# Thread: Find the indicated probability?

1. ## Find the indicated probability?

P(B)=0.5, P(A or B)=0.3, and P(A and B)=0.6, find P(A)

How would you do this?

The answer is {P(A)= 0.4} but how in the world do you get that? How do you do it?

Thanks!

2. Hello !

Do you know this formula ... ?

$\displaystyle \mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)$

Just draw a diagram

3. What kind of diagram? Like a Venn Diagram? Where would I put all the information though?

I don't get it

4. Originally Posted by hellotarina
P(B)=0.5, P(A or B)=0.3, and P(A and B)=0.6, find P(A)
The problem contains inconsistent data.
For all events $\displaystyle X~\&~Y$ this is true: $\displaystyle X \subseteq Y\; \Rightarrow \;P(X) \leqslant P(Y)$.

But look at the given: $\displaystyle B \subseteq A \cup B\;\text{but} \;P(B) > P(A \cup B).$
So is impossible.

5. For some reason my professor tells me the answer is P(A) = 0.4 and it is for a test which I have tomorrow and I didn't have a chance to ask about it.

I have made sure I've typed it correctly to no avail.

6. Originally Posted by hellotarina
For some reason my professor tells me the answer is P(A) = 0.4 and it is for a test which I have tomorrow and I didn't have a chance to ask about it.
Perhaps you copied it down incorrectly.
If it were $\displaystyle P(B)=0.5,~{\color{red}P(A\cup B)=0.6,~\&~ P(A\cap B)=.3}\text{ then }P(A)=0.4$ is correct.

7. Well thank you guys anyway.

8. Originally Posted by Plato
Perhaps you copied it down incorrectly.
If it were $\displaystyle P(B)=0.5,~{\color{red}P(A\cup B)=0.6,~\&~ P(A\cap B)=.3}\text{ then }P(A)=0.4$ is correct.
I suspect the question has been copied correctly but the professor who set the question has not thought past the following equation:

$\displaystyle \Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$

$\displaystyle \Rightarrow 0.3 = 0.5 + \Pr(B) - 0.6$

$\displaystyle \Rightarrow \Pr(B) = 0.4$.

@OP: Ask your professor how comfortable s/he is that it follows from the given data that $\displaystyle \Pr(B \cap A') = {\color{red}-} 0.1$.