If a random variable X has the following probability distribution values
| $\mathrm{X}$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| $\mathrm{P(X):}$ | 0 | $\mathrm{k}$ | $\mathrm{2k}$ | $\mathrm{2k}$ | $\mathrm{3k}$ | $\mathrm{k^2}$ | $\mathrm{2k^2}$ | $\mathrm{7k^2+k}$ |
Then $P(X \geq 6)$ has the value
A random variable X takes the values $0,1,2,3$ and its mean is 1.3 . If $\mathrm{P}(\mathrm{X}=3)=2 \mathrm{P}(\mathrm{X}=1)$ and $P(X=2)=0.3$, then $P(X=0)$ is
Three persons $\mathrm{P}, \mathrm{Q}$ and R independently try to hit a target. If the probabilities of their hitting the target are $\frac{3}{4}, \frac{1}{2}$ and $\frac{5}{8}$ respectively, then the probability that the target is hit by P or Q but not by $R$, is
A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one by one, with replacement, then the variance of the number of green balls drawn is
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