P(a∩b): Given P (A) ≥ 0.99 and P (B) ≥ 0.97, show that P (A∩B) ≥ 0.96

koguter

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P(a∩b): Given P (A) ≥ 0.99 and P (B) ≥ 0.97, show that P (A∩B) ≥ 0.96

hi guys, i'm brand new to these probability themes and so confused about many of it's problems. So i have a question and hope you guys could help.

In a finite probability space (Ω, P) let A and B be events with P (A) ≥ 0.99 and P (B) ≥ 0.97. Show that P (A∩B) ≥ 0.96.

as i know when A and B independent are, P (A∩B) would = P (A) * P (B) and that would be easy to show. But since the task doesn't mention anything about it, how could i know if A and B independent or not or which way should i use to calculate P (A∩B) ? Thank you!
 

tkhunny

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Consider \(\displaystyle p(A^{c})\)
Consider \(\displaystyle p(B^{c})\)
 

koguter

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Consider \(\displaystyle p(A^{c})\)
Consider \(\displaystyle p(B^{c})\)
i haven't really understood can you please explain a little bit more about that ?
 

stapel

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i haven't really understood can you please explain a little bit more about that ?
Do you know what the notation means?
 

koguter

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Do you know what the notation means?
yeah it means the complement of events A and B and in this case would be 0.01 and 0.03. But i don't see how does it come to the solution ?
 

koguter

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Do you know what the notation means?
i mean if there's a chance that they're dependent then P (A∩B) = P (A|B) * P (B) so i would need to find P (A|B) somehow to solve that right ?
 

tkhunny

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No, it doesn't mean 0.01 or 0.03. It means AT MOST, 0.01 or 0.03.

What are the THREE possible arrangements of two sets in a specified universe?

1) They might intersect to some extent, but not entirely.
2) ??
3) ??
 

koguter

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No, it doesn't mean 0.01 or 0.03. It means AT MOST, 0.01 or 0.03.

What are the THREE possible arrangements of two sets in a specified universe?

1) They might intersect to some extent, but not entirely.
2) ??
3) ??
they intersect entirely or they don't intersect at all ?
 

tkhunny

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Great.

Word of Explanation: "intersect entirely" means one is a subset of the other.

Okay, now answer the question three times.
 

koguter

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Great.

Word of Explanation: "intersect entirely" means one is a subset of the other.

Okay, now answer the question three times.
So it means in the first case
P (A∩B) = P (A) * P (B)
second case
P (A∩B) = either P (A) or P(B)
and third case
P (A∩B) = 0 ? Am i right ?
 

tkhunny

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1) P(A) + P(B) > P(A Union B) {SOME intersection}
3) P(A) + P(B) = P(A Union B) {NO Intersection: Can't happen}
2) P(A Union B) = P(A) {Complete Intersection: B is a subset of A}

Case #2

P(A) > 0.99
P(A Intersect B) = P(B) = 0.97 > 0.96

Case #3 - A and B are too big for this.

Case #1
0.97 <= P(B) < 0.99 <= P(A)
P(~A) <= 0.01 -- This is the important point. Where can B hide?
 
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koguter

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1) P(A) + P(B) > P(A Union B) {SOME intersection}
3) P(A) + P(B) = P(A Union B) {NO Intersection: Can't happen}
2) P(A Union B) = P(A) {Complete Intersection: B is a subset of A}

Case #2

P(A) > 0.99
P(A Intersect B) = P(B) = 0.97 > 0.96

Case #3 - A and B are too big for this.

Case #1
0.97 <= P(B) < 0.99 <= P(A)
P(~A) <= 0.01 -- This is the important point. Where can B hide?
P(A Intersect B) would be 1 - things that don't belong to A and to B and would be 1 - 0.01 - 0.03 ? :D
 
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