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Miscellaneous 7 (I) · Q45
Q.

If the a d.r.v. X has the following probability distribution :

X1234567
P (X = x)k2k2k3kk²2k²7k² + k

then k =

(A) 1/7

(B) 1/8

(C) 1/9

(D) 1/10

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k+2k+2k+3k+k2+2k2+(7k2+k)=1⇒9k+10k2=1⇒10k2+9k−1=0k+2k+2k+3k+k^2+2k^2+(7k^2+k)=1\Rightarrow9k+10k^2=1\Rightarrow10k^2+9k-1=0. Solving, k=−9+1120=0.1k=\dfrac{-9+11}{20}=0.1 (rejecti …

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