# Conditional probability problem

Could someone please help explain how to calculate $P(A)$ given:

$P(B) = 0.2$

$P(A|B) = 0.6$

$P(A|{\rm not} \ B) = 0.6$

Any explaination/help would be greatly appreciated.

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What is $P(A|'B)$? –  krlmlr Feb 23 '12 at 13:23
@user946850 it is the probability of A given 'Not B'. –  cubesqrd Feb 23 '12 at 13:49
Assuming $P(A|'B)$ means $P(A|{\rm not } \ B)$, then you know that $P(A|B) = P(A|{\rm not } \ B) = .6$. The outcomes $B$ and ${\rm not } \ B$ partition the sample space, so what can you deduce about the unconditional probability, $P(A)$? –  Macro Feb 23 '12 at 13:49
I disagree with Xi'an's statement that this derivation is called Bayes' Theorem.: it is the law of total probability. Bayes theorem would be involved if the problem asked for $P(B|A)$ which "turns" the conditioning around. $P(B)$ is playing the part of the prior probability, $P(A|B)$ corresponds to the likelihood, and $P(B|A)$ to the posterior probability, and the Reverend Thomas Bayes gets involved. In this instance, the Reverend is not to blame. –  Dilip Sarwate Feb 23 '12 at 15:40
Hint: Since the probability of $A$ does not depend on $B$, do you even care about knowing the probability of $B$? If not, then what do you think the probability of $A$ is? –  whuber Feb 23 '12 at 18:01
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$P(A)=P(A\cap B)+P(A\cap B^c)=P(A\mid B)P(B)+P(A\mid B^c)P(B^c)$
And to continue $$P(A\mid B)P(B)+P(A\mid B^c)P(B^c) = 0.6P(B)+0.6(1-P(B))=0.6$$ –  Henry Feb 23 '12 at 22:54