# Another event probability question

For two events A and B,

P(A) = 0.2,
P(B) = 0.3,
P(~A | ~B) = 0.8


I need to calculate P(~B | A), but I'm not sure if what I'm doing is correct.

So knowing P(~A | ~B), I can say

P(A | ~B) = 0.2.


Now if we try to calculate P(~B|A) using Bayes, we get

P(~B | A) = P(A|~B)*P(~B) / (P(A|~B)*P(~B) + P(A|B)P(~A)).


The only thing that we don't know if P(A|B), but that's equal to P(A&B)/P(B).

P(A) = P(A&B) + P(A&~B)
P(A&~B) = P(A|~B)*P(~B) = 0.2*0.7 = 0.14
So, P(A&B) = 0.2 - 0.14 = 0.06

P(A|B) is then 0.06/0.3 = 0.2

So putting P(A|B) back in the formula we get 0.2*0.7 / (0.2*0.7 + 0.2*0.8)
= .14 / .3 = .46


That 0.46 doesn't seem correct though. Can someone point me in the right direction please?

• you may want to check your Bayes formula..
– Nana
May 16, 2012 at 0:40
• Oh, yes...the last bit is wrong. It's P(B) instead of P(~A), in which case it's 0.3*0.6. So the total is 0.14/0.2 = 0.7, right? May 16, 2012 at 0:46
• right. ${}{}{}{}{}{}$
– Nana
May 16, 2012 at 0:50

Since you know both unconditional probabilities ($P(A)$ and $P(B)$) and one conditional probability, you can derive one joint probability easily: $$0.8 = P(\neg A \vert \neg B) = \frac{P(\neg A \wedge \neg B)}{P(\neg B)}=\frac{P(\neg A \wedge \neg B)}{0.7} \implies P(\neg A \wedge \neg B)=0.8\cdot0.7=0.56.$$ The other joint probabilities follow by subtraction: $$\begin{eqnarray} P(A \wedge \neg B) &=& P(\neg B) - P(\neg A \wedge \neg B) = 0.7 - 0.56=0.14; \\ P(\neg A \wedge B) &=& P(\neg A) - P(\neg A \wedge \neg B) = 0.8 - 0.56=0.24; \\ P(A \wedge B) &=& P(A) - P(A \wedge \neg B) = 0.2 - 0.14 = 0.06. \end{eqnarray}$$ You can conclude that $$P(\neg B \vert A) = \frac{P(\neg B\wedge A)}{P(A)}=\frac{0.14}{0.2}=0.7$$ (and any of the other conditional probabilities could be written down at this point in the same way, since you now have all the pieces).
$$P(A\text{ and ~ }B)) = P(\text{~} B\text{ and }A)$$ $$P(A | \text{~}B)*P(\text{~}B) = P(\text{~}B | A)*P(A)$$ $$0.2 * 0.7 = P(\text{~}B | A)*0.2$$ $$0.7 = P(\text{~}B | A)$$