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 is greater than 9. E: (6,4), (4,6), (5,5), (5,6), (6,5), (6,6). The total possible outcomes = 6.

$\Rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{6}{36} = \frac{1}{6}$

Let F be set of events where the black die rolled a 5. F: (1,5), (2,5), (3,5), (4,5), (5,5), (6,5). The total possible outcomes = 6.

$\Rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{6}{36} = \frac{1}{6}$

For our set of events $E \cap F =(5,5), (6,6). Total number of outcomes = 2.

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

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

$\Rightarrow P(E/F) = \Large\frac {\Large \frac{2}{36}}{\Large\frac{1}{6}}$ = $\large\frac{1}{3}$

Ask Question

Tag:MathPhyChemBioOther

Take Test

...