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
The following is the probability distribution of X
$$ \begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1+\mathrm{p}}{5} & \frac{2-2 \mathrm{p}}{5} & \frac{2-\mathrm{p}}{5} & \frac{2 \mathrm{p}}{5} \\ \hline \end{array} $$
$$ \text { For a minimum value of } p \text {, the value of } 5 E(X) \text { is } $$
A random variable X takes values $0,1,2,3$, ........ with probabilities. $\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1)\left(\frac{1}{2}\right)^x, \mathrm{k}$ is a constant, then $P(X=1)=$
The probability that in a random arrangement of the letters of the word 'UNIVERSITY', the two 'I's do not come together is