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Question

The random variable X has a probability distribution P(X) of the following form, where k is some number
⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪K, if X=02k, if X=13k, if X=20, if otherwise

Determine the value of k

⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪K, if X=02k, if X=13k, if X=20, if otherwise

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Solution

Given distribution of X is

X 0 1 2otherwiseP(X)k 2k 3k 0

Since, P(X)=1, therefore P(0)+P(1)+P(2)+P(otherwise)=1

k+2k+3k+0=16k=1k=16

Given distribution of X is

X 0 1 2otherwiseP(X)k 2k 3k 0​​​

P(X<2)=P(0)+P(1)=k+2k=3k=3×16=12

P(X2)=P(0)+P(1)+P(2)=k+2k+3k=6k=6×16=1

and P(X2)=P(2)+P(otherwise)=3k+0=3×16=12


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