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 $F)=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}$ $=\frac{2+4-1}{8}=\frac{5}{8}$
The probabilities of three mutually exclusive events are $\frac{2}{3} , \frac{1}{4}$ and $\frac{1}{6}$. The statement is
Given two independent events $A$ and $B$ such $P(A)=0.3,\,P(B)=0.6 .$ Find $P($ neither $A$or $B)$
If $P(A) = \frac{1}{2},\,\,P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{7}{{12}},$ then the value of $P\,(A' \cap B')$ is
If ${A_1},\,{A_2},...{A_n}$ are any $n$ events, then
For an event, odds against is $6 : 5$. The probability that event does not occur, is