1. Good post? |

## Re: sets

realized I screwed up, ignore this post

2. Good post? |

## Re: sets

realized I screwed up, ignore this post

3. Good post? |

## Re: sets

I was looking for a way to draw a venn diagram. I do not know how draw that here. I got this link which has venn diagram. Please check this link.

http://www.urch.com/forums/showthrea...85%25+surveyed

4. 1 out of 1 members found this post helpful. Good post? |

## Re: sets

After checking out Scoot's link (and following the links from there), I have found the following general formulas for three-component set problems:

If there are three sets A, B, and C, then
P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)

Number of people in exactly one set =
P(A) + P(B) + P(C) – 2P(AnB) – 2P(AnC) – 2P(BnC) + 3P(AnBnC)

Number of people in exactly two of the sets =
P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC)

Number of people in exactly three of the sets =
P(AnBnC)

Number of people in two or more sets =
P(AnB) + P(AnC) + P(BnC) – 2P(AnBnC)

5. Good post? |

## Re: sets

[QUOTE=If there are three sets A, B, and C, then
P(AnBnC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)

=/QUOTE]

shouldnt this be a UNION of sets???

6. Good post? |

## Re: sets

sorry about that ish, I just fixed it

7. Good post? |

## Re: sets

these formulas are really good.Thanks for the trouble.

8. Good post? |

## Re: sets

Let X be P(AnB) + P(AnC) + P(BnC).
Using the formulas above, we have

85 = 50+30+20-X+5

X = 20

Number of people having two or more = X - 2(AnBnC) = 20 - 10 = 10

9. Good post? |

## Re: sets

To add to TwinSplitter's forumla list:

No of people in atleast 1 set =
P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)

This formula can also be derived from:
No of people in atleast 1 set =
no of people in exactly 1 set + no of people in exactly 2 set + no of people in exactly 3 set

~~~~~~~~
More you look, more you see!

10. Good post? |
Originally Posted by hem
To add to TwinSplitter's forumla list:

No of people in atleast 1 set =
P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)

This formula can also be derived from:
No of people in atleast 1 set =
no of people in exactly 1 set + no of people in exactly 2 set + no of people in exactly 3 set
Shouldnt this be

P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + P(AnBnC)
When I add no of people in exactly 1 set with no of people in exactly 2 set I see that 3P(AnBnC) gets cancelled and then I add P(AnBnc) for exactly 3 set.