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Question

X 1 2 3 4 5 6
P(X=x) k 2k 3k 4k 5k 6k
The probability distribution of discrete r.v. X is
Then P(X4) =

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Solution

Sum of the probabilities is equal to 1

k+2k+3k+4k+5k+6k=1

21k=1

k=121

P(X4)

=P(1)+P(2)+P(3)+P(4)

=k+2k+3k+4k

=121+221+321+421

=1+2+3+421

=1021

P(X4)=0.476

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