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If the probability distribution of a random variable X is as given below :

X-10123
P(X)K15\frac{1}{5}2K310\frac{3}{10}K

Then the value of K is :

Solution

✅ Correct Option: 4

Sum of probabilities =1= 1: K+15+2K+310+K=1K + \frac{1}{5} + 2K + \frac{3}{10} + K = 1. So 4K=1−12=124K = 1 - \frac{1}{2} = \frac{1}{2}, giving K=18K = \frac{1}{8}.

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