How to find P(B)?

[Harry_the_cat]

I think my brain is fried. I have a problem which I thought would be easy but am missing something.

The problem boils down to knowing that P(A) = 0.25, P(B!A) = 0.8, P(B!A') = 0.1.

I have calculated P(A and B) = 0.2 and P(A' and B')=0.675.

Also calculated P(A' and B) = 0.075 and P(A and B') = 0.05.

Can someone please point me to the path to find P(B)?

 

[Harry_the_cat]

P(B and A) + P(B and A') = P(B) right?

Yes it does.

So P(B) = 0.2 + 0.075 = 0.275. Correct?

I think my brain works again! ??

 

[JeffM]

$$P(B \ | \ A) = \dfrac{P(B \ \& \ A)}{P(A)} \implies P(B \ \& \ A) = 0.25 * 0.8 = 0.2.\\ P(\neg A) = 1 - P(A) = 1 - 0.25 = 0.75.\\ P(B \ | \ \neg A) = \dfrac{P(B \ \& \ \neg A)}{P(\neg A)} \implies P(B \ \& \ \neg A) = 0.75 * 0.1 = 0.075.\\ P(B) = P(B \ | \ A) + P(B \ | \ \neg A) = 0.2 + 0.075 = 0.275.$$
You deserve a saucer of cream.

 

[pka]

The problem boils down to knowing that P(A) = 0.25, P(B!A) = 0.8, P(B!A') = 0.1.

Here is my short way.
$\mathcal{P}(B|A)=\dfrac{\mathcal{P}(B\cap A)}{\mathcal{P}(A)}=0.8$. Because $\mathcal{P}(A)=0.25$ we get $\mathcal{P}(A\cap B)=0.2$.
Likewise from $\mathcal{P}(B|A^c)=0.1$ we get $\mathcal{P}(A^c\cap B)=0.075$.
So $\mathcal{P}(B)=\mathcal{P}(A\cap B)+\mathcal{P}(A^c\cap B)=0.2+0.075=0.275$

$$$$$$$\mathcal{P}$

 

[Harry_the_cat]

Thanks guys for confirming my brain still works. ?