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If the probability distribution of a discrete random variable XX is given as

X01234
P(X)0.52k3k3k2k

Then the value of kk is:

Solution

✅ Correct Option: 2

Sum of probabilities =1= 1: 0.5+2k+3k+3k+2k=10.5 + 2k + 3k + 3k + 2k = 1, so 0.5+10k=10.5 + 10k = 1, giving 10k=0.510k = 0.5, hence k=120k = \frac{1}{20}.

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