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The random variable XX has a probability distribution

P(X=x)={5k,x=02k,x=13k,x=20,otherwise.P(X=x) = \begin{cases} 5k, & x=0 \\ 2k, & x=1 \\ 3k, & x=2 \\ 0, & \text{otherwise.} \end{cases}

Then, the value of E(X)E(X) is :

Solution

✅ Correct Option: 1

Sum of probabilities: 5k+2k+3k=10k=15k+2k+3k=10k=1, so k=0.1k=0.1. E(X)=0(0.5)+1(0.2)+2(0.3)=0.2+0.6=0.8E(X) = 0(0.5) + 1(0.2) + 2(0.3) = 0.2 + 0.6 = 0.8.

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