Question

It can be shown that for any three events, A,B, and C, the probability that at
least one of them will occur is given by
P(AUBUC=P(A)+P(B)+P(C)-P(AintersectionB)-P(AintersectionC)-P(BintersectionC)+P(AintersectionBintersectionC))
Verify that this formula holds the probabilities of the follow venn diagram.

Answer 1

The top left circle is A the next one to the right is B and the botttom one is C. I'll post up what I got so far.

Answer 2

I've tried solving this two ways but can't seem to figure out what my mistake is:
The union of A, B,C is .86 so the other side must equal the same.


attempt 1:

.89-.06+.04-.16+.04-.04+.11+.04=.86

where,.89 is the P(A)+P(B)+P(C), .06+.04 is P(AintersectionB), .16+.04
P(AintersectionC), .04 + .11 is P(BintersectionC), and .04 is
P(AintersectionBintersectionC)).

attempt 2:

.89-.06-.16-.11+.04=.6

Where is my mistake?

Answer 3

at attempt 2 why do u leave out other probability (on intersection)

Answer 4

I wasn't quite sure about the values of intersection between each attempt,
in particular P(A intersection B),P(A intersection C) and P(B intersection C).
Am I just looking at what is contained in both subsets in general or in all cases? For example, would P(A intersection B) be .06, and .04 or just .06?

Answer 5

check this this clear tell us that attemp two is correct
http://www.brynmawr.edu/math/people/anmyers/PAPERS/Venn.pd

Answer 6

If that is correct, then I'm still unsure of how that right side is suppose to be .89. I'm sure I've done everything correctly.