If $E$ and $F$ are events such that $P ( E )=\frac{1}{4}$, $P ( F )=\frac{1}{2}$ and $P(E$ and $F )=\frac{1}{8},$ find : $P ( E$ or $F )$
Here, $P ( E )=\frac{1}{4}$, $P ( F )=\frac{1}{2},$ and $P ( E$ and $F )=\frac{1}{8}$
We know that $P ( E$ and $F )= P ( E )+ P ( F )- P ( E$ and $F )$
$\therefore P(E $ or $F)=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}$ $=\frac{2+4-1}{8}=\frac{5}{8}$
Events $\mathrm{A}$ and $\mathrm{B}$ are such that $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{7}{12}$ and $\mathrm{P}$ $($ not $ \mathrm{A}$ or not $\mathrm{B})=\frac{1}{4} .$ State whether $\mathrm{A}$ and $\mathrm{B}$ are independent?
Check whether the following probabilities $P(A)$ and $P(B)$ are consistently defined $P ( A )=0.5$, $ P ( B )=0.7$, $P ( A \cap B )=0.6$
Twelve tickets are numbered $1$ to $12$. One ticket is drawn at random, then the probability of the number to be divisible by $2$ or $3$, is
If $A$ and $B$ are two mutually exclusive events, then $P\,(A + B) = $
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