A random variable X has the following probability distribution

Then the values of 'a' and P(0 < x < 5) respectively are

$\frac{1}{81},\frac{8}{27}$

The correct answer is Option (1) → $\frac{1}{81},\frac{8}{27}$

Given probability distribution

$X:0,1,2,3,4,5,6,7,8$

$P(X):a,3a,5a,7a,9a,11a,13a,15a,17a$

Sum of probabilities equals $1$

$a(1+3+5+7+9+11+13+15+17)=1$

$1+3+5+7+9+11+13+15+17=81$

$81a=1$

$a=\frac{1}{81}$

The reqd probability $P(0<\text{ x }<5)$ corresponds to $X=1,2,3,4$

$P(0<\text{ x }<5)=3a+5a+7a+9a$

$=24a$

$=\frac{24}{81}=\frac{8}{27}$

The values of $a$ and $P(0