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The random variable X has a probability distribution P(X) of the following form, where k is some number :P(X) = {(k,ifx=0),(2k,ifx=1),(3k,ifx=2),(0otherwise)(a) Determine k.(b) Find P (X<2) , P (X≤2) , P(X≥2) Select the correct answer from above options

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(i)As, we know, Sum of all probablities in a probablity distribution is 1. So, in this question, k+2k+3k+0=1⇒6k=1 ⇒k= 1 6 which is the required value. (ii)(a)P(X<2)=P(X=0)+P(X=1)<br> =k+2k=3k=3( 1 6 )= 1 2 (b)P(X≤2)=P(X=0)+P(X=1)+P(X=2)
=k+2k+3k=6k=6( 1 6 )=1 (b)P(X≥2)=P(X=2)+0 =3k= 1 2

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