An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability, that the three balls have different colours, is

A random variable X takes values $-1,0,1,2$ with probabilities $\frac{1+3 \mathrm{p}}{4}, \frac{1-\mathrm{p}}{4}, \frac{1+2 \mathrm{p}}{4}, \frac{1-4 \mathrm{p}}{4}$ respectively, where p varies over $\mathbb{R}$. Then the minimum and maximum values of the mean of X are respectively.

If the mean and the variance of Binomial variate $X$ are 2 and 1 respectively, then the probability that X takes a value greater than or equal to one is

If $\mathrm{P}(\mathrm{X}=2)=0.3, \mathrm{P}(\mathrm{X}=3)=0.4, \mathrm{P}(\mathrm{X}=4)=0.3$, then the variance of random variable X is