P(A u B) =0.5 and P(A u (B^c))=0.6. A and B are independent. Find p(A).

[crybloodwing]

So I already did this question but without the condition that A and B are independent. In this, p(A)=.1.

I know if A and B are independent, then p(A and B) = p(A)p(B).

Therefore, p(A u B)= p(A)+p(B)-p(A)p(B).
So, 0.5 = p(A)+p(B)-p(A)p(B).

I have tried multiple ways to figure out what p(A) is, but I am unable to figure out what the p(B) is, which I think I need to solve this. I have tried drawing pictures also. I know that the complement of p(A u (B^c)) is .4, which I got from 1-p(A u (B^c)).

 

[Steven G]

Can you please show us at least one of you multiple attempts at solving this problem so we can guide you to the correct solution?

 

[pka]

So I already did this question but without the condition that A and B are independent. In this, p(A)=.1. I know if A and B are independent, then p(A and B) = p(A)p(B). Therefore, p(A u B)= p(A)+p(B)-p(A)p(B). So, 0.5 = p(A)+p(B)-p(A)p(B). I have tried multiple ways to figure out what p(A) is, but I am unable to figure out what the p(B) is, which I think I need to solve this. I have tried drawing pictures also. I know that the complement of p(A u (B^c)) is .4, which I got from 1-p(A u (B^c)).

First I agree that \(\displaystyle \mathcal{P}(A)=0.1\). If we look at the above: because \(\displaystyle \mathcal{P}(A\cup B)=0.5\) The complement (see diagram 1) is also \(\displaystyle 0.5\). Look at this link. \(\displaystyle \mathcal{P}(A\cap B^c)=0.4\) and \(\displaystyle \mathcal{P}(A\cup B)=0.5\) from the above it is clear that \(\displaystyle \mathcal{P}(A)=0.1\).
Now as to your comment about the independence of \(\displaystyle A~\&~B\). I wonder if this is just part of a larger question?
This is the case: if \(\displaystyle A~\&~B\) are independent then so are \(\displaystyle A~\&~B^c\) as are \(\displaystyle A^c~\&~B\) as well as \(\displaystyle A^c~\&~B^c\). If you have not proven that for yourself it is a very informative excise.

 

[Dr.Peterson]

As pka has implied, you don't need the condition of independence for this part of the problem, so that may be there for a subsequent part. (Adding a condition wouldn't change the answer, except possibly to make it invalid.)

But it is possible, knowing A and B are independent, to find P(B), which I expect them to ask eventually. You may need a little algebra. Suppose that P(B) = x. What equation can you write, using independence?

 

[crybloodwing]

As pka has implied, you don't need the condition of independence for this part of the problem, so that may be there for a subsequent part. (Adding a condition wouldn't change the answer, except possibly to make it invalid.)

But it is possible, knowing A and B are independent, to find P(B), which I expect them to ask eventually. You may need a little algebra. Suppose that P(B) = x. What equation can you write, using independence?

Would you suggest doing like a system of equations? Also, it was 2 questions only, nothing after this.

1. Given P(A u B) =0.5 and P(A u (B^c))=0.6. Find P(A).

This was simple and P(A) was 0.1.

2. Given P(A u B) =0.5 and P(A u (B^c))=0.6. A and B are independent. Find P(A).

Nothing after that.

 

[crybloodwing]

View attachment 14196
First I agree that \(\displaystyle \mathcal{P}(A)=0.1\). If we look at the above: because \(\displaystyle \mathcal{P}(A\cup B)=0.5\) The complement (see diagram 1) is also \(\displaystyle 0.5\). Look at this link. \(\displaystyle \mathcal{P}(A\cap B^c)=0.4\) and \(\displaystyle \mathcal{P}(A\cup B)=0.5\) from the above it is clear that \(\displaystyle \mathcal{P}(A)=0.1\).
Now as to your comment about the independence of \(\displaystyle A~\&~B\). I wonder if this is just part of a larger question?
This is the case: if \(\displaystyle A~\&~B\) are independent then so are \(\displaystyle A~\&~B^c\) as are \(\displaystyle A^c~\&~B\) as well as \(\displaystyle A^c~\&~B^c\). If you have not proven that for yourself it is a very informative excise.

1. Given P(A u B) =0.5 and P(A u (B^c))=0.6. Find P(A).

This was simple and P(A) was 0.1.

2. Given P(A u B) =0.5 and P(A u (B^c))=0.6. A and B are independent. Find P(A).

Nothing after that.

I knew the independence of all of the things you mentioned and did try to use that to find P(B) or P(A) but could not come up with anything.

 

[pka]

Would you suggest doing like a system of equations? Also, it was 2 questions only, nothing after this.
1. Given P(A u B) =0.5 and P(A u (B^c))=0.6. Find P(A).
This was simple and P(A) was 0.1.
2. Given P(A u B) =0.5 and P(A u (B^c))=0.6. A and B are independent. Find P(A).
Nothing after that.

I will use the older Moore notation: \(\displaystyle AB\) is \(\displaystyle A\cap B\) and \(\displaystyle B'\) is the complement of \(\displaystyle B\)
\(\displaystyle \mathcal{P}(A\cup B)=\mathcal{P}(A)+\mathcal{P}( B)-\mathcal{P}(A)\mathcal{P}( B)=0.5\)
\(\displaystyle \mathcal{P}(A\cup B')=\mathcal{P}(A)+\mathcal{P}( B')-\mathcal{P}(A)\mathcal{P}( B')=0.6\)
\(\displaystyle 2\mathcal{P}(A)+\mathcal{P}( B)+\mathcal{P}( B')-\mathcal{P}(A)(\mathcal{P}( B)+\mathcal{P}( B'))=1.1\)
\(\displaystyle 2\mathcal{P}(A)+1-\mathcal{P}(A)(1)=1.1\)
\(\displaystyle 2\mathcal{P}(A)-\mathcal{P}(A)(1)=0.1\)
\(\displaystyle \mathcal{P}(A)=0.1\)

 

[crybloodwing]

Can you please show us at least one of you multiple attempts at solving this problem so we can guide you to the correct solution?

I tried messing around with these equations and combining them in different ways. Letting x be the section of A that excludes the intersection of A and B.



0.5=P(A)+P(B)-P(A)P(B)
x= P(A)-P(A)P(B)
0.5= P(B)+x
P(A|B)=P(A)
P(A intersection B)/P(B) = P(A)
P(B)=.4+P(A)P(B)

 

[pka]

I tried messing around with these equations and combining them in different ways. Letting x be the section of A that excludes the intersection of A and B.
0.5=P(A)+P(B)-P(A)P(B)
x= P(A)-P(A)P(B)
0.5= P(B)+x
P(A|B)=P(A)
P(A intersection B)/P(B) = P(A)
P(B)=.4+P(A)P(B)

Have you gone through carefully reply #7. That proves that \(\displaystyle \bf{\mathcal{P}(A)=0.1}\).
Moreover, \(\displaystyle \bf{\mathcal{P}(A'B)=0.4}\) so that means \(\displaystyle {\mathcal{P}(A\cup B)=\mathcal{P}(A)+\mathcal{P}(A'B)=0.5}\)

 

[Steven G]

Yes, pka showed you all the work. Is there anything that you did not understand?