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Set A consists of 25 distinct numbers. We pick n numbers

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

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Last edited by Bunuel on 09 Jul 2013, 15:29, edited 2 times in total.
Added the OA.
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New post 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) :roll:
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New post 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) :roll:


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
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New post 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) :roll:


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

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New post 11 Jan 2010, 02:04
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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.
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New post 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.
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Re: Probability of picking numbers in ascending order. [#permalink]

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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
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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.

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

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New post 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)}]
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Re: Probability of picking numbers in ascending order. [#permalink]

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

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

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cagdasgurpinar wrote:
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 \(P=\frac{1}{3!}=\frac{1}{6}\).
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Re: Probability of picking numbers in ascending order. [#permalink]

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

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New post 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 = \(C^5_{25}\).

\(P=\frac{C^5_{25}}{P^5_{25}}=\frac{1}{5!}=\frac{1}{120}\)
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Re: Set A consists of 25 distinct numbers. We pick n numbers [#permalink]

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

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New post 26 May 2014, 03:13
PiyushK wrote:
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.
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Re: Set A consists of 25 distinct numbers. We pick n numbers [#permalink]

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

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

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New post 25 Jul 2016, 09:43
Berlin77 wrote:
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.


We should know how many numbers we are picking from 25.
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Re: Set A consists of 25 distinct numbers. We pick n numbers   [#permalink] 25 Jul 2016, 09:43
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