Q.

If a random variable X has the probability distribution given by P(X=0)=3C3, P(X=2)=5C-10C2 and P(X=4)=4C-1. then the variance of that distribution is 
 

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a

12881

b

689

c

229

d

61281

answer is D.

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Detailed Solution

Given that 

Given P(X=0)=3C3, P(X=2)=5C-10C2, P(X=4)=4C-1

Since   P(X=x)=1

 3C3+5C-10C2+4C-1=1  3C3-10C2+9C-2=0  (C-1) (3C2-7C+2)=0  (C-1) (3C-1) (C-2)=0  C=1, 13, 2  C=13   0P1

X=x024
P(X=x)195913

Mean (x)=xi P(X= xi) 

                =0+109+43=229

Now Variance (a2)= (xi-x)2.P(X=xi)

                                 =0-2292 19+2-2292 59+4-2292 13 =1152729=12881

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