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Given that $P(A \mid B)$ is known, and $P(B)$ is known, how can $P(A)$ be calculated? I tried using the definition of conditional probability - $P(A \mid B) = P(A \cap B)/ P(B)$ - to solve for $P(A \cap B)$ but I don't know where to go from there. Any hints?

senshin
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baba
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3 Answers3

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In general, you can't calculate $P(A)$ from just $P(A \mid B)$ and $P(B)$ — you also need to know $P(A \mid \lnot B)$.

If you do know that, the calculation is easy enough:

$$\begin{aligned}P(A) &= P(A \land B) + P(A \land \lnot B) \\ &= P(A \mid B)P(B) + P(A \mid \lnot B)P(\lnot B) \\ &= P(A \mid B)P(B) + P(A \mid \lnot B)(1 - P(B)) \end{aligned}$$

Ilmari Karonen
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  • I know P(¬B) but not P(A∣¬B). Anything I can do? – baba Jun 01 '14 at 22:45
  • Well, assuming $P(A\mid\lnot B)=0$ gives you a lower bound for $P(A)$, and assuming $P(A\mid\lnot B)=1$ gives you an upper bound. That's pretty much it. – Ilmari Karonen Jun 01 '14 at 22:46
  • just realized that A and ¬B are disjoint, so P(A)=P(A∧B) which I already calculated using conditional probability equation. Thanks for the help! – baba Jun 01 '14 at 23:06
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    No problem. If you found these answers helpful, you may want to mark one of them as accepted by clicking the check mark icon to the left of it. – Ilmari Karonen Jun 01 '14 at 23:46
  • Ok! I just did. Sorry, I am new here. This is a great community – baba Jun 01 '14 at 23:56
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You don't have enough information to get $P(A).$

Consider a dice with six faces. Consider $A_1=\{1,2\},A_2=\{2\},B=\{2,4,6\}.$ It is $P(A_i/B)=\frac13$ and $P(B)=\frac{1}{2}.$ However $P(A_1)=\frac13$ and$P(A_2)=\frac16$

mfl
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It cannot be. To see this, let's look at a couple of examples:

First: suppose that you flip a fair coin twice, and let $A$ be the event that the first is heads and $B$ the event that the second is heads. The coin flips are independent; so, in this case, $$ P(A)=.5\qquad P(B)=.5\qquad P(A\mid B)=.5 $$

Second: say we again flip two fair coins, and again let $B$ be the event that the second is heads. However, let $A$ be the event that BOTH are heads. Then $$ P(A)=.25\qquad P(B)=.5\qquad P(A\mid B)=.5 $$

If there was a way to determine $P(A)$ from $P(B)$ and $P(A\mid B)$, then these two examples would need to have the same value for $P(A)$; but, they don't!

Nick Peterson
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