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Question

Mathematics Question on Probability

Let the probability distribution of random variable X be

X-2-1123
P(X=x)k2k2kk3k

Then, the value of E(X2) is

A

199\frac{19}{9}

B

133\frac{13}{3}

C

359\frac{35}{9}

D

113\frac{11}{3}

E

73\frac{7}{3}

Answer

133\frac{13}{3}

Explanation

Solution

The correct option is (B): 133\frac{13}{3}