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If XX is a random variable such that XX can take values 0,1, 2 or 3. Then the expectation of XX for the following data is :

XX0123
P(X)P(X)kkk2k^21−5k21-5k^2k2k^2

(where k>0k>0)

Solution

✅ Correct Option: 3

Total probability: k+k2+1−5k2+k2=1k+k^2+1-5k^2+k^2=1 gives k−3k2=0k-3k^2=0, so k=13k=\dfrac13.

Probabilities become 13,19,49,19\dfrac13,\dfrac19,\dfrac49,\dfrac19.

E(X)=0+19+89+39=129=43E(X)=0+\dfrac19+\dfrac89+\dfrac39=\dfrac{12}{9}=\dfrac43.

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