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Miscellaneous 7 (I) · Q44
Q.

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

X−2−10123
P (X = x)0.1k0.22k0.3k

then P (X = −1) =

(A) 1/10

(B) 2/10

(C) 3/10

(D) 4/10

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Summing to 1: 0.1+k+0.2+2k+0.3+k=1⇒0.6+4k=1⇒k=0.10.1+k+0.2+2k+0.3+k=1\Rightarrow0.6+4k=1\Rightarrow k=0.1. So $P(X=-1 …

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