac{1}{12}$
- $\frac{1}{6}$
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Concept
For any discrete probability distribution, the sum of all probabilities must equal 1: $\sum P(X=x) = 1$.
Step 2: Meaning
Summing the probabilities for all possible values of $X$ will yield an equation in terms of $k$ that we can solve.
Step 3: Analysis
Summing the probabilities: \[ \sum P(X=x) = k + 3k + 5k + 7k + 9k + 11k \] \[ \sum P(X=x) = 36k \] Since the sum must be 1: \[ 36k = 1 \implies k = \frac{1}{36} \]
Step 4: Conclusion
The value of $k$ is $\frac{1}{36}$. Final Answer: (A)
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