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I found this question on probability.
Given that $$P(a \cup b) = \frac{2}{3}$$ $$P(a \cap \neg b) = \frac{1}{3}$$ $$P(a|b) = \frac{1}{6}$$ I have to find $P(a)$ and $P(b)$. I tried a lot but felt it was missing data. Thanks.

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2 Answers 2

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$$P(A|B) = P(A \cap B) / P(B)$$

$$P(B) = P(A \cup B) - P(A \cap B^\prime)$$

So then if we re-arrange, we get

$$P(A|B) = P(A \cap B) / (P(A \cup B) - P(A \cap B^\prime))$$

Where

$P(A|B) = \frac{1}{6}$, $P(A \cup B) = \frac{2}{3}$, and $P(A \cap B^\prime) = \frac{1}{3}$

Now you can solve for $P(A \cap B)$ and from there you can solve for $P(B)$ and then $P(A)$

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  • $\begingroup$ -1 Why does $P(B)$ equal $P(A\cup B) - P(A\cap B)$? $\endgroup$ Commented Mar 5, 2013 at 14:58
  • $\begingroup$ It's $P(B) = P(A \cup B)-P(A \cap B')$ - $B'$ here meaning not $B$ $\endgroup$
    – mardat
    Commented Mar 5, 2013 at 21:55
  • $\begingroup$ I missed that little quote mark (doesn't show on the screen when I read math.SE using the Firefox browser). I have reversed my down vote. $\endgroup$ Commented Mar 6, 2013 at 0:33
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Draw a picture (Venn diagram) and everything will become clear. We follow tradition by using capital letters for events. If $E$ is an event, then $E^c$ denotes the complement of $E$.

Since $\Pr(A\cap B^c)=\frac{1}{3}$, the part of $A$ outside $B$ has probability $\frac{1}{3}$. But $A\cup B$ is the disjoint union of $A\cap B^c$ and $B$. It follows that $\Pr(B)=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$.

Note that $\Pr(A| B)=\dfrac{\Pr(A\cap B)}{\Pr(B)}$. Thus $\Pr(A\cap B)=\frac{1}{6}\cdot\frac{1}{3}=\frac{1}{18}$.

Finally, we can use the formula $\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)$ to find $\Pr(A)$. We get $\Pr(A)=\frac{2}{3}-\frac{1}{3}+\frac{1}{18}=\frac{7}{18}$.

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