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Intern  Joined: 10 Jul 2010
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If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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2
17 00:00

Difficulty:   25% (medium)

Question Stats: 62% (01:42) correct 38% (01:44) wrong based on 96 sessions

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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?

A. 1/8
B. 36/175
C. 117/175
D. 139/175
E. 7/8
Math Expert V
Joined: 02 Sep 2009
Posts: 64322
Re: If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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4
2
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 vise-versa.

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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Manager  Joined: 03 May 2010
Posts: 72
WE 1: 2 yrs - Oilfield Service
Re: If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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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
Manager  Joined: 03 Jun 2010
Posts: 115
Location: United States (MI)
Concentration: Marketing, General Management
Re: If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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4C39/5C40.
Where 5C40 - total # of outcomes.
4C39 means that 4 winning tickets were taken out by 39 persons.
Manager  Joined: 11 Jul 2010
Posts: 150
Re: If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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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 set-up using combination's formula? thanks.
Math Expert V
Joined: 02 Sep 2009
Posts: 64322
Re: If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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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 set-up 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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Manager  Joined: 26 Mar 2010
Posts: 75
Re: If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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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 vise-versa.

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 64322
Re: If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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2
1
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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Intern  Joined: 12 Jan 2012
Posts: 21
GMAT 1: 720 Q49 V39
Re: If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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1
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
Intern  Joined: 20 Dec 2012
Posts: 16
Re: If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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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?

Math Expert V
Joined: 02 Sep 2009
Posts: 64322
Re: If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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1
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?

The factorial of a non-negative 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.
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Intern  Joined: 20 Dec 2012
Posts: 16
Re: If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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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?

The factorial of a non-negative 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?
Intern  Joined: 22 Dec 2012
Posts: 16
GMAT 1: 720 Q49 V39
Re: If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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Hi,

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 vise-versa.

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 64322
Re: If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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Hiho wrote:
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?

It is tested.

Check here: math-combinatorics-87345.html and here: math-probability-87244.html
_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 64322
Re: If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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SpotlessMind wrote:
Hi,

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 vise-versa.

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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Manager  G
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If 40 people get the chance to pick a card from a canister that contai  [#permalink]

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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?

A. 1/8
B. 36/175
C. 117/175
D. 139/175
E. 7/8

Bunuel is it the correct way to approach?
we have only 1 way to select the 40th person, and we can give him or her 5 free passes in 5C1 ways
so total ways=1*5c1
And we have 40 ways to select one person=40c1
So required probability=(1*5c1)/40c1
=1/8
A:) If 40 people get the chance to pick a card from a canister that contai   [#permalink] 05 Apr 2020, 21:42

# If 40 people get the chance to pick a card from a canister that contai  