# Math Help - Mr Rayon's easy probability questions

1. ## Mr Rayon's easy probability questions

If $Pr(A)= 0.6$, $Pr(B)= 0.5$ and $Pr({A}\cap B') = 0.36$, find:

a) $Pr({A}'\cup B)$
b) $Pr(A\mid {B}')$
c) $Pr(B\mid{A}')$

As you guys may notice, I have trouble figuring these types of questions out if one of the probability sets is a A' and the other is B or when one is B' and the other is A'. It's SO confusing. I just don't see the logic. Would anybody be able to draw some venn diagrams for me to help me understand this basic concept? I'm really embarrassed to ask this question in school because I'm afraid it's too basic.

Anyway, any help would be appreciated dearly!

$P(A')=1-P(A)~\&~P(B')=1-P(B)$

$P(A)=P(A\cap B)+P(A\cap B')$

$P(B)=P(A\cap B)+P(A'\cap B)$

$P(A|B')=\dfrac{P(A\cap B')}{P(B')}$

3. Originally Posted by Plato
$P(A)=P(A\cap B)+P(A\cap B')$

$P(B)=P(A\cap B)+P(A'\cap B)$

$P(A|B')=\dfrac{P(A\cap B')}{P(B')}$
I don't understand what these mean conceptually. I just don't see the logic. For example for finding Pr(A) and Pr(B) do the B and the B' cancel each other out somehow for getting Pr(A)? Could somebody draw venn diagrams because I cannot visually imagine these even though I roughly know what the textbook definition of them is asking me.

4. I really think the sort of help that you require is beyond what you can reasonably expect to get from a site such as this one.
You need to sit down with a live tutor.

5. Originally Posted by Mr Rayon
If $Pr(A)= 0.6$, $Pr(B)= 0.5$ and $Pr({A}\cap B') = 0.36$, find:

a) $Pr({A}'\cup B)$
b) $Pr(A\mid {B}')$
c) $Pr(B\mid{A}')$

As you guys may notice, I have trouble figuring these types of questions out if one of the probability sets is a A' and the other is B or when one is B' and the other is A'. It's SO confusing. I just don't see the logic. Would anybody be able to draw some venn diagrams for me to help me understand this basic concept? I'm really embarrassed to ask this question in school because I'm afraid it's too basic.

Anyway, any help would be appreciated dearly!
Draw a Karnaugh table that represents the given information:

$\begin{tabular}{l | c | c | c} & A & A' & \\ \hline B & & & 0.5 \\ B' & 0.36 & & \\ \hline & 0.6 & & 1 \\ \end{tabular}
$

Fill in the gaps.

Read off the required probabilities. I will give the answer to (c) (which of course relies on you having filled in the gaps):

0.26/0.4 = 26/40 = 13/20 = 0.65.

You should be able to see where this has come from.