Question

A fair die is rolled. consider events E = {1, 3,5} F = {2,3} and G = {2,3,4,5}. Find
$P\left(\frac{E}{F}\right)andP\left(\frac{F}{E}\right)$

$P\left(\frac{E}{G}\right)andP\left(\frac{G}{E}\right)$

$P\left(\frac{E\cup F}{G}\right)andP\left(\frac{E\cap F}{G}\right)$

Answer 1

Here, the sample space si $S=1,2,3,4,5,6Given,E=\{1,3,5\},F=\{2,3|andG=|2,3,4,5\}\Rightarrow E\cap F=\left\{3\right\},E\cap G=\{3,5\},E\cap F=\{1,2,3,5\},(E\cup F)\cap G=(2,3,5)and(E\cap F)\cap G=\left\{3\right\}.\Rightarrow n\left(S\right)=6,n\left(E\right)=3,n\left(F\right)=2,n\left(G\right)=4n(E\cap F)=1,n(E\cap G)=2,n(E\cap F)=4n\left[\right(E\cup F)\cap G]=3andn\left[\right(E\cap F)\cap G]=1$
$\therefore$ by the formula, Probability = $\frac{Numberoffavourableevents}{Totalnumberofevents}$
We have $P\left(E\right)=\frac{3}{6}=\frac{1}{2},P\left(F\right)=\frac{2}{6}=\frac{1}{3},P\left(G\right)=\frac{4}{6}=\frac{2}{3}P(E\cap F)=\frac{1}{6}P(E\cap G)=\frac{2}{6}=\frac{1}{3}p\left[\right(E\cup F)\cap G]=\frac{3}{6}=\frac{1}{2}andP\left[\right(E\cap F)\cap G]=\frac{1}{6}$
$P\left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{1}{6}}{\frac{1}{3}}=\frac{1}{2}andP\left(\frac{F}{E}\right)=\frac{P(F\cap E)}{P\left(E\right)}=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1}{3}$

Here, the sample space si $S=1,2,3,4,5,6Given,E=\{1,3,5\},F=\{2,3|andG=|2,3,4,5\}\Rightarrow E\cap F=\left\{3\right\},E\cap G=\{3,5\},E\cap F=\{1,2,3,5\},(E\cup F)\cap G=(2,3,5)and(E\cap F)\cap G=\left\{3\right\}.\Rightarrow n\left(S\right)=6,n\left(E\right)=3,n\left(F\right)=2,n\left(G\right)=4n(E\cap F)=1,n(E\cap G)=2,n(E\cap F)=4n\left[\right(E\cup F)\cap G]=3andn\left[\right(E\cap F)\cap G]=1$
$\therefore$ by the formula, Probability = $\frac{Numberoffavourableevents}{Totalnumberofevents}$
We have $P\left(E\right)=\frac{3}{6}=\frac{1}{2},P\left(F\right)=\frac{2}{6}=\frac{1}{3},P\left(G\right)=\frac{4}{6}=\frac{2}{3}P(E\cap F)=\frac{1}{6}P(E\cap G)=\frac{2}{6}=\frac{1}{3}p\left[\right(E\cup F)\cap G]=\frac{3}{6}=\frac{1}{2}andP\left[\right(E\cap F)\cap G]=\frac{1}{6}$
$P\left(\frac{E}{G}\right)=\frac{P(E\cap G)}{P\left(G\right)}=\frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{2}andP\left(\frac{G}{E}\right)=\frac{P(G\cap E)}{p\left(E\right)}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$

Here, the sample space si $S=1,2,3,4,5,6Given,E=\{1,3,5\},F=\{2,3|andG=|2,3,4,5\}\Rightarrow E\cap F=\left\{3\right\},E\cap G=\{3,5\},E\cap F=\{1,2,3,5\},(E\cup F)\cap G=(2,3,5)and(E\cap F)\cap G=\left\{3\right\}.\Rightarrow n\left(S\right)=6,n\left(E\right)=3,n\left(F\right)=2,n\left(G\right)=4n(E\cap F)=1,n(E\cap G)=2,n(E\cap F)=4n\left[\right(E\cup F)\cap G]=3andn\left[\right(E\cap F)\cap G]=1$
$\therefore$ by the formula, Probability = $\frac{Numberoffavourableevents}{Totalnumberofevents}$
We have $P\left(E\right)=\frac{3}{6}=\frac{1}{2},P\left(F\right)=\frac{2}{6}=\frac{1}{3},P\left(G\right)=\frac{4}{6}=\frac{2}{3}P(E\cap F)=\frac{1}{6}P(E\cap G)=\frac{2}{6}=\frac{1}{3}p\left[\right(E\cup F)\cap G]=\frac{3}{6}=\frac{1}{2}andP\left[\right(E\cap F)\cap G]=\frac{1}{6}$
$P\left(\frac{E\cup F}{G}\right)=\frac{P\left[\right(E\cup F\left)capG\right]}{P\left(G\right)}=\frac{\frac{1}{2}}{\frac{2}{3}}\frac{3}{4}andP\left(\frac{E\cap F}{G}\right)=\frac{P\left[\right(E\cap F)\cap G]}{P\left(G\right)}=\frac{\frac{1}{6}}{\frac{2}{3}}=\frac{1}{4}$