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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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    $\begingroup$ What is $P(A|'B)$? $\endgroup$
    – krlmlr
    Commented Feb 23, 2012 at 13:23
  • $\begingroup$ @user946850 it is the probability of A given 'Not B'. $\endgroup$
    – cubesqrd
    Commented Feb 23, 2012 at 13:49
  • $\begingroup$ 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, $P(A)$? $\endgroup$
    – Macro
    Commented Feb 23, 2012 at 13:49
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    $\begingroup$ 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. $\endgroup$ Commented Feb 23, 2012 at 15:40
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    $\begingroup$ 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? $\endgroup$
    – whuber
    Commented Feb 23, 2012 at 18:01

1 Answer 1

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$P(A)=P(A\cap B)+P(A\cap B^c)=P(A\mid B)P(B)+P(A\mid B^c)P(B^c)$

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  • $\begingroup$ And to continue $$P(A\mid B)P(B)+P(A\mid B^c)P(B^c) = 0.6P(B)+0.6(1-P(B))=0.6$$ $\endgroup$
    – Henry
    Commented Feb 23, 2012 at 22:54
  • $\begingroup$ @Henry giving away too much I think ... $\endgroup$
    – tdc
    Commented Feb 24, 2012 at 9:46

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