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 $F)=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}$ $=\frac{2+4-1}{8}=\frac{5}{8}$
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Let $A$ and $B$ be events for which $P(A) = x$, $P(B) = y,$$P(A \cap B) = z,$ then $P(\bar A \cap B)$ equals
Let $S$ be a set containing n elements and we select $2$ subsets $A$ and $B$ of $S$ at random then the probability that $A \cup B = S$ and $A \cap B = \phi $ is
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Prove that if $E$ and $F$ are independent events, then so are the events $\mathrm{E}$ and $\mathrm{F}^{\prime}$.