Question

If a random variable \( X \) has the probability distribution:
\[ P(X = x) = \begin{cases} k, & \text{if } x = 0 \\ 2k, & \text{if } x = 1 \text{ or } 2 \\ 0, & \text{otherwise} \end{cases} \] then the value of \( k \) is

• A. $\dfrac{1}{3}$
• B. $\dfrac{1}{5}$
• C. $\dfrac{1}{6}$
• D. $\dfrac{1}{4}$

Hint:

Always ensure that the sum of all probabilities in a distribution equals 1 to solve for unknown constants.

Answer 1

The Correct Option is C

The total probability of all outcomes of a random variable must equal 1.
Here, possible values of $X$ are 0, 1, and 2 with corresponding probabilities $k$, $2k$, and $2k$.
So, total probability = $k + 2k + 2k = 5k$
Set $5k = 1 \Rightarrow k = \dfrac{1}{5}$
Wait — that gives option (B), but let’s double-check:
Oh! $2k$ was only for $x=1$ or $x=2$ — i.e., $2k$ each.
Then total = $k + 2k + 2k = 5k$ is incorrect.
Actually, it means: $P(0)=k$, $P(1)=2k$, $P(2)=2k$
So total = $k + 2k + 2k = 5k$ still correct.
Thus, $5k = 1 \Rightarrow k = \dfrac{1}{5}$
Correct answer is (B): $\dfrac{1}{5}$