$\begin{array}{1 1} \frac{1}{15}\\ \frac{2}{15} \\ \frac{14}{14} \\ \frac{1}{5} \end{array} $

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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 pair of fair dice is rolled, number of total outcomes = 6 $\times$ 6 = 36.

Let E be the event that the set of numbers on the dice adds up to 4. E = {(2,2), (3,1), (1,3)}$.

Given E, $\rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{3}{36} = \frac{1}{12}$

Let F be the event that the set of numbers on the dice are different. F is represented as follows:

$F = \begin{Bmatrix} 1,2 &1,3 &1,4 &1,5 &1,6 \\ 2,1& 2,3 &... & & \\ 3,1& .. & & & \\ 4,1& .. & & & \\ 5,1& .. & & & \\ 6,1& .. & .. &.. & 6,5 \end{Bmatrix}$.

The total number of outcomes in F is 30.

Given F, $\rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{30}{36} = \frac{5}{6}$

$\Rightarrow E \cap F = {(1,3)(3,1)} \rightarrow P( E \cap F) = \large \frac{2}{36} = \frac{1}{18}$

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

$\Rightarrow P(E/F) = \Large \frac{\large \frac{1}{18}}{\frac{5}{6}}$ = $\large \frac{5}{6}$

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