A fair die is rolled. Consider events E = 1, 3, 5 , F= 2, 3 and G= 2, 3, 4, 5 . Find: P(E|F) and P(F|E)

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(F)=2 n(G)=4 P (E )= n (E ) n (S ) =3 6 P (F )= n (F ) n (S ) =2 6 E = (3 ) n (E )=1 P (E )= n (E F ) n (S ) =1 6 P (E|F )= P (E F ) P (F ) = 1 6 2 6 =1 2 and P (F|E )= P (E F ) P (E ) = 1 6 3 6 =1 3

A fair die is rolled. Consider events E = 1, 3, 5 , F= 2, 3 and G…

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Question

A fair die is rolled. Consider events E = $\{1,\ 3,\ 5\},\ \text{F}=\{2,\ 3\}\ \text{and}\ \text{G}=\{2,\ 3,\ 4,\ 5\}.\ \text{Find}:$
$\text{P}(\text{E}|\text{F})\ \text{and}\ \text{P}(\text{F}|\text{E})$

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Answer

$\text{S}=(1,\ 2,\ 3,\ 4,\ 5,\ 6)\ \Rightarrow\ \ \ \ \ \ \text{n}(\text{S})=6$
$\text{E}=(1,\ 3,\ 5)\ \ \ \ \ \ \ \ \text{F}=(2,\ 3)\ \ \ \ \ \ \ \ (\text{G})=(2,\ 3,\ 4,\ 5)$
$\Rightarrow\ \ \ \ \ \text{n}(\text{E})=3\ \ \ \ \ \ \ \ \text{n}(\text{F})=2\ \ \ \ \ \ \ \text{n}(\text{G})=4$
$\text{P}\left(\text{E}\right)=\frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}=\frac{3}{6}\ \ \ \ \ \ \ \ \ \text{P}\left(\text{F}\right)=\frac{\text{n}\left(\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{2}{6}$
$\text{E}\cap\text{F}=\left(3\right)\ \Rightarrow\ \ \ \ \text{n}\left(\text{E}\cap\text{F}\right)=1$
$\text{P}\left(\text{E}\cap\text{F}\right)=\frac{\text{n}\left(\text{E}\ \cap\ \text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{1}{6}$
$\text{P}\left(\text{E}|\text{F}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{F}\right)}{\text{P}\left(\text{F}\right)}=\frac{\frac{1}{6}}{\frac{2}{6}}=\frac{1}{2}\ \ \ \ \ \ \ \ \\ \text{and}\ \ \ \text{P}\left(\text{F}|\text{E}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{F}\right)}{\text{P}\left(\text{E}\right)}=\frac{\frac{1}{6}}{\frac{3}{6}}=\frac{1}{3}$

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