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

 It is currently 22 May 2013, 07:03

# Probability of picking numbers in ascending order.

Author Message
TAGS:
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11565
Followers: 1795

Kudos [?]: 9570 [0], given: 826

Probability of picking numbers in ascending order. [#permalink]  11 Jan 2010, 00:29
00:00

Question Stats:

27% (02:44) correct 72% (00:40) wrong based on 2 sessions
This is my question, so no OA just my solution.

Set A consists of 25 distinct numbers. We pick n numbers from the set A one-by-one (n<=25). What is the probability that we pick numbers in ascending order?

(1) Set A consists of even consecutive integers;
(2) n=5.
_________________
Joined: 20 Aug 2009
Posts: 314
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)
Followers: 7

Kudos [?]: 68 [0], given: 69

Re: Probability of picking numbers in ascending order. [#permalink]  11 Jan 2010, 01:28
P = (25-n+1)/25 * 1/24 * 1/23 * . . . * 1/(25-n)

If that's right, the only thing probability depends on is n. St1 is Insufficient and answer is (B)
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11565
Followers: 1795

Kudos [?]: 9570 [0], given: 826

Re: Probability of picking numbers in ascending order. [#permalink]  11 Jan 2010, 01:35
shalva wrote:
P = (25-n+1)/25 * 1/24 * 1/23 * . . . * 1/(25-n)

If that's right, the only thing probability depends on is n. St1 is Insufficient and answer is (B)

The actual probability is much higher. Formula is not correct.
_________________
Joined: 20 Aug 2009
Posts: 314
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)
Followers: 7

Kudos [?]: 68 [0], given: 69

Re: Probability of picking numbers in ascending order. [#permalink]  11 Jan 2010, 01:50
Bunuel wrote:
shalva wrote:
P = (25-n+1)/25 * 1/24 * 1/23 * . . . * 1/(25-n)

If that's right, the only thing probability depends on is n. St1 is Insufficient and answer is (B)

The actual probability is much higher. Formula is not correct.

yes, of course, the probability is much higher. I meant the picked numbers to be consecutive, it may not be so
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11565
Followers: 1795

Kudos [?]: 9570 [0], given: 826

Re: Probability of picking numbers in ascending order. [#permalink]  11 Jan 2010, 01:55
shalva wrote:
Bunuel wrote:
shalva wrote:
P = (25-n+1)/25 * 1/24 * 1/23 * . . . * 1/(25-n)

If that's right, the only thing probability depends on is n. St1 is Insufficient and answer is (B)

The actual probability is much higher. Formula is not correct.

yes, of course, the probability is much higher. I meant the picked numbers to be consecutive, it may not be so

So, if B is the correct answer what's the probability then?
_________________
Joined: 20 Aug 2009
Posts: 314
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)
Followers: 7

Kudos [?]: 68 [0], given: 69

Re: Probability of picking numbers in ascending order. [#permalink]  11 Jan 2010, 02:04
Bunuel wrote:
So, if B is the correct answer what's the probability then?

I've absolutely no idea

Last edited by shalva on 11 Jan 2010, 02:42, edited 5 times in total.
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11565
Followers: 1795

Kudos [?]: 9570 [0], given: 826

Re: Probability of picking numbers in ascending order. [#permalink]  11 Jan 2010, 02:15
shalva wrote:
Bunuel wrote:
So, if B is the correct answer what's the probability then?

I've absolutely no idea

Somehow the probability depends on the first number we pick and changes with each next pick

let's suppose our Set consists of numbers x1, x2 . . . x25, where numbers are sorted in ascending order.

Probability of picking x1 and x21 as first number is the same. But - probability of picking 5 number in ascending order beginning with x1 is much much higher then beginning with x21

That is not so. But the issue you mentioned is the key part to answer the question.
_________________
Manager
Joined: 06 Jan 2010
Posts: 74
Followers: 2

Kudos [?]: 6 [1] , given: 15

Re: Probability of picking numbers in ascending order. [#permalink]  11 Jan 2010, 16:45
1
KUDOS
I think B is the right answer

this is my reasoning for the actual probability

Given 5 nos (doesn't matter what they are)
for the first choice- there is exactly one option out of 5 options. i.e we should pick the smallest of the nos. so probability is 1/5
for the second choice - again there is exactly one option i.e smallest no of the remanining for nos. so probability is 1/4
for the 3rd choice- only one smallest no of the remaining 3 nos. so probability is 1/3

and so

so total probability is 1/5*1/4*1/3*1/2*1 = 1/120
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11565
Followers: 1795

Kudos [?]: 9570 [0], given: 826

Re: Probability of picking numbers in ascending order. [#permalink]  11 Jan 2010, 17:06
janani wrote:
I think B is the right answer

this is my reasoning for the actual probability

Given 5 nos (doesn't matter what they are)
for the first choice- there is exactly one option out of 5 options. i.e we should pick the smallest of the nos. so probability is 1/5
for the second choice - again there is exactly one option i.e smallest no of the remanining for nos. so probability is 1/4
for the 3rd choice- only one smallest no of the remaining 3 nos. so probability is 1/3

and so

so total probability is 1/5*1/4*1/3*1/2*1 = 1/120

+1.

We should understand following two things:
1. The probability of picking any n numbers from the set of 25 distinct numbers is the same. For example if we have set of numbers from 1 to 25 inclusive, then the probability we pick n=5 numbers {3,5,1,23,25} is the same as that of we pick n=5 numbers {9,10,4,6,18}. So picking any 5 numbers \{x_1,x_2,x_3,x_4,x_5\} from the set is the same.

2. Now, imagine we have chosen the set \{x_1,x_2,x_3,x_4,x_5\}, where x_1<x_2<x_3<x_4<x_5. We can pick this set of numbers in 5!=120 # of ways and only one of which, namely \{x_1,x_2,x_3,x_4,x_5\} is in ascending order. So 1 out of 120. P=\frac{1}{n!}=\frac{1}{5!}=\frac{1}{120}.

According to the above the only thing we need to know is the size of the set (n) we are choosing from the initial set A.

_________________
Intern
Joined: 25 Aug 2010
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 10

Re: Probability of picking numbers in ascending order. [#permalink]  30 Aug 2010, 05:05
If we pick n numbers, probability of picking in ascending order will always be [1/{(25*24*...(25-n)}].
Explanation:
No of ways of picking n numbers from 25 (25*24*...(25-n) .. And out of that in only one all will be in ascending order.
So probability will be
[1/{(25*24*...(25-n)}]
Manager
Joined: 19 Sep 2010
Posts: 187
Followers: 2

Kudos [?]: 60 [0], given: 18

Re: Probability of picking numbers in ascending order. [#permalink]  22 Sep 2010, 06:27
let the numbers be a1 , a2 , a3 , ............, a25

Clearly option one is insufficient...Now as per option 2 (n=5)
There are 25C5 ways to select a set of 5 different numbers.
Now if we consider all the permutations of these 5 diff numbers , then only one satisfies
our criteria . Therefore out of 5! cases , only 1 is favorable and hence the probability is 1/5! = 1/120.

Therefore , the ans is B according to me.

had it been some other n<=25 , the probability would be 1/n!
Intern
Joined: 26 Jun 2010
Posts: 2
Followers: 0

Kudos [?]: 1 [1] , given: 24

Re: Probability of picking numbers in ascending order. [#permalink]  28 Sep 2010, 15:45
1
KUDOS
We have a set consists of 6 numbers {1,2,3,4,5,6} and the probability that we pick 3 numbers in ascending order :
If I use your approach I will get [1][/3] * [1][/2] = [1][/6]
but check it out
{1,2,3},{1,2,4},{1,3,4},{1,4,5},{2,3,4},{2,5,6},{3,4,5}...........
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11565
Followers: 1795

Kudos [?]: 9570 [0], given: 826

Re: Probability of picking numbers in ascending order. [#permalink]  29 Sep 2010, 00:04
cagdasgurpinar wrote:
We have a set consists of 6 numbers {1,2,3,4,5,6} and the probability that we pick 3 numbers in ascending order :
If I use your approach I will get [1][/3] * [1][/2] = [1][/6]
but check it out
{1,2,3},{1,2,4},{1,3,4},{1,4,5},{2,3,4},{2,5,6},{3,4,5}...........

1/6 would be a correct answer for your example: if you continue to write 3 numbers sequences in ascending order from a set {1, 2, 3, 4, 5, 6} you'll get 20 possibilities and total # of picking 3 numbers from 6 when order matters is P^3_6=120 --> P=\frac{20}{120}=\frac{1}{6}.

Let's consider smaller set {1, 2, 3, 4}. What is the probability that we pick 3 numbers in ascending order?

P=Favorable scenarios/Total # of possible scenarios.

# of favorable scenarios is 4: {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, ;
Total # of possible scenarios is 24: P^3_4=24;

P=\frac{# \ of \ favorable \ scenarios}{Total \ # \ of \ possible \ scenarios}=\frac{4}{24}=\frac{1}{6} or P=\frac{1}{3!}=\frac{1}{6}.
_________________
Manager
Joined: 04 Sep 2010
Posts: 51
Followers: 2

Kudos [?]: 50 [1] , given: 1

Re: Probability of picking numbers in ascending order. [#permalink]  02 Oct 2010, 12:42
1
KUDOS
According to my understanding, probability for option b can be calculated by:
As we have to choose 5 among 25 so 25c5....(1)
then we can arrange those five in 5! ways..
so the outcome will be 25c2 * 5! = 25p5.
so finally out of all sets only 1 will be in ascending order, so ans = 1/(25p5)
plz correct me if i m somewhere wrong..
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11565
Followers: 1795

Kudos [?]: 9570 [0], given: 826

Re: Probability of picking numbers in ascending order. [#permalink]  02 Oct 2010, 13:07
sudhanshushankerjha wrote:
According to my understanding, probability for option b can be calculated by:
As we have to choose 5 among 25 so 25c5....(1)
then we can arrange those five in 5! ways..
so the outcome will be 25c2 * 5! = 25p5.
so finally out of all sets only 1 will be in ascending order, so ans = 1/(25p5)
plz correct me if i m somewhere wrong..

No, P=\frac{1}{n!}=\frac{1}{5!}=\frac{1}{120} (please see the solution above).

If we do the way you are proposed then:

Total # of outcomes = P^5_{25} - total # of ways to pick any 5 numbers out of 25 when order matters;
Favorable outcomes = C^5_{25}.

P=\frac{C^5_{25}}{P^5_{25}}=\frac{1}{5!}=\frac{1}{120}
_________________
Re: Probability of picking numbers in ascending order.   [#permalink] 02 Oct 2010, 13:07
Similar topics Replies Last post
Similar
Topics:
There are four prime numbers written in ascending order of 14 09 Jan 2004, 12:45
Set S consists of n numbers arranged in ascending order. A 4 05 Sep 2008, 21:59
Does Gmat Club Level of Exams increase in ascending order 0 28 Apr 2009, 15:52
picking numbers 3 05 Jul 2010, 11:22
Probability of a number when order doesn't matter 1 06 Apr 2013, 23:24
Display posts from previous: Sort by