P(A'nB) / P(A'uB)

Hi,

I DON'T want to know the formula. I want to understand the concept.

P(A'nB)
I've searched and searched, but I just don't understand it.
Firstly, the meaning. I know it's trying to say "the probability of not A AND B" - Right? So this essentially means P(A') + P(B)?
So, on a Venn Diagram, you'd shade in everything apart from the "ONLY A" section.
If so, it doesn't seem to be giving me the right answers.
For example,

This (http://postimg.org/image/p075pgf5x/) is a Venn diagram to represent the random selection of cards from a pack of 52. A represents Ace and D represents Diamond.

I understand everything apart from working out the probability of P(A'nD)
As stated above, I'd shade in everything apart from the "ONLY A" section which gives me 3/52.

But the book states that P(A'nD) is the "event the card chosen is not Ace and is a Diamond"

So is the concept I stated above completely wrong???

P(A'uB)
What does it mean?
How would it be represented on the Venn diagram?
In this question, what is P(A'uB)?

I just don't understand it because the book says one thing, and previous TSR threads say something else.

Thanks in advv!

Here is why I consider UNION to be the best description

You want all the Not A group to join with the B groups

So, now you have all the Not Aces and all the diamonds

This is all the cards apart from AC, AS, AH

Thanks, I understand the first part.

Isn't there a contradiction? All the Not Aces and all the diamonds (because the diamonds overlap with Aces)

The overlap is irrelevant

This is a union of 2 groups, if you happen to be in both groups that is ok


Or are you asking about the Ace of Diamonds, he gets in because ALL of the diamonds are allowed in this union

The overlap is irrelevant

This is a union of 2 groups, if you happen to be in both groups that is ok


Or are you asking about the Ace of Diamonds, he gets in because ALL of the diamonds are allowed in this union

Is this about right?

not the best images but, from what I can see, yes

not the best images but, from what i can see, yes

p(aub) = p(bua)?
P(a'ub) = p(b'ua)?

yes
no

P(A'uB) = P (BuA')
P(A'nB) = P (BnA')

I think they're right, just wanted to double check

Yes

$P(A \cup B) = P(B \cup A)$

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

No matter what sets A and B are

I understand it, apart from the "interpolate" part.

PS: It's to do with cumulative distribution functions