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stolyar, paul, stoolfi, praetorian and akamaibrah are [#permalink]
11 Feb 2004, 16:21

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0% (00:00) correct
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stolyar, paul, stoolfi, praetorian and akamaibrah are friends and hosting individual newyear parties. Folks from GMAT club are invited to join.
Stolyar can accomodate 25
Paul can accomodate 30
Stoolfi can acomodate 40
Praetorian can accomodate 15
Akamaibrah can accomodate 10

They are yet to send out personal invitations. Every person who receives invitation will attend 100%.

They together (5 of them ) consider it as successfull year only if atleast 3 of them can attract as many members as they can accomodate. What is the minimum number of invitations that should be sent out.

Hmmm, I wonder if this question has some kind of trap. I also got 50 but is it really that simple? If yes then the explanation is that, provided that every card sent has 100% positive response, we only need to satisfy Akamaibrah, Praetorian and Stolyar's accomodation capacity. Therefore, 10+15+25=50. I would be very surprised if it were this simple though... _________________

Oh, I see now. Then I think the answer is:
minimum # of people needed to make the event successful is Akamaibrah, Praetorian and Stolyar's accomodation capacity. Therefore, 10+15+25=50. Probability of anyone attending 1 of the 5 houses is 1/5. Then 50 / [1/5] = 250 invitations should be sent out at the minimum _________________

Oh, I see now. Then I think the answer is: minimum # of people needed to make the event successful is Akamaibrah, Praetorian and Stolyar's accomodation capacity. Therefore, 10+15+25=50. Probability of anyone attending 1 of the 5 houses is 1/5. Then 50 / [1/5] = 250 invitations should be sent out at the minimum

maximum capacity (all 5 hosts) is 120.
So, it can't be 250.

Yes, could not find out an answer right now but 250 is wrong. I again made a stupid mistake and interpreted the premise as probability of people accepting is 1 out of 5. Very bad. Sleep now. I'll think about it tomorrow unless someone comes up with the answer before that _________________

Gee, I'm back to this problem when I have to be sleeping now. I can't think of an answer but yet can't sleep without figuring it out. This is gonna haunt me tonight. _________________

What Kpadma said makes sense. I slept with this problem in mind and came up with 10*5 + 4*5 + 3*10 = 100 invitations sent out provided that there is an even probability on the invitation cards of each one being chosen. But 118 seems right. _________________