Complements, Intersections, and Unions

Statistics
Tutor: None Selected Time limit: 1 Day

Sep 28th, 2015

Thank you for the opportunity to help you with your question!

Hi ihalfpen,

First off, P(a) + P(b) +  P(c) +  P(d) = 0.32 + 0.17 + 0.28 + 0.23 = 0.49 + 0.51 = 1.

The "points" a, b,c, and d are distributed across the sets A, B and (A or C)' where the prime indicates complement, like the upper case c shown in the diagram. I'm using 'or' instead of the U shape and I'm using 'and' 
instead of the upside-down U because I can't easily display them.

a. P(A) = P(b) + P(c) = 0.17 + 0.28 = 0.45

b. P(B) = P(b) +  P(c) +  P(d) = 0.17 + 0.28 + 0.23 = 0.68

c. 1) P(A') = P(a) + P(d) = 0.32 + 0.23 = 0.55

2) P(A') = 1 - P(A) = 1 - 0.45 = 0.55

d. P(A and B) = P(b) + P(c) = 0.45

e)

1) P(A or B) = P(b) + P(c) + P(d)

2) P(A or B) = P(A) + P(B) - P(A and B) = 0.45 + 0.68 - 0.45 = 0.68

Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 29th, 2015

ihalfpen,

I think you also had a problem involving a normal distribution. I was working on it and lost my connection. If you want to post it again, I'll be on the lookout. Sorry.

Sep 29th, 2015

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Sep 28th, 2015
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Sep 28th, 2015
Dec 9th, 2016
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