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Q.8 A random variable X has the following probability distribution:

ii) P(X<3)

Answers (1)

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The sum of probabilities of the probability distribution of the random variables is 1.

0+k+2k+2k+3k+k2+2k2+7k2+k=1 
10k2+9k1=0
(10k1)(k+1)=0
k=110 and k=1
k=1 is not possible.

So, k=110 P(X<3)=P(X=0)+P(X=1)+P(X=2)

=0+K+2K
=3K
=3×110
=310

Posted by

seema garhwal

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