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Q.8 A random variable $X$ has the following probability distribution:
ii) $P(X<3)$
Answers (1)
The sum of probabilities of the probability distribution of the random variables is 1.
$\therefore 0+k+2 k+2 k+3 k+k^2+2 k^2+7 k^2+k=1$
$10 k^2+9 k-1=0$
$(10 k-1)(k+1)=0$
$k=\frac{1}{10}$ and $k=-1$
$k=-1$ is not possible.
So, $k=\frac{1}{10}$ $P(X<3)=P(X=0)+P(X=1)+P(X=2)$
$=0+K+2 K$
$=3 K$
$=3 \times \frac{1}{10}$
$=\frac{3}{10}$
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