Question

If $E\text{and}F$ are events such that
$P\left(E\right)=\frac{1}{4},P\left(F\right)=\frac{1}{2}$ and
$P\left(E\text{and}F\right)=\frac{1}{8}$. Find

$\left(i\right)$ $P\left(E\text{or}F\right)$

$\left(ii\right)$ $P\left(\text{not}E\text{and not}F\right)$.

Answer 1

$\left(i\right)$ It is given that
$P\left(E\right)=\frac{1}{4},P\left(F\right)=\frac{1}{2}\text{and}P(E\cap F)=\frac{1}{8}$ $....\left(i\right)$

As we know that,
$P(E\cup F)=P\left(E\right)+P\left(F\right)-P(E\cap F)$ $....\left(ii\right)$

Substituting the values $\left(i\right)$ in $\left(ii\right)$

$P(E\cup F)=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}$

$P(E\cup F)=\frac{2+4-1}{8}=\frac{5}{8}$

Hence, $P(E\cup F)=\frac{5}{8}$

$\left(ii\right)$ It is given that
$P\left(E\right)=\frac{1}{4},P\left(F\right)=\frac{1}{2}\text{and}P(E\cap F)=\frac{1}{8}$

As we know that,
$P(E\cup F)=P\left(E\right)+P\left(F\right)-P(E\cap F)$

$P(E\cup F)=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}$

$P(E\cup F)=\frac{5}{8}$ $...\left(i\right)$

$P\left(\text{not}E\text{and not}F\right)=P(E^{\prime }\cap F^{\prime })$.

$P(E^{\prime }\cap F^{\prime })=P(E\cup F{)}^{\prime }$ (Demorgan's law)

$P(E\cup F)+P(E\cup F{)}^{\prime }=1$ $...\left(ii\right)$

Substituting value $\left(i\right)$ in $\left(ii\right)$

$P(E\cup F{)}^{\prime }=1-P(E\cup F)$ $($By Demorgan’s law$)$

$P(E\cup F{)}^{\prime }=1-\frac{5}{8}=\frac{3}{8}$

Hence, $P(E\cup F{)}^{\prime }=\frac{3}{8}$