This is a multi part question answered separately on Clay6.com

- 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 P(E), P(F), P(E $\cap$ F), P(E $\cup$ F) = P(E) + P(F) - P(E $\cap$ F)

Given one black die and one red die are rolled. The sample space of equally likely events = 6 $\times$ 6 = 36.

Let E: set of events where the sum = 8 $\rightarrow$ E = (2,6), (6,2), (3,5), (5,3), (4,4). The total number of outcomes = 5.

$\Rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{5}{36}$

Let F: set of events where the red die rolls less than 4 $\rightarrow$ F: (1,1), (2,1), (3,1), (4,1) (5,1), (6,1), (1,2),....(6,3). The total number of outcomes = 18.

$\Rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{18}{36}$

\(n(E\cap\;F)=2\) $\rightarrow$ $\Rightarrow P(E\;\cap\;F) = \large \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in S}} = \frac{2}{36}$.

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

$\Rightarrow P(E/F) = \Large\frac {\Large \frac{2}{36}}{\Large\frac{18}{36}}$ = $\large\frac{1}{9}$

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