Last visit was: 18 Nov 2025, 19:52 It is currently 18 Nov 2025, 19:52
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
555-605 Level|   Probability|      
avatar
mpevans
avatar
Joined: 10 Jul 2010
Last visit: 12 Jul 2010
Posts: 1
Own Kudos:
81
 [81]
Posts: 1
Kudos: 81
 [81]
If 40 people get the chance to pick a card from a canister that contai 10 Jul 2010, 06:08
4
Kudos
Add Kudos
77
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

25% (medium)

Question Stats:

76% (01:30) correct 24% (02:10) wrong based on 527 sessions

History

Date
Time
Result
Not Attempted Yet
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
ShowHide Answer
Official Answer
A
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,086
 [20]
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,086
 [20]
If 40 people get the chance to pick a card from a canister that contai 10 Jul 2010, 08:54
11
Kudos
Add Kudos
9
Bookmarks
Bookmark this Post
mpevans
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?
Show more

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
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
Show more

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.
avatar
geneticsgene
avatar
Joined: 12 Jan 2012
Last visit: 24 Jan 2021
Posts: 17
Own Kudos:
29
 [7]
Given Kudos: 10
GMAT 1: 720 Q49 V39
GMAT 1: 720 Q49 V39
Posts: 17
Kudos: 29
 [7]
If 40 people get the chance to pick a card from a canister that contai 21 Dec 2012, 23:35
6
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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
General Discussion
User avatar
AbhayPrasanna
User avatar
Joined: 04 May 2010
Last visit: 05 Jan 2025
Posts: 61
Own Kudos:
354
 [2]
Given Kudos: 7
GPA: 3.8
WE 1: 2 yrs - Oilfield Service
Products:
My Reviews
Posts: 61
Kudos: 354
 [2]
If 40 people get the chance to pick a card from a canister that contai 10 Jul 2010, 08:17
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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
avatar
ulm
avatar
Joined: 03 Jun 2010
Last visit: 20 Aug 2019
Posts: 95
Own Kudos:
616
 [1]
Given Kudos: 40
Location: United States (MI)
Concentration: Marketing, General Management
Schools: Ross '15 (A) Darden '14 (D) Duke '15 (D)
WE:Business Development (Consumer Packaged Goods)
Products:
My Reviews
Schools: Ross '15 (A) Darden '14 (D) Duke '15 (D)
Posts: 95
Kudos: 616
 [1]
If 40 people get the chance to pick a card from a canister that contai 10 Jul 2010, 22:13
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
I made the same.
4C39/5C40.
Where 5C40 - total # of outcomes.
4C39 means that 4 winning tickets were taken out by 39 persons.
User avatar
gmat1011
User avatar
Joined: 11 Jul 2010
Last visit: 22 Dec 2012
Posts: 139
Own Kudos:
256
Given Kudos: 20
Posts: 139
Kudos: 256
If 40 people get the chance to pick a card from a canister that contai 11 Jul 2010, 06:03
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,086
 [3]
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,086
 [3]
If 40 people get the chance to pick a card from a canister that contai 11 Jul 2010, 07:41
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
gmat1011
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.
Show more

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.
User avatar
utin
User avatar
Joined: 27 Mar 2010
Last visit: 26 Sep 2011
Posts: 63
Own Kudos:
39
Given Kudos: 17
Posts: 63
Kudos: 39
If 40 people get the chance to pick a card from a canister that contai 12 Jul 2010, 11:41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
mpevans
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
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.
Show more


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.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,086
 [5]
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,086
 [5]
If 40 people get the chance to pick a card from a canister that contai 12 Jul 2010, 11:54
3
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
utin

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.
Show more

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.
avatar
Hiho
avatar
Joined: 20 Dec 2012
Last visit: 22 Jul 2016
Posts: 16
Own Kudos:
6
Given Kudos: 3
Posts: 16
Kudos: 6
If 40 people get the chance to pick a card from a canister that contai 22 Dec 2012, 06:43
Kudos
Add Kudos
Bookmarks
Bookmark this Post
geneticsgene
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
Show more

Hello!

I am unfamiliar with ! in math, what does it mean?

Thanks in advance
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,086
 [1]
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,086
 [1]
If 40 people get the chance to pick a card from a canister that contai 22 Dec 2012, 06:52
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hiho
geneticsgene
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
Show more

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: \(4!=1*2*3*4=24\).

For more check here: everything-about-factorials-on-the-gmat-85592.html

Hope it helps.
avatar
Hiho
avatar
Joined: 20 Dec 2012
Last visit: 22 Jul 2016
Posts: 16
Own Kudos:
6
Given Kudos: 3
Posts: 16
Kudos: 6
If 40 people get the chance to pick a card from a canister that contai 22 Dec 2012, 07:22
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Hiho
geneticsgene
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 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.
Show more

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?
avatar
SpotlessMind
avatar
Joined: 22 Dec 2012
Last visit: 27 Nov 2013
Posts: 11
Own Kudos:
40
Given Kudos: 19
GMAT 1: 720 Q49 V39
GMAT 1: 720 Q49 V39
Posts: 11
Kudos: 40
If 40 people get the chance to pick a card from a canister that contai 22 Dec 2012, 11:38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
mpevans
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
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.
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,086
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,086
If 40 people get the chance to pick a card from a canister that contai 23 Dec 2012, 06:18
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hiho
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?
Show more

It is tested.

Check here: math-combinatorics-87345.html and here: math-probability-87244.html
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,086
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,086
If 40 people get the chance to pick a card from a canister that contai 23 Dec 2012, 06:22
Kudos
Add Kudos
Bookmarks
Bookmark this Post
SpotlessMind
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
mpevans
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
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.
Show more

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.
User avatar
satya2029
User avatar
Joined: 10 Dec 2017
Last visit: 29 Sep 2025
Posts: 231
Own Kudos:
249
Given Kudos: 138
Location: India
Posts: 231
Kudos: 249
If 40 people get the chance to pick a card from a canister that contai 05 Apr 2020, 22:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mpevans
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
Show more
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:)
User avatar
gmatuser523
User avatar
Joined: 17 Apr 2025
Last visit: 01 Jul 2025
Posts: 4
Posts: 4
Kudos: 0
If 40 people get the chance to pick a card from a canister that contai 20 Jun 2025, 15:58
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ok maybe I did this wrong but I got to the right answer so I'm not sure what I did here, but my logic was that each person draws from 40 cards with a 1/8 chance to win. The 40th person's outcome is independent of everyone else so their win or loss rate does not matter - therefore each person's outcome is the same which means this guy, like everyone else has a 1/8 chance of winning. Question makes no mention of replacement or no replacement, so this is the same as 40 people drawing from 40 boxes at the same time.
Moderators:
Bunuel
Math Expert
105355 posts
Krunaal
Tuck School Moderator
805 posts