Let $\mathrm{E}$ and $\mathrm{F}$ be events with $\mathrm{P}(\mathrm{E})=\frac{3}{5}, \mathrm{P}(\mathrm{F})$ $=\frac{3}{10}$ and $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{5} .$ Are $\mathrm{E}$ and $\mathrm{F}$ independent ?

It is given that $P(E)=\frac{3}{5}, \,P(F)=\frac{3}{10}$ and $P(E F)=P(E \cap F)=\frac{1}{5}$

$P(E) .P(F)=\frac{3}{5} \times \frac{3}{10}=\frac{9}{50} \neq \frac{1}{5}$

$\Rightarrow P(E). P(F) \neq P(E F)$

Therefore, $\mathrm{E}$ and $\mathrm{F}$ are not independent.

If $A$ and $B$ are arbitrary events, then

In a class of $60$ students, $30$ opted for $NCC$ , $32$ opted for $NSS$ and $24$ opted for both $NCC$ and $NSS$. If one of these students is selected at random, find the probability that The student has opted $NSS$ but not $NCC$.

The probability of happening at least one of the events $A$ and $B$ is $0.6$. If the events $A$ and $B$ happens simultaneously with the probability $0.2$, then $P\,(\bar A) + P\,(\bar B) = $

- [IIT 1987]

The probability of solving a question by three students are $\frac{1}{2},\,\,\frac{1}{4},\,\,\frac{1}{6}$ respectively. Probability of question is being solved will be

$A, B, C$ are any three events. If $P (S)$ denotes the probability of $S$ happening then $P\,(A \cap (B \cup C)) = $