Given two independent events $A$ and $B$ such that $P(A) $ $=0.3, \,P(B)=0.6$ Find $P(A$ and $B)$.
It is given that $P(A)=0.3,\,P(B)=0.6$..
Also, $A$ and $B$ are independent events.
$\mathrm{P}(\mathrm{A}$ and $\mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$
$\Rightarrow $ $ \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.3 \times 0.6=0.18$
For the three events $A, B$ and $C, P$ (exactly one of the events $A$ or $B$ occurs) = $P$ (exactly one of the events $B$ or $C$ occurs)= $P$ (exactly one of the events $C$ or $A$ occurs)= $p$ and $P$ (all the three events occur simultaneously) $ = {p^2},$ where $0 < p < 1/2$. Then the probability of at least one of the three events $A, B$ and $C$ occurring is
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 ?
If odds against solving a question by three students are $2 : 1 , 5:2$ and $5:3$ respectively, then probability that the question is solved only by one student is
Three persons $P, Q$ 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
If $A$ and $B$ are two mutually exclusive events, then $P\,(A + B) = $