Probability Question - CBSE

Question

The probability distribution of a random variable $X$ is given below:
$X = 0, 1, 2$
$P(X) = k, 2k, 3k$
The value of $k$ is:
(a) $\dfrac{1}{6}$
(b) $\dfrac{1}{3}$
(c) $\dfrac{1}{2}$
(d) $1$

Accepted Answer

$k + 2k + 3k = 1 \Rightarrow 6k = 1 \Rightarrow k = 1/6$. [1.0 Mark]

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🎯 Official CBSE Marking Scheme:
Evaluating $k = 1/6$: 1.0 Mark

Question:

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