Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 29 Jun 2016, 12:08

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

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

There are 1111 cars in the parking lot each with number 1 to

Author Message
VP
Joined: 29 Dec 2005
Posts: 1348
Followers: 8

Kudos [?]: 52 [0], given: 0

There are 1111 cars in the parking lot each with number 1 to [#permalink]

Show Tags

13 Jun 2006, 22:55
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

There are 1111 cars in the parking lot each with number 1 to 1111 and no two cars have the same number. At 5pm when they all leave the parking lot one by one, what would be the probability that numbers of the first four cars to leave are in increasing order?

seems laxi is in short break for a while. everybody is appreciated for his/her inputs.
SVP
Joined: 30 Mar 2006
Posts: 1737
Followers: 1

Kudos [?]: 65 [0], given: 0

Show Tags

14 Jun 2006, 00:02
Ok I know

the first car can have numbers 1 - 1108
The second car can have numbers 2 - 1109
The third car can have numbers 3 - 1110
The fourth car can have numbers 4 - 1111.

Total number of ways cars can be arranged = 1111!

The second part I am still trying to figure it out.
SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 18

Kudos [?]: 230 [0], given: 0

Show Tags

14 Jun 2006, 02:44
is it 1 / ( 1108*1109*1110*1111) ?

Hi, buddy, thank you for remembering me This is a tricky one, i would say, especially, probability is not my strong point . Anyway, i tried to solve and got that result. In fact, I'm sunk in tons of tasks right now, considering my up-coming G-day and 8 final papers early next month

Last edited by laxieqv on 14 Jun 2006, 04:27, edited 2 times in total.
Manager
Joined: 09 May 2006
Posts: 99
Location: Bangalore, India
Followers: 2

Kudos [?]: 27 [0], given: 0

Show Tags

14 Jun 2006, 03:36
Am assuming increasing order to mean consecutive

Total number of ways for first four cars = 1111P4 = 1111*1110*1109*1108

Number of ways in which four cosecutive numbers can be chosen from 1-1111 = number of ways the first number can be chosen = 1108

P = 1108/(1111*1110*1109*1108)
= 1/1111*1110*1109

Am pretty sure the assumption stated upfront is invalid. Waiting for the OA!
VP
Joined: 21 Sep 2003
Posts: 1065
Location: USA
Followers: 3

Kudos [?]: 57 [0], given: 0

Show Tags

14 Jun 2006, 05:48
A shot in the dark. Is it 1/24?

The favorable event (that is the cars leaving the lot in increasing order) can be viewed as choosing 4 numbers from 1111 numbers in a certain order.
Hence favorable arrangements = 1111C4
All arrangements = 1111P4

Reqd probability = 1111C4/1111P4= 1/4! = 1/24
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 18

Kudos [?]: 230 [0], given: 0

Show Tags

14 Jun 2006, 06:24
giddi77 wrote:
A shot in the dark. Is it 1/24?

The favorable event (that is the cars leaving the lot in increasing order) can be viewed as choosing 4 numbers from 1111 numbers in a certain order.
Hence favorable arrangements = 1111C4
All arrangements = 1111P4

Reqd probability = 1111C4/1111P4= 1/4! = 1/24

I also think like that:

The problem can be drawn to getting the probability of a set of 4.
Because, for whichever the set of 4 is, there must be one and only one increasing order because the 4 numbers are different!
The probability for each increasing order is : 1/4! ( because such order is only one among 4! ways of arranging the set of 4)

But then i wrongly divided this by C(1111,4) . This is wrong , I did feel so. Now that i saw Giddi buddy has same direction; I can eliminate the ambiguousity

Continue:
There're C(1111,4) sets of 4 ---> the needed probability is:

1/4!* C(1111,4) / C(1111,4)
Current Student
Joined: 29 Jan 2005
Posts: 5239
Followers: 23

Kudos [?]: 293 [0], given: 0

Show Tags

14 Jun 2006, 07:59
Seems like a dependant probability question:

1/1111*1/1110*1/1109*1/1108 = 1/ some huge number
SVP
Joined: 03 Jan 2005
Posts: 2243
Followers: 16

Kudos [?]: 286 [0], given: 0

Show Tags

14 Jun 2006, 08:08
No matter what group of four cars you choose, there is 1/4! chance that they are in increasing order. So the number 1111 really doesn't matter, isn't it?
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5062
Location: Singapore
Followers: 29

Kudos [?]: 282 [0], given: 0

Show Tags

14 Jun 2006, 08:36
Total number of arrangements = 1111C4 * 4!
Only 1 combination from every set is in an increasing order.
Total number of increasing sets = 1111C4
Probability = 1/4!
------------------------------------------
Here's how I tested:

4 numbers: 1,2,3,4
# of ways to pick 3 numbers from the 4 = 4C3 = 4 --> (1,2,3) (1,3,4) (2,3,4) (1,2,4)

# of ways to arrange each set = 3! = 6
(1,2,3) = (1,2,3) (1,3,2) (2,1,3) (2,3,1) (3,1,2) (3,2,1)

Total number of ways to arrange = 24 = 6 * 4 = 3! * 4C3

Note: each set only has 1 combination in increasing order (1,2,3) (1,3,4) (2,3,4) (1,2,4) --> That's your 4C3

So P = 4C3/(4C3 * 6)

I just substituted 4C3 with 111C3 and 3! with 4!
VP
Joined: 21 Sep 2003
Posts: 1065
Location: USA
Followers: 3

Kudos [?]: 57 [0], given: 0

Show Tags

14 Jun 2006, 08:48
HongHu wrote:
No matter what group of four cars you choose, there is 1/4! chance that they are in increasing order. So the number 1111 really doesn't matter, isn't it?

As usual your explanation is perfect! I somehow tend to think only in terms of nCr, nPr, n!.. formulas
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

VP
Joined: 29 Dec 2005
Posts: 1348
Followers: 8

Kudos [?]: 52 [0], given: 0

Show Tags

14 Jun 2006, 13:06
laxieqv wrote:
is it 1 / ( 1108*1109*1110*1111) ?

Hi, buddy, thank you for remembering me This is a tricky one, i would say, especially, probability is not my strong point . Anyway, i tried to solve and got that result. In fact, I'm sunk in tons of tasks right now, considering my up-coming G-day and 8 final papers early next month

thats , laxi and goodluck to you..

thanx everybody. giddi, honghu, and ywilfred, all, did the goal.

it is 1/4! = 1/24
Display posts from previous: Sort by