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11.11 A fair die is rolled. Consider events $E=\left \{ 1,3,5 \right \},F=\left \{ 2,3 \right \}$ and $G=\left \{ 2,3,4,5 \right \}$ Find

(iii) $P((E\cup F)\mid G)$ and $P((E\cap F)\mid G)$

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A fair die is rolled.

Total outcomes $=\left \{ 1,2,3,4,5,6 \right \}=6$

$E=\left \{ 1,3,5 \right \},F\left \{ 2,3 \right \}$ and $G=\left \{ 2,3,4,5 \right \}$

$E\cap G=\left \{ 3,5 \right \}$ , $F\cap G=\left \{ 2,3\right \}$

$(E\cap G)\cap G =\left \{ 3 \right \}$

$P[(E\cap G)\cap G] =\frac{1}{6}$ $P(E\cap G) =\frac{2}{6}$ $P(F\cap G) =\frac{2}{6}$

$P((E\cup F)|G) = P(E|G)+P(F|G) - P[(E\cap F)|G]$

$=\frac{P(E\cap G)}{P(G)}+\frac{P(F\cap G)}{P(G)}-\frac{P((E\cap F)\cap G)}{P(G)}$

$=\frac{\frac{2}{6}}{\frac{4}{6}}+\frac{\frac{2}{6}}{\frac{4}{6}}-\frac{\frac{1}{6}}{\frac{4}{6}}$

$=\frac{2}{4}+\frac{2}{4}-\frac{1}{4}$

$=\frac{3}{4}$

$P((E\cap F)|G)=\frac{P((E\cap F)\cap G)}{P(G)}$

$P((E\cap F)|G)=\frac{\frac{1}{6}}{\frac{4}{6}}$

$P((E\cap F)|G)=\frac{1}{4}$

Posted by

seema garhwal

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