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Set theory/ Venn diagrams

Question 5d. https://imgur.com/a/iGwgu

I got the answer to this as 0.8, adding together the 0.4 from set A and the 0.4 outside of either set. According to the official solution, the answer is 0.9, which implies it includes the 0.1 from the intersection of a and b. I can't get my head around why this is, because to be NOT B, surely an intersection of A and B can't be possible? An intersection of A and B satisfies the criteria of A and B occurring, so if it occurs, B has occurred, and therefore it surely wouldn't count as NOT B?

We're looking for things in set A OR not in set B, right?



They've counted the intersection for A and B as being classed as in set A, but not in set B, is what it looks like to me. I cannot work this out.

Someone explain please :s-smilie:
(edited 6 years ago)
Reply 1
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Illidan2
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Think i've answered my own question. The intersection is in set A and that's fine, because it doesn't have to not be in set B, because it's "OR" not in set B. So I can count it.
Reply 2
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Illidan2
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Wow, can't believe I missed that :L i'd delete this thread but I don't know if it's possible for me to do so and if it is, I don't know how to do so :L
If P(B) = 0.2 then P(B’) = 1 - 0.2 = 0.8, regardless of any intersections.
Reply 4
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Illidan2
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Sorry, I edited my original post. I meant question 5d, I got 5b right the first time aha :L

Original post by old_engineer
If P(B) = 0.2 then P(B’) = 1 - 0.2 = 0.8, regardless of any intersections.
Original post by Illidan2
Sorry, I edited my original post. I meant question 5d, I got 5b right the first time aha :L


Can you express P(AuB) in terms of an intersection and other sets? It's a standard result, you should know.

Now just replace B with B' in that formula.

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