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- In any problem that involves tossing a coin or a die and determing the outcomes, we can write down the sample space and cout the number of favorable outcomes. Then we find P(E), P(F), P(E$\cap$F) using the set of outcomes.
- Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$

Given that a fair die is rolled, number of total outcomes = 6.

Given E: {1, 3, 5} $\rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{3}{6} = \frac{1}{2}$

Given F: {2,3} $\rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{2}{6} = \frac{1}{3}$

Given G: {2,3,4,5} $\rightarrow P(G) = \large \frac{\text{Number of favorable outcomes in G}}{\text{Total number of outcomes in S}} = \frac{4}{6} = \frac{2}{3}$

It also follows that we can calculate the various "Union" and "Intersection" of the sets and their probabilities as follows:

$\Rightarrow E \cup F = {1,2,3,5} \rightarrow P( E \cup F) = \large \frac{4}{6} = \frac{2}{3}$

$\Rightarrow E \cap F = {3} \rightarrow P( E \cap F) = \large \frac{1}{6}$

$\Rightarrow E \cup G = {1,2,3,4,5} \rightarrow P( E \cup G) = \large \frac{5}{6}$

$\Rightarrow E \cap G = {3,5} \rightarrow P( E \cap G) = \large \frac{2}{6} = \frac{1}{3}$

Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$

Therefore P($\large \frac{E}{F}$) = $\Large \frac{\Large \frac{1}{6}}{\Large \frac{1}{3}}$ = $\large \frac{1}{2}$

and P($\large \frac{F}{E}$) = $\Large \frac{\Large \frac{1}{6}}{\Large \frac{1}{2}}$ = $\large \frac{1}{3}$

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