Probability clarification

B

Baron

Junior Member
Joined
Oct 3, 2010
Messages
73
I understand probability when it is written in words, but the symbols sometimes have me confused.

Please click on the attachment for the question

The answer to the P(\(\displaystyle \overline{ A }\)) is easy 6/11. But if the question was the P(\(\displaystyle \overline{ A}\) or \(\displaystyle \overline{ B}\)) what will the answer be? And apparently P(\(\displaystyle \overline{ A}\) or \(\displaystyle \overline{ B}\)) is not equal to the P(\(\displaystyle \overline{ A or B }\))

What is the difference between these two signs?

Is the probability of P(\(\displaystyle \overline{ A}\) or \(\displaystyle \overline{ B}\)) equal to the P(\(\displaystyle \overline{ A}\)) or P(\(\displaystyle \overline{ B}\))?

Thanks for any clarification.
 

Attachments

  • Untitled1.jpg
    Untitled1.jpg
    30.5 KB · Views: 153
There are 5 dots in A, 11 dots total. P(A) = 5/11.

There are 4 dots in B. P(B) = 4/11.

If you combine A and B, they only contain 8 dots, so the P(A or B) = 8/11.

We cannot simply add the probabilities of A and B because they overlap; they contain 1 dot in common. That dot would get counted twice if we simply added the two probabilities. Therefore,

P(A or B) = P(A) + P(B) - P(A and B)

Make sense?
 
It helps if the elements are distinct:

A contains 1, 2, 3
B contains 3, 4, 5
Universe conatans A and B and 6, 7

(Not A) contains 4, 5, 6, 7
(Not B) contains 1, 2, 6, 7
((Not A) and (Not B)) contains 6, 7

(A and B) contains 3
(Not (A and B)) contains 1, 2, 4, 5, 6, 7

You do it with "or".
 
tkhunny said:
It helps if the elements are distinct:

A contains 1, 2, 3
B contains 3, 4, 5
Universe conatans A and B and 6, 7

(Not A) contains 4, 5, 6, 7
(Not B) contains 1, 2, 6, 7
((Not A) and (Not B)) contains 6, 7

(A and B) contains 3
(Not (A and B)) contains 1, 2, 4, 5, 6, 7

You do it with "or".
Click to expand...

I don't understand your numbering of the elements. There are 4 elements in A

A contains 1, 2, 3, 4, 5
B contains 5, 6, 7, 8
Universe contains A and B and 9, 10, 11

(Not A) contains 6, 7, 8, 9, 10, 11
(Not B) contains 1, 2 ,3 , 4, 9, 10, 11
((Not A) or (Not B)) contains 1, 2, 3, 4, 6, 7, 8, 9, 10, 11 ??

(A or B) contains 1, 2, 3, 4, 5, 6, 7, 8,
(Not (A and B)) contains 9, 10, 11 ??

So is the probability of P(\(\displaystyle \overline{ A}\) or \(\displaystyle \overline{ B}\)) equal to the P(\(\displaystyle \overline{ A}\)) or P(\(\displaystyle \overline{ B}\))?
 
I don't understand your numbering of the elements. There are 4 elements in A

A contains 1, 2, 3, 4, 5
B contains 5, 6, 7, 8
Universe contains A and B and 9, 10, 11

(Not A) contains 6, 7, 8, 9, 10, 11
(Not B) contains 1, 2 ,3 , 4, 9, 10, 11
((Not A) or (Not B)) contains 1, 2, 3, 4, 6, 7, 8, 9, 10, 11 ??

(A or B) contains 1, 2, 3, 4, 5, 6, 7, 8,
(Not (A and B)) contains 9, 10, 11 ??

So is the probability of P(A or B) equal to the P(A) or P(B)?
Click to expand...

First statement: wrong; there are 5 elements in A.

The rest of your statements until the final statement are correct.

What part of P(A or B) = P(A) + P(B) - P(A and B) did you not understand? Please consult your book for verification.
 
Baron said:
I don't understand your numbering of the elements. There are 4 elements in A
Click to expand...

It's called an example and is useful for exploration. Knowledge gained can than be transferred to the problem at hand.
 
wjm11 said:
What part of P(A or B) = P(A) + P(B) - P(A and B) did you not understand? Please consult your book for verification.
Click to expand...

I understand that statement. I'm wondering about complements. The complement to P(A or B) is P(\(\displaystyle \overline{ A or B }\)) not P(\(\displaystyle \overline{ A}\) or \(\displaystyle \overline{ B}\))

Like I said:

Baron said:
And apparently P(\(\displaystyle \overline{ A}\) or \(\displaystyle \overline{ B}\)) is not equal to the P(\(\displaystyle \overline{ A or B }\))
Click to expand...

Is the P(\(\displaystyle \overline{ A}\) or \(\displaystyle \overline{ B}\)) equal to P(\(\displaystyle \overline{ A}\)) or P(\(\displaystyle \overline{ B}\)). Is it equal to P(\(\displaystyle \overline{ A and B }\)) too?

Just because the P(A or B) = P(A) + P(B) - P(A and B) does not mean the P(A or B) cannot equal to something else.

tkhunny said:
It's called an example and is useful for exploration. Knowledge gained can than be transferred to the problem at hand.
Click to expand...

It would have been helpful if you said it was an example. Confusion can then be minimized. It is called clarification.
 
Yeah I did. I think the problem I'm having is that I don't understand how the symbols of probability apply to word problems. I can solve word problems just fine through going through cases but give me a Venn diagram with symbols and I'm lost.

Does P(\(\displaystyle \overline{ A}\) or \(\displaystyle \overline{ B}\)) mean "the probability it is not A or the probability it is not B" in english?
 
Maybe you're leaning a little too much on teh English and not enough on the technical definition:

\(\displaystyle A\) - The collection of elements in set A
\(\displaystyle B\) - The collection of elements in set B
\(\displaystyle \bar{A}\) - The collection of elements NOT in set A
\(\displaystyle \bar{B}\) - The collection of elements NOT in set B
\(\displaystyle \bar{B}\;AND\;\bar{A};\) - The collection of elements in neither set A nor set B
\(\displaystyle \bar{B}\;OR\;\bar{A};\) - The collection of elements either not in A or not in B

The more elements included, the more difficult to understand without disection. DeMorgan can help us.

\(\displaystyle \bar{B}\;\lor\;\bar{A}\) - The collection of elements either not in A or not in B

\(\displaystyle \overline{B\;\land\;A}\) - The collection of elements not contained in both A and B - everything but the intersection.
 
You must log in or register to reply here.
Top