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# Stats 1 help watch

1. Can someone help me with questions (iii). Shouldn't it just be P(A) + P(B) since these two probabilities include the event that they both occur?

2. (Original post by SWISH99)

Can someone help me with questions (iii). Shouldn't it just be P(A) + P(B) since these two probabilities include the event that they both occur?

this is normally given as P ( A ∪ B )

if you add P( A ) and P ( B ) you are counting the intersection ( overlap ) twice...
3. (Original post by the bear)
this is normally given as P ( A ∪ B )

if you add P( A ) and P ( B ) you are counting the intersection ( overlap ) twice...
Why would you be counting it twice. I thought the rule was the P(A or B) = P(A) + P(B) - P(A and B)
4. (Original post by SWISH99)
Why would you be counting it twice. I thought the rule was the P(A or B) = P(A) + P(B) - P(A and B)
It's counting it twice because

Source: https://math.dartmouth.edu/archive/m...Section6-2.pdf
5. (Original post by SWISH99)
Why would you be counting it twice. I thought the rule was the P(A or B) = P(A) + P(B) - P(A and B)
RDKGames has clearly explained.
6. (Original post by RDKGames)
It's counting it twice because

Source: https://math.dartmouth.edu/archive/m...Section6-2.pdf
OK thanks I understand this part now. Any idea on how to do (iv)
Im guessing its P(A' and B') / P(B')
7. (Original post by SWISH99)
OK thanks I understand this part now. Any idea on how to do (iv)
Im guessing its P(A' and B' ) / P(B' )
Yes.

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Updated: October 23, 2017
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