# If E and F are events such that P(E) =, P(F) = and P(E and F) =, find:(i) P(E or F), (ii) P(not E and not F).

Question:

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:(i) $P(E$ or $F)$, (ii) $P($ not $E$ and not $F)$.

Solution:

Here, $P(E)=\frac{1}{4}, P(F)=\frac{1}{2}$, and $P(E$ and $F)=\frac{1}{8}$

(i) We know that $P(E$ or $F)=P(E)+P(F)-P(E$ and $F)$

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

(ii) From (i), $P(E$ or $F)=P(E \cup F)=\frac{5}{8}$

We have $(E \cup F)^{\prime}=\left(E^{\prime} \cap F^{\prime}\right)$ [By De Morgan's law]

$\therefore \mathrm{P}\left(\mathrm{E}^{\prime} \cap \mathrm{F}^{\prime}\right)=\mathrm{P}(\mathrm{E} \cup \mathrm{F})^{\prime}$

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

$\therefore P\left(E^{\prime} \cap F^{\prime}\right)=\frac{3}{8}$

Thus, $\mathrm{P}($ not $\mathrm{E}$ and not $\mathrm{F})=\frac{3}{8}$