Prove that if $E$ and $F$ are independent events, then so are the events $\mathrm{E}$ and $\mathrm{F}^{\prime}$.
since $\mathrm{E}$ and $\mathrm{F}$ are independent, we have
$\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{F})$ ......... $(1)$
From the venn diagram in Fig it is clear that $E \cap \mathrm{F}$ and $\mathrm{E} \cap \mathrm{F}^{\prime}$ are mutually exclusive events and also $\mathrm{E}=(\mathrm{E} \cap \mathrm{F}) \cup\left(\mathrm{E} \cap \mathrm{F}^{\prime}\right)$
Therefore $\quad P(E)=P(E \cap F)+P\left(E \cap F^{\prime}\right)$
or $P\left(E \cap F^{\prime}\right)=P(E)-P(E \cap F)$
$=\mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{F})$ (by $(1))$
$=\mathrm{P}(\mathrm{E})(1-\mathrm{P}(\mathrm{F}))$
$=\mathrm{P}(\mathrm{E})$ . $\mathrm{P}\left(\mathrm{F}^{\prime}\right)$
Hence, $\mathrm{E}$ and $\mathrm{F}^{\prime}$ are independent
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 ?
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) = $
Let $E$ and $F$ be two independent events. The probability that both $E$ and $F$ happens is $\frac{1}{{12}}$ and the probability that neither $E$ nor $F$ happens is $\frac{1}{2},$ then
Events $E$ and $F$ are such that $P ( $ not $E$ not $F )=0.25,$ State whether $E$ and $F$ are mutually exclusive.
A bag contains $9$ discs of which $4$ are red, $3$ are blue and $2$ are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be either red or blue.