Question

A discrete random variable X has the following probability distribution:

X	1	2	3	4	5	6
P(X)	\(\frac{2}{k}\)	\(\frac{4}{k}\)	\(\frac{1}{k}\)	\(\frac{2}{k}\)	\(\frac{3}{k}\)	\(\frac{5}{k}\)

The value of k is:

• A. \(\frac{2}{17}\)
• B. 17
• C. 5
• D. \(\frac{1}{17}\)

Answer 1

The Correct Option is B

The sum of the probabilities for a discrete random variable must be equal to 1. Given the probabilities:

X	1	2	3	4	5	6
P(X)	\(\frac{2}{k}\)	\(\frac{4}{k}\)	\(\frac{1}{k}\)	\(\frac{2}{k}\)	\(\frac{3}{k}\)	\(\frac{5}{k}\)

We find:
\[\frac{2}{k}+\frac{4}{k}+\frac{1}{k}+\frac{2}{k}+\frac{3}{k}+\frac{5}{k}=1\]
Combining all terms:
\[\frac{17}{k}=1\]
To solve for \(k\), multiply both sides by \(k\):
\[17=k\]
Therefore, the value of \(k\) is 17.