I have a probability problem where I have to calculate the total probability and a Bayes probability from two events. The chances as given are:

$P(A) = 0.1, P(A^c) = 0.9$

$P(B|A) = 0.9, P(B|A^c) = 0.05$

Using these probabilities, I was able to work out the total probability:

$P(B) = P(A)P(B|A) + P(A^c)P(B|A^c) = (0.1*0.9) + (0.9*0.05) = 0.135$

And the chance of A given B using Bayes:

$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{0.9*0.1}{0.135} = \frac{2}{3}$

Pretty standard stuff so far. However, the last part of the question adds an event C. The chance of C given A and B is 0.1. Knowing this, I have to calculate the chance of an outcome having A, B, and C. Is this really as easy as $P(A \cap B \cap C) = P(A)*P(B)*P(C) = 0.1*0.9*0.1 = 0.009?$


Use: $P(A \cap B \cap C) = P(C | A \cap B) P(A \cap B)$

Your answer assumes that the events $A, B, C$ are mutually independent which is not true. For example, $P(B) \ne P(B|A)$. In other words, knowledge of the event $A$ causes us to revise our probability of event $B$.

  • $\begingroup$ Thank you for your reply. I know how to calculate P(A and B), but how would I construct the P(C|A and B) sum? $\endgroup$ – John Smith Oct 23 '13 at 15:15
  • $\begingroup$ Isn't that already given to you? $\endgroup$ – response Oct 23 '13 at 15:49
  • $\begingroup$ How so? My definition of a conditional probability P(A|B) is P(A)P(B)/P(B). If a second variable is included like in P(C|A and B), does that make it P(C)P(A)P(B) / P(A)P(B)? $\endgroup$ – John Smith Oct 23 '13 at 19:07
  • $\begingroup$ "The chance of $C$ given $A$ and $B$ is $0.1$" $\endgroup$ – response Oct 23 '13 at 23:25

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