In a Binomial distribution consisting of 5 independent trials, probabilities of exactly 1 and 2 successes are 0.4096 and 0.2048 respectively, then the probability, of getting exactly 4 successes, is

A random variable X has the following probability distribution

For the events $\mathrm{E}=\{\mathrm{X}$ is prime number $\}$

$$\mathrm{F}=\{\mathrm{X}<4\}$$

Then $P(E \cup F)=$

There are three events $\mathrm{A}, \mathrm{B}, \mathrm{C}$, one of which must and only one can happen. The odds are 8:3 against $\mathrm{A}, 5: 2$ against B and the odds against C is $43: 17 \mathrm{k}$, then value of k is

Four persons can hit a target correctly with probabilities $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and $\frac{1}{5}$ respectively. If all hit at the target independently, then the probability that the target would be hit, is