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# A black and a red dice are rolled. Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

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

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