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The Order of events are as follows $A$ or $A^{'}$ than $B$ or $B^{'}$ lastly $C$ or $C^{'}$

I am given the following: $P(A)=0.75, P(B|A)=0.9, P(B|A^{'})=0.8, P(C|A \cap B)=0.8, P(C|A^{'} \cap B)=0.7, P(C| A \cap B^{'})=0.6, P(C| A^{'} \cap B^{'})=0.3$.

I'm asked to find $P(B \cap C)$. I know that this is equal to P(B|C)P(C) or P(C|B)P(B) but I do not know how to calculate these values from the given information.

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Apply the law of total probability: en.wikipedia.org/wiki/Law_of_total_probability#Statement –  Hoda Feb 4 at 21:58
So $P(B|A)+P(B|A^{'})=P(B)$ So $P(B)=1.7$? A remarkable phenomenon that is. –  adam Feb 4 at 22:06
:) $P(B|A)\cdot P(A)+P(B|A′)\cdot P(A') =P(B)$. –  Hoda Feb 4 at 22:10
For $P(B)$ I get 0.875 and $P(C)$ I get 0.74. Multiplying the two gives me 0.6475. The answer is 0.68 though. –  adam Feb 4 at 22:41
Multiplication works only for independent events. You can obtain $P(C|B)$ from $P(C|A\cap B)$ and $P(C|A'\cap B)$ using the law of total probability. –  Hoda Feb 4 at 22:49

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