In a Binomial distribution with $$\mathrm{n}=4$$, if $$2 \mathrm{P}(\mathrm{X}=3)=3 \mathrm{P}(\mathrm{X}=2)$$, then the variance is

$$\mathrm{A}, \mathrm{B}, \mathrm{C}$$ are three events, one of which must and only one can happen. The odds in favor of $$\mathrm{A}$$ are $$4: 6$$, the odds against $$B$$ are $$7: 3$$. Thus, odds against $$\mathrm{C}$$ are

The probability mass function of random variable X is given by

$$P[X=r]=\left\{\begin{array}{ll} \frac{{ }^n C_r}{32}, & n, r \in \mathbb{N} \\ 0, & \text { otherwise } \end{array} \text {, then } P[X \leq 2]=\right.$$

Three fair coins numbered 1 and 0 are tossed simultaneously. Then variance Var (X) of the probability distribution of random variable $$\mathrm{X}$$, where $$\mathrm{X}$$ is the sum of numbers on the uppermost faces, is