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A random variable X has the following probability distribution:Determine (i) k (ii) `P(X<3)`(iii) `P(X >6)` (iv) `P(0<X<3)` Select the correct answer from above options

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As we know, for any probablity distribution, `sumP(X) = 1` So, for given distribution, `0+k+2k+2k+3k+k^2+2k^2+7k^2+k=1` `10k^2+9k-1=0` `10k^2+10k-k-1=0` `10k(k+1)-1(k+1)=0` `(10k-1)(k+1) = 0` As, k can not be negative, `k = 1/10` (ii) `P(X<3) = P(X=0)+P(X=1)+P(X=2)`<br> ` = 3k = 3/10` (iii) `P(X>6) = P(X=7)` `=7k^2+k = 7/100+1/10 = 17/100` (iv)` P(0ltXlt3)= P(X=1)+P(X=2)` `=3k = 3/10`

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