Proof of $P(A'\mid B)=1-P(A\mid B)$

Question

Could someone help me understand how to prove $P(A'\mid B) = 1-P(A \mid B)$? I tried to make it so:

$P(A'\mid B)= \cfrac{P(A'\cap B)}{P(B)}$ but I'm not sure how to continue.

(I see that there is a question asking to prove $$P(A|B) = 1-P(A'|B) $$ How is it possible that P(Not A |B) would yield the same proof as P(A|B)?

prove what? $P(A' \mid B)$ is not a formula – balddraz Jul 25 '15 at 06:56 — balddraz, Jul 25 '15 at 06:56

wythagoras · Accepted Answer · 2015-07-25T07:24:05.747

4

By the definition of conditional probability, we have $$P(C\mid B) = \frac{P(C \cap B)}{P(B)}$$

Taking $C=A'$, we have

$$P(A'\mid B) = \frac{P(A' \cap B)}{P(B)}=\frac{P(B)-P(A \cap B)}{P(B)}=1-\frac{P(A \cap B)}{P(B)}=1-P(A\mid B)$$

Where we have $P(A' \cap B)+P(A \cap B)=P(B)$ because we always have either $A$ or $A'$ happening in addition to $B$, but never both. Rewriting gives $P(A' \cap B)=P(B)-P(A \cap B)$.

This proves the asked.

For the edit, this is just rewriting.

$$P(A'\mid B)=1-P(A\mid B)$$

$$P(A'\mid B)+P(A\mid B)=1$$

$$P(A\mid B)=1-P(A'\mid B)$$

edited Jul 25 '15 at 07:24

answered Jul 25 '15 at 07:02

wythagoras

25,026

would it be possible use $$ \frac{P(A')-P(A' \cap B')}{P(B)}$$ to solve it? – nodel Jul 25 '15 at 07:19
@risdel $P(A')-P(A' \cap B') = P(A' \cap B)$. Not sure what we can achieve with this. – wythagoras Jul 25 '15 at 07:22

Proof of $P(A'\mid B)=1-P(A\mid B)$

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