Let $$\mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3}$$ be three mutually exclusive events such that $$\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{2+3 \mathrm{p}}{6}, \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{2-\mathrm{p}}{8}$$ and $$\mathrm{P}\left(\mathrm{E}_{3}\right)=\frac{1-\mathrm{p}}{2}$$. If the maximum and minimum values of $$\mathrm{p}$$ are $$\mathrm{p}_{1}$$ and $$\mathrm{p}_{2}$$, then $$\left(\mathrm{p}_{1}+\mathrm{p}_{2}\right)$$ is equal to :

If $$A$$ and $$B$$ are two events such that $$P(A)=\frac{1}{3}, P(B)=\frac{1}{5}$$ and $$P(A \cup B)=\frac{1}{2}$$, then $$P\left(A \mid B^{\prime}\right)+P\left(B \mid A^{\prime}\right)$$ is equal to

If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :

If the numbers appeared on the two throws of a fair six faced die are $$\alpha$$ and $$\beta$$, then the probability that $$x^{2}+\alpha x+\beta>0$$, for all $$x \in \mathbf{R}$$, is :