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Q.

The probability distribution of a random
variable is given below:
 

X = x01234567
P(X = x)0K2K2K3KK22K27K2+K

Then P(0 < X < 5) 
 

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a

810

b

310

c

710

d

110

answer is C.

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Detailed Solution

Given 

X = x01234567
P(X = x)0K2K2K3KK22K27K2+K

P(X=x)=1

0+k+2k+2k+3k+k2+2k2+7k2+k=1 10k2+9k-1=0 10k2+10k-k-1=0 (k+1)(10k-1)=0 k=-1 (or) k=110 k=110 0P(x)1

Now P(0<x<5)=P(x=1)+P(x=2)+P(α=3)+P(X=4)

         =k+2k+k+3k =8k =810

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The probability distribution of a randomvariable is given below: X = x01234567P(X = x)0K2K2K3KK22K27K2+KThen P(0 < X < 5)