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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

62% (02:13) correct
38% (01:01) wrong based on 168 sessions

HideShow timer Statistics

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.

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.
_________________

Re: Probability of picking numbers in ascending order. [#permalink]

Show Tags

11 Jan 2010, 16:45

1

This post received 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

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.

Re: Probability of picking numbers in ascending order. [#permalink]

Show Tags

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)}]

Re: Probability of picking numbers in ascending order. [#permalink]

Show Tags

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!

Re: Probability of picking numbers in ascending order. [#permalink]

Show Tags

28 Sep 2010, 15:45

1

This post received KUDOS

How about this 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}...........

How about this 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 \(P=\frac{1}{3!}=\frac{1}{6}\).
_________________

Re: Probability of picking numbers in ascending order. [#permalink]

Show Tags

02 Oct 2010, 12:42

1

This post received 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..

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 = \(C^5_{25}\).

Re: Set A consists of 25 distinct numbers. We pick n numbers [#permalink]

Show Tags

25 May 2014, 11:54

Why cant we calculate probability of picking 2 or 3 or 4 or 5 or ... 25 numbers in ascending order and add them all. These are mutually exclusive cases and we can have a value for picking numbers in ascending order for any value of n. P(2 numbers in ascending) + P(3 numbers in ascending) + P(4 numbers in ascending) ... + P(25 numbers in ascending)

I think question stem is its self sufficient to answer this question.
_________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Why cant we calculate probability of picking 2 or 3 or 4 or 5 or ... 25 numbers in ascending order and add them all. These are mutually exclusive cases and we can have a value for picking numbers in ascending order for any value of n. P(2 numbers in ascending) + P(3 numbers in ascending) + P(4 numbers in ascending) ... + P(25 numbers in ascending)

I think question stem is its self sufficient to answer this question.

When a DS question asks to find the value, then the statement is sufficient ONLY if you can get the single numerical value. For different n's the probability is different, thus we need to know the value of n, to get the single numerical value of the probability.
_________________

Re: Set A consists of 25 distinct numbers. We pick n numbers [#permalink]

Show Tags

23 Jul 2016, 13:12

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Set A consists of 25 distinct numbers. We pick n numbers [#permalink]

Show Tags

25 Jul 2016, 09:40

Hello,

Why can't it be D?

St.1 We know that the set consists of 25 numbers. These numbers are consecutive even numbers no matter what their exact value is. So the chances of picking the numbers in ascending order should be 1/25 * 1/24 * 1/23....1/2*1= 1/(25!)

It seems that in Statement 1 , we are not given the number of n. We could pick 4 or 5 or 25 numbers as well.

St.1 We know that the set consists of 25 numbers. These numbers are consecutive even numbers no matter what their exact value is. So the chances of picking the numbers in ascending order should be 1/25 * 1/24 * 1/23....1/2*1= 1/(25!)

It seems that in Statement 1 , we are not given the number of n. We could pick 4 or 5 or 25 numbers as well.

We should know how many numbers we are picking from 25.
_________________

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...