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Home  >>  CBSE XII  >>  Math  >>  Probability
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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.

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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.
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) $= \large \frac{P(E \;\cap \;F)}{P(F)}$
$\Rightarrow P(E/F) = \Large\frac {\Large \frac{2}{36}}{\Large\frac{1}{6}}$ = $\large\frac{1}{3}$
answered Jun 18, 2013 by balaji.thirumalai
 

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