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:
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 $36k = 1 \implies k = 1/36$.