A random variable, $X$ has p.m.f. $\mathrm{P}(\mathrm{X}=x)=\frac{{ }^4 \mathrm{C}_x}{2^4}, x=0,1,2,3,4$ and $\mu$ and $\sigma^2$ are mean and variance respectively of random variable X , then

$$ \text { The c.d.f. of a discrete random variable } \mathrm{X} \text { is } $$

$$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \mathrm{X} & -3 & -1 & 0 & 1 & 3 & 5 & 7 & 9 \\ \hline \mathrm{~F}(\mathrm{X}=x) & 0.1 & 0.3 & 0.5 & 0.65 & 0.75 & 0.85 & 0.90 & 1 \\ \hline \end{array} $$

Then $\frac{P[X=-3]}{P[X<0]}=$

If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are mutually exclusive and exhaustive events of a sample space $S$ such that $P(B)=\frac{3}{2} P(A)$ and $P(C)=\frac{1}{2} P(B)$, then $P(A)=$

If a random variable X follows the Binomial distribution $B(10, \quad p)$ such that $5 \mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\mathrm{X}=1)$, then the value of $\frac{\mathrm{P}(\mathrm{X}=5)}{\mathrm{P}(\mathrm{X}=6)}$ is equal to