$\begin{array}{1 1} 0 \\ \frac{3}{20} \\ \frac{1}{4} \\ \frac{1}{10} \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 if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. The sample space can be represented as follows:

$S = \begin{Bmatrix} 3,1 &3,2 &3,3 &3,4 &3,5 &3,6 \\ 6,1 &6,2 &6,3 & 6,4 &6,5 &6,6 \\ 1,H &2,H &4,H &5,H & & \\ 1,T& 2,T& 4,T& 5,T & & \end{Bmatrix}$

The total number of outcomes in S = 20.

Let E be the event that the coin shows a tail. E = {(1,T), (2,T), (4,T), (5,T)}. The number of outcomes = 4.

Given E, $\rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{4}{20} = \frac{1}{5}$

Let F be the event such that at least one die shows up 3. F = {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (6,3)}, The number of outcomes is 7.

Given F, $\rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{7}{20} $

$\Rightarrow$ because there is no common element in E and F, $ E \cap F = \phi \rightarrow P( E \cap F)$ = 0

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{0}{\frac{5}{6}}$ = 0.

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