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Q.

A random variable has the following distribution
 

X = xi1234
PX = xik2k3k4k

The value of k and P(x < 3) are equal to 
 

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a

k=310, P(x<3)=110

b

k=110, P(x<3)=310

c

k=110, P(x<3)=512

d

k=110, P(x<3)=35

answer is B.

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Detailed Solution

detailed_solution_thumbnail

Given 

X = xi1234
PX = xik2k3k4k

Since P(X=xi)=1

k+2k+3k+4k=1 10k=1 k=110

Now

 P(x<3)=P(x=1)+P(x=2)              =k+2k              =3k              =310

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