A random variable X has the following probability distribution:

The value of $P(4 <x<7)$ is equal to

$\frac{3}{100}$

The correct answer is Option (1) → $\frac{3}{100}$

Given probability distribution:

$X: 0,1,2,3,4,5,6,7$

$P(X): 0, k, 2k, 2k, 3k, k^{2}, 2k^{2}, 7k^{2}+k$

Sum of probabilities is $1$.

$k+2k+2k+3k+k^{2}+2k^{2}+(7k^{2}+k)=1$

$9k+10k^{2}=1$

$10k^{2}+9k-1=0$

$k=\frac{-9+\sqrt{81+40}}{20}=\frac{-9+11}{20}=\frac{1}{10}$

Required probability:

$P(4<\text{ X }<7)=P(5)+P(6)$

$=k^{2}+2k^{2}=3k^{2}$

$=3\left(\frac{1}{10}\right)^{2}=\frac{3}{100}$

Final answer: $\frac{3}{100}$