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

(iv) $P(0<X<3)$

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

\therefore \, \, \, \, 0+k+2k+2k+3k+k^{2}+2k^{2}+7k^{2}+k=1

           10k^{2}+9k-1=0

         (10k-1)(k+1)=0

k=\frac{1}{10}\, \, and\, \, k=-1

k = -1 is not possible. So, k=\frac{1}{10}\

 

P(0< X< 3)=P(X=1)+P(X=2)

                                =k+2k

                                  =3k

                                  =3\times \frac{1}{10}

                                = \frac{3}{10}

Posted by

seema garhwal

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