Let $$X$$ be a binomially distributed random variable with mean 4 and variance $$\frac{4}{3}$$. Then, $$54 \,P(X \leq 2)$$ is equal to :

The mean and variance of a binomial distribution are $$\alpha$$ and $$\frac{\alpha}{3}$$ respectively. If $$\mathrm{P}(X=1)=\frac{4}{243}$$, then $$\mathrm{P}(X=4$$ or 5$$)$$ is equal to :

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 :