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!
 
Consider \(\displaystyle p(A^{c})\)
Consider \(\displaystyle p(B^{c})\)
 
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) ??
 
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 ?
 
Great.

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

Okay, now answer the question three times.
 
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 ?
 
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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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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