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If 40 people get the chance to pick a card from a canister [#permalink]
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10 Jul 2010, 06:08
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If 40 people get the chance to pick a card from a canister that contains 5 free passes to an amusement park mixed in with 35 blank cards what is the probability that the 40th person who picks will win?



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Re: probability of picking last [#permalink]
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10 Jul 2010, 08:17
35  lose, 5  win
Pick 5 people to win => 40C5 = total number of outcomes.
Favorable outcome is : First pick the 40th person, then pick any other 4. => 1*40C4
So, probability = 40C5 / 40C4
= 40!*36!*4! / (35!*5!*40!)
= 36/(35*5)
= 36 / 175



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Re: probability of picking last [#permalink]
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10 Jul 2010, 08:54
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mpevans wrote: if 40 people get the chance to pick a card from a canister that contains 5 free passes to an amusement park mixed in with 35 blank cards what is the probability that the 40th person who picks will win? I guess we have the situation when 40 people standing in a row and picking the cards one after another and check them in the end. We are asked what is the probability that 40th person win the pass? If so, then probability of picking the pass will be the same for all 40 people  \(\frac{5}{40}\), (initial probability of picking the pass (\(\frac{5}{40}\)) will be the same for any person in a row). AbhayPrasanna wrote: 35  lose, 5  win
Pick 5 people to win => 40C5 = total number of outcomes.
Favorable outcome is : First pick the 40th person, then pick any other 4. => 1*40C4
So, probability = 40C5 / 40C4
= 40!*36!*4! / (35!*5!*40!)
= 36/(35*5)
= 36 / 175 This way is also valid and must give the same result. The problem is that you calculated favorable outcomes incorrectly: when you pick 40th person to win, then you have only 39 left to pick 4 from, so # of favorable outcomes is \(1*C^4_{39}\). Also \(probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}\) and you wrote viseversa. So \(P=\frac{1*C^4_{39}}{C^5_{40}}=\frac{36*37*38*39}{4!}*\frac{5!}{36*37*38*39*40}=\frac{5}{40}\). Hope it helps.
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Re: probability of picking last [#permalink]
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10 Jul 2010, 22:13
I made the same. 4C39/5C40. Where 5C40  total # of outcomes. 4C39 means that 4 winning tickets were taken out by 39 persons.



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Re: probability of picking last [#permalink]
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11 Jul 2010, 06:03
The number of passes here is 40 (35 +5) And the number of people is also 40
How will this problem change if there are 10 passes available and 45 blank passes mixed in and there are 40 people?
will the probability of 40th person picking the pass be 10/55 = 2/11?
Can someone explain the favorable outcomes/total outcomes setup using combination's formula? thanks.



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Re: probability of picking last [#permalink]
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11 Jul 2010, 07:41
gmat1011 wrote: The number of passes here is 40 (35 +5) And the number of people is also 40
How will this problem change if there are 10 passes available and 45 blank passes mixed in and there are 40 people?
will the probability of 40th person picking the pass be 10/55 = 2/11?
Can someone explain the favorable outcomes/total outcomes setup using combination's formula? thanks. Yes, if there are 10 passes and 45 blank cards and only 40 people are to pick the cards the probability that 40th person will pick the pass will still be 10/55. Consider another example the deck of 52 cards. If we put them in a line, what is the probability that 40th card will be an ace? As there are 4 aces then probability that any particular card in a line is an ace is 4/52. Hope it helps.
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Re: probability of picking last [#permalink]
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12 Jul 2010, 11:41
Bunuel wrote: mpevans wrote: if 40 people get the chance to pick a card from a canister that contains 5 free passes to an amusement park mixed in with 35 blank cards what is the probability that the 40th person who picks will win? I guess we have the situation when 40 people standing in a row and picking the cards one after another and check them in the end. We are asked what is the probability that 40th person win the pass? If so, then probability of picking the pass will be the same for all 40 people  \(\frac{5}{40}\), (initial probability of picking the pass (\(\frac{5}{40}\)) will be the same for any person in a row). AbhayPrasanna wrote: 35  lose, 5  win
Pick 5 people to win => 40C5 = total number of outcomes.
Favorable outcome is : First pick the 40th person, then pick any other 4. => 1*40C4
So, probability = 40C5 / 40C4
= 40!*36!*4! / (35!*5!*40!)
= 36/(35*5)
= 36 / 175 This way is also valid and must give the same result. The problem is that you calculated favorable outcomes incorrectly: when you pick 40th person to win, then you have only 39 left to pick 4 from, so # of favorable outcomes is \(1*C^4_{39}\). Also \(probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}\) and you wrote viseversa. So \(P=\frac{1*C^4_{39}}{C^5_{40}}=\frac{36*37*38*39}{4!}*\frac{5!}{36*37*38*39*40}=\frac{5}{40}\). Hope it helps. Hi... thanks for the explanation Bunuel But I do not understand how can the probability of selecting a free pass =5/40 for all in case we assume the people are picking the cards and keeping it with them. won't it keep reducing as 4/39 for the second successfull fick of a free card... Please explain it seems i am missing some logic somewhere.



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Re: probability of picking last [#permalink]
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12 Jul 2010, 11:54
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utin wrote: Hi... thanks for the explanation Bunuel
But I do not understand how can the probability of selecting a free pass =5/40 for all in case we assume the people are picking the cards and keeping it with them.
won't it keep reducing as 4/39 for the second successfull fick of a free card... Please explain it seems i am missing some logic somewhere. Consider this: put 40 cards on the table and 40 people against them. What is the probability that the card which is against the 40th person is the winning one? 5/40, it's the same probability as for the first, second, ... for any. When we pick the cards from a canister and not knowing the results till the end then it's basically the same scenario.
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Re: If 40 people get the chance to pick a card from a canister [#permalink]
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21 Dec 2012, 23:35
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Consider 40 places to arrange the 40 cards with 35 blank and 5 passes = 40!/(35!*5!) Favorable outcome is when the 40th place contains a pass, so we have 39 places to arrange 35 blanks and 4 passes = 39!/(4!*35!) =(39!*35!*5!)/(40!*35!*4!) = 1/8



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Re: If 40 people get the chance to pick a card from a canister [#permalink]
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22 Dec 2012, 06:43
geneticsgene wrote: Consider 40 places to arrange the 40 cards with 35 blank and 5 passes = 40!/(35!*5!) Favorable outcome is when the 40th place contains a pass, so we have 39 places to arrange 35 blanks and 4 passes = 39!/(4!*35!) =(39!*35!*5!)/(40!*35!*4!) = 1/8 Hello! I am unfamiliar with ! in math, what does it mean? Thanks in advance



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Re: If 40 people get the chance to pick a card from a canister [#permalink]
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22 Dec 2012, 06:52



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Re: If 40 people get the chance to pick a card from a canister [#permalink]
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22 Dec 2012, 07:22
Bunuel wrote: Hiho wrote: geneticsgene wrote: Consider 40 places to arrange the 40 cards with 35 blank and 5 passes = 40!/(35!*5!) Favorable outcome is when the 40th place contains a pass, so we have 39 places to arrange 35 blanks and 4 passes = 39!/(4!*35!) =(39!*35!*5!)/(40!*35!*4!) = 1/8 Hello! I am unfamiliar with ! in math, what does it mean? Thanks in advance The factorial of a nonnegative integer \(n\), denoted by \(n!\), is the product of all positive integers less than or equal to \(n\). For example: \(4!=1*2*3*4=24\). Hope it helps. Yes, it does. Thanks. I understand the concept, but not the use of it in this particular case. Is it tested on the GMAT, or is it just additional help on some questions for those who are familiar with it?



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Re: probability of picking last [#permalink]
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22 Dec 2012, 11:38
Hi, Please help me out here... we have 40 cards with 5 valid passes and rest junks we have 40 people ... The probability of 1st person picking junk is 35/40 and then he doesnt replace the card rite.. he takes it with him.. so now we are left with 39 cards.. The probability of 2nd person taking a junk card is 34/39 right??? so wont it be 35/40 x 34/39 x 33/38 x ...... 1/5??? what am I missing here pls? Bunuel wrote: mpevans wrote: if 40 people get the chance to pick a card from a canister that contains 5 free passes to an amusement park mixed in with 35 blank cards what is the probability that the 40th person who picks will win? I guess we have the situation when 40 people standing in a row and picking the cards one after another and check them in the end. We are asked what is the probability that 40th person win the pass? If so, then probability of picking the pass will be the same for all 40 people  \(\frac{5}{40}\), (initial probability of picking the pass (\(\frac{5}{40}\)) will be the same for any person in a row). AbhayPrasanna wrote: 35  lose, 5  win
Pick 5 people to win => 40C5 = total number of outcomes.
Favorable outcome is : First pick the 40th person, then pick any other 4. => 1*40C4
So, probability = 40C5 / 40C4
= 40!*36!*4! / (35!*5!*40!)
= 36/(35*5)
= 36 / 175 This way is also valid and must give the same result. The problem is that you calculated favorable outcomes incorrectly: when you pick 40th person to win, then you have only 39 left to pick 4 from, so # of favorable outcomes is \(1*C^4_{39}\). Also \(probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}\) and you wrote viseversa. So \(P=\frac{1*C^4_{39}}{C^5_{40}}=\frac{36*37*38*39}{4!}*\frac{5!}{36*37*38*39*40}=\frac{5}{40}\). Hope it helps.



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Re: If 40 people get the chance to pick a card from a canister [#permalink]
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23 Dec 2012, 06:18



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Re: probability of picking last [#permalink]
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23 Dec 2012, 06:22
SpotlessMind wrote: Hi, Please help me out here... we have 40 cards with 5 valid passes and rest junks we have 40 people ... The probability of 1st person picking junk is 35/40 and then he doesnt replace the card rite.. he takes it with him.. so now we are left with 39 cards.. The probability of 2nd person taking a junk card is 34/39 right??? so wont it be 35/40 x 34/39 x 33/38 x ...... 1/5??? what am I missing here pls? Bunuel wrote: mpevans wrote: if 40 people get the chance to pick a card from a canister that contains 5 free passes to an amusement park mixed in with 35 blank cards what is the probability that the 40th person who picks will win? I guess we have the situation when 40 people standing in a row and picking the cards one after another and check them in the end. We are asked what is the probability that 40th person win the pass? If so, then probability of picking the pass will be the same for all 40 people  \(\frac{5}{40}\), (initial probability of picking the pass (\(\frac{5}{40}\)) will be the same for any person in a row). AbhayPrasanna wrote: 35  lose, 5  win
Pick 5 people to win => 40C5 = total number of outcomes.
Favorable outcome is : First pick the 40th person, then pick any other 4. => 1*40C4
So, probability = 40C5 / 40C4
= 40!*36!*4! / (35!*5!*40!)
= 36/(35*5)
= 36 / 175 This way is also valid and must give the same result. The problem is that you calculated favorable outcomes incorrectly: when you pick 40th person to win, then you have only 39 left to pick 4 from, so # of favorable outcomes is \(1*C^4_{39}\). Also \(probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}\) and you wrote viseversa. So \(P=\frac{1*C^4_{39}}{C^5_{40}}=\frac{36*37*38*39}{4!}*\frac{5!}{36*37*38*39*40}=\frac{5}{40}\). Hope it helps. You are finding the probability that the first 34 people will not win and the 35th person wins, which is clearly not what we were asked to get.
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