For what value of `k` the following distribution is a probability distribution? `X=x_i :\ 0 \ \ 1 \ \ 2 \ \ 3` `P(X=x_i):2k^4 \ \ 3k^2-5k^3 \ \ 2k-3k^2 \ \ 3k-1`

Answer

The values of k are:`-1,1,2,1/2`

Explanation

To have distribution of probabilities becomes a probability distribution, then the Sum of probability distribution is `1`, then,


`Sigma P(x_i)=1`


`P(X=0)+P(X=1
)+P(X=2)+P(X=3)=1`


`rArr (2k^4) + (3k^2-5k^3) + (2k-3k^2) + (3k-1)=1`


`2k^4-5k^3+5k-1=1`


`2k^4-5k^3+5k-2=0`


Thus, `k=1` satisfy the equation, then on simplifying,

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