Q.8 A random variable X has the following probability distribution:
ii) P(X<3)
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+9k−1=0 (10k−1)(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
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