Question

A random variable \(X\) has the following probability mass function as follows:

Then the value of \(a\) is

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

For any probability mass function, always use the condition: \[ \sum P(X=x)=1 \]

Answer 1

The Correct Option is B

Concept:
For a probability mass function, the sum of all probabilities must be equal to \(1\). \[ \sum P(X=x)=1 \] Here, the probabilities are: \[ \frac{a}{6},\quad \frac{a}{4},\quad \frac{a}{12} \] So, their sum must be: \[ \frac{a}{6}+\frac{a}{4}+\frac{a}{12}=1 \]

Step 1: Add all probabilities.
\[ \frac{a}{6}+\frac{a}{4}+\frac{a}{12}=1 \] Take LCM of \(6,4,12\), which is \(12\). \[ \frac{2a}{12}+\frac{3a}{12}+\frac{a}{12}=1 \] \[ \frac{2a+3a+a}{12}=1 \] \[ \frac{6a}{12}=1 \] \[ \frac{a}{2}=1 \]

Step 2: Solve for \(a\).
\[ \frac{a}{2}=1 \] Multiplying both sides by \(2\): \[ a=2 \] Hence, the correct answer is: \[ \boxed{(B)\ 2} \]