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Moonlight352
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#1
How is P(AUB) = P (A) + p(B) - p(AnB).
I dont get why because the minus p(AnB) means ur taking away the intersection part?
I dont get why because the minus p(AnB) means ur taking away the intersection part?
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Skiwi
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#2
(Original post by Moonlight352)
How is P(AUB) = P (A) + p(B) - p(AnB).
I dont get why because the minus p(AnB) means ur taking away the intersection part?
How is P(AUB) = P (A) + p(B) - p(AnB).
I dont get why because the minus p(AnB) means ur taking away the intersection part?
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Moonlight352
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#3
(Original post by Skiwi)
P(A U B) is the entire inside of the venn diagram. If you add P(A) and the P(B) then you are essentially adding the intersection part twice as P(A) and P(B) both contain the intersection bit. To make up for this you can take away one lot of the intersection P(AnB) to get what P(A U B) is.
P(A U B) is the entire inside of the venn diagram. If you add P(A) and the P(B) then you are essentially adding the intersection part twice as P(A) and P(B) both contain the intersection bit. To make up for this you can take away one lot of the intersection P(AnB) to get what P(A U B) is.
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