Question

A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G={2, 3, 4, 5}. Find:

1. P(E|F) and P(F|E)
2. P(E|G) and P(G|E)
3. P[(E∪F)G] and P[(E∩F)G]

Answer 1

$S = (1, 2, 3, 4, 5, 6)$
$⇒ n(S) = 6$

$E = (1 ,3, 5)$
$F = (2, 3)$
$G = (2, 3, 4, 5)$
$⇒ n(E) = 3$
$⇒ n(E) = 3$
$⇒ n(G) = 4$

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$(i)$ $P(E)=\frac {n(E)}{n(S)}$ $=\frac 36$

$P(F)=\frac {n(F)}{n(S)} =\frac {2}{6}$

$E∩F(=3)$
$⇒n(E∩F)=1$

$P(E∩F)=\frac {n(E∩F)}{n(S)} =\frac 16$

$P(E|F)=\frac {P(E∩F)}{P(F) }=\frac {1/6}{2/6} =\frac 12$

$P(F|E)=\frac {P(E∩F)}{P(E)} =\frac {1/6}{3/6} =\frac 13$

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$(ii)$ $P(E)=\frac {n(E)}{n(S}=\frac {3}{6}$

$P(G)=\frac {n(G)}{n(S)}=\frac 46$

$E∩G=(3,5)$
$⇒n(E∩G)=2$

$P(E∩G)=\frac {n(E∩G)}{n(S)} =\frac 26$

$P(E|G)=\frac {P(E∩G)}{P(G) }$

$P(E|G)=\frac {2/6}{4/6 }$

$P(E|G)=\frac 24$

$P(E|G)=\frac 12$

$P(G|E)=\frac {P(E∩G)}{PE)}$

$P(G|E)=\frac {2/6}{3/6}$

$P(G|E)=\frac 23$

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$(iii)$ $P(G)=\frac {n(G)}{n(S)} =\frac 46$

$E∪F=1,2,3,5 \ and \ G=(2,3,4,5)$
$(E∪F)∩G=(2,3,5)$

$⇒n[(E∪F)∩G]=3$
$P[(E∪F)∩G]=\frac 36$

$P(E∪F|G)=\frac {P[(E∪F)∩G]}{P(G)}$

$P(E∪F|G)=\frac {3/6}{4/6}$

$P(E∪F|G)=\frac 34$

$Again \ E∩F=(3)$
$(E∩F)∩G=3$

$⇒n[(E∩F)∩G]=1$
$P[(E∩F)∩G]=\frac 16$

$P(E∩F|G)=\frac {P[(E∩F)∩G]}{P(G)}$

$P(E∩F|G)=\frac {1/6}{4/6}$

$P(E∩F|G)=\frac 14$