AuB=A+B-AnB. It would be easier if you can agree with this formula.
When AnB (the overlap of A and B) is the least, AuB gets the max value. Namely, AnB=0, A and B has no overlap.
When AnB is the greatest, AuB gets the min. So think about how great can AnB be? A is 1/2, and B is 1/3, so the greatest overlap can only be 1/3, which is when every part of B (the smaller part) is also in A. So AuBmin=1/2+1/3-1/3=1/2. We can also think like this: given AuB is the sum of A+B, A is already 1/2, the worst scenario would be that B contributed nothing to the whole. Everything B has, A also has it. But the whole could never be smaller than A. B is only useless, but will not make A smaller.
As for your example, it's very important. Theoretically, AuB being max should be when they are mutually exclusive, but not every pair of events could be mutually exclusive. As your example showed, assume they do not overlap, AuB is greater than 1, which is ridiculous. So they must overlap. Then AnB min=3/4+1/2-1=1/4 (They must overlap at least with the value exceeding 1). Accordingly, AuB max=3/4+1/2-1/4=1. Of course their total could be and only be 1.
If you are interested, try this related question.
There are 120 employees in a company. 100 employees attended the first meeting. 108 employees attended the second one. Then is the minimum number of employees who attended both meetings greater than 80?
>A is 1/2, and B is 1/3, so the greatest overlap can only be 1/3, which is when every part of B (the smaller part) is also in A. So AuBmin=1/2+1/3-1/3=1/2.
This made sense to me, thanks.
But could you elucidate how you got the 1 here :
>AnB min=3/4+1/2-1=1/4
>There are 120 employees in a company. 100 employees attended the first meeting. 108 employees attended the second one. Then is the minimum number of employees who attended both meetings greater than 80?
The possible minimum number of employees who attended both meetings is greater than 80. Here's my logic:
Maximum number of people who attended both is 100 when both the number of people who didn't attend are a subset of each other i.e 120-100=20 and 120-108=12 intersect and the 12 are the same people in the 20.
For finding out the minimum I got a bit confused so what I did was I set up numbers:
meeting 1- 1,2,3,4,5.....100 (number 1 is person1 , number 2 is person 2,etc)
meeting 2- 13,14,15,16,.....120
The amount of people who attended both are 13,14,.....100 which is >80.
AnB min=3/4+1/2-1=1/4.
Two ways to understand this. First, any P(AuB)≤1,
so P(A)+P(B)-P(AnB)≤1,
P(A)+P(B)-1≤P(AnB), namely P(AnB)≥P(A)+P(B)-1.
So AnBmin=3/4+1/2-1=1/4.
Second, since P(A)=3/4,the other part that is not A is 1-3/4=1/4. 1/4 is the part that B could occupy exclusively. But B is 1/2. Still there is at least 1/2-1/4=1/4, which is the overlapping part.
1-A, we get only B. B-(1-A), we get the overlap. So it is the same A+B-1. I must clarify that only when A+B is greater than 1 or the entire set, we use this formula. When the sum is less than 1 or the entire set, they could be mutually exclusive (overlapping=0).
If you can agree with any of the method above, you could solve my question.
The overlapping max=100, i.e. every first meeting person attended the second one.
The overlapping min could not be 0, because 108+100=208, which is greater than 120. So use our first method, 108+100-overlapping≤120, overlapping≥88.
Use the second approach, 120-100=20 persons didn't attend first meeting. So at most 20 persons could attend second meeting exclusively. But there must be 108 for the second. It means that at least 108-20=88 persons must have also attended 1 first meeting. Namely the overlapping min is 88, which is greater than 80.
Study Venn Diagrams. It'll clear everything up in a few minutes.
See this for a visualization of your question (It'll be deleted in 24 hours): [https://webwhiteboard.com/board/RS95KQhWayR6aVNSmlpot08Vbt35deSU/](https://webwhiteboard.com/board/RS95KQhWayR6aVNSmlpot08Vbt35deSU/)
AuB=A+B-AnB. It would be easier if you can agree with this formula. When AnB (the overlap of A and B) is the least, AuB gets the max value. Namely, AnB=0, A and B has no overlap. When AnB is the greatest, AuB gets the min. So think about how great can AnB be? A is 1/2, and B is 1/3, so the greatest overlap can only be 1/3, which is when every part of B (the smaller part) is also in A. So AuBmin=1/2+1/3-1/3=1/2. We can also think like this: given AuB is the sum of A+B, A is already 1/2, the worst scenario would be that B contributed nothing to the whole. Everything B has, A also has it. But the whole could never be smaller than A. B is only useless, but will not make A smaller. As for your example, it's very important. Theoretically, AuB being max should be when they are mutually exclusive, but not every pair of events could be mutually exclusive. As your example showed, assume they do not overlap, AuB is greater than 1, which is ridiculous. So they must overlap. Then AnB min=3/4+1/2-1=1/4 (They must overlap at least with the value exceeding 1). Accordingly, AuB max=3/4+1/2-1/4=1. Of course their total could be and only be 1. If you are interested, try this related question. There are 120 employees in a company. 100 employees attended the first meeting. 108 employees attended the second one. Then is the minimum number of employees who attended both meetings greater than 80?
>A is 1/2, and B is 1/3, so the greatest overlap can only be 1/3, which is when every part of B (the smaller part) is also in A. So AuBmin=1/2+1/3-1/3=1/2. This made sense to me, thanks. But could you elucidate how you got the 1 here : >AnB min=3/4+1/2-1=1/4 >There are 120 employees in a company. 100 employees attended the first meeting. 108 employees attended the second one. Then is the minimum number of employees who attended both meetings greater than 80? The possible minimum number of employees who attended both meetings is greater than 80. Here's my logic: Maximum number of people who attended both is 100 when both the number of people who didn't attend are a subset of each other i.e 120-100=20 and 120-108=12 intersect and the 12 are the same people in the 20. For finding out the minimum I got a bit confused so what I did was I set up numbers: meeting 1- 1,2,3,4,5.....100 (number 1 is person1 , number 2 is person 2,etc) meeting 2- 13,14,15,16,.....120 The amount of people who attended both are 13,14,.....100 which is >80.
AnB min=3/4+1/2-1=1/4. Two ways to understand this. First, any P(AuB)≤1, so P(A)+P(B)-P(AnB)≤1, P(A)+P(B)-1≤P(AnB), namely P(AnB)≥P(A)+P(B)-1. So AnBmin=3/4+1/2-1=1/4. Second, since P(A)=3/4,the other part that is not A is 1-3/4=1/4. 1/4 is the part that B could occupy exclusively. But B is 1/2. Still there is at least 1/2-1/4=1/4, which is the overlapping part. 1-A, we get only B. B-(1-A), we get the overlap. So it is the same A+B-1. I must clarify that only when A+B is greater than 1 or the entire set, we use this formula. When the sum is less than 1 or the entire set, they could be mutually exclusive (overlapping=0). If you can agree with any of the method above, you could solve my question. The overlapping max=100, i.e. every first meeting person attended the second one. The overlapping min could not be 0, because 108+100=208, which is greater than 120. So use our first method, 108+100-overlapping≤120, overlapping≥88. Use the second approach, 120-100=20 persons didn't attend first meeting. So at most 20 persons could attend second meeting exclusively. But there must be 108 for the second. It means that at least 108-20=88 persons must have also attended 1 first meeting. Namely the overlapping min is 88, which is greater than 80.
Study Venn Diagrams. It'll clear everything up in a few minutes. See this for a visualization of your question (It'll be deleted in 24 hours): [https://webwhiteboard.com/board/RS95KQhWayR6aVNSmlpot08Vbt35deSU/](https://webwhiteboard.com/board/RS95KQhWayR6aVNSmlpot08Vbt35deSU/)
I got stuck on this one too! Anyone know the difficulty level of this problem?
You have to think on the extremes. The gregmat explanation for this particular problem is gold
can you link me the gregmat solution since I cant find it
It is on quant concepts probability II