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 sum of probabilities for all possible outcomes of a random variable must be 1.
Given that the possible values of $X$ are 0, 1, and 2, with probabilities $k$, $2k$, and $2k$ respectively.
Therefore, the total probability is $k + 2k + 2k = 5k$.
Setting the total probability to 1: $5k = 1 \Rightarrow k = \dfrac{1}{5}$.
This corresponds to option (B). Let's verify.
The probabilities are $P(X=0)=k$, $P(X=1)=2k$, and $P(X=2)=2k$.
The sum $k + 2k + 2k = 5k$ remains correct.
Thus, $5k = 1 \Rightarrow k = \dfrac{1}{5}$.
The correct answer is (B): $\dfrac{1}{5}$.