Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 25 May 2017, 19:08

### 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

# In a 5 digit ID number, what is the probability of exactly t

Author Message
TAGS:

### Hide Tags

Intern
Joined: 18 Feb 2010
Posts: 20
Followers: 0

Kudos [?]: 12 [0], given: 2

### Show Tags

04 Mar 2010, 23:24
Step 1: Total number of ways = 10*10*10*10*10 = 10^5
Step 2: First choose three places for three 2s which can be done in 5C3 ways. (No point arranging these 2s amongst themselves) = 10 ways
Step 3: Now for the remaining two positions there are 9 numbers (as 2 is already used) each to be placed which can be done in 9*9 ways.
Step 4: favorable ways = 10*9*9 = 810

Thus probability = favorable ways / total ways = 810/10^5=0.0081 or 0.81%

Hope the solution helps
_________________

If you like my post, shower kudos

Believe that something extraordinary is possible

Current Student
Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 960
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)
Followers: 171

Kudos [?]: 1611 [0], given: 229

### Show Tags

01 Jun 2014, 07:30
X X X X X

5C3 = 10, 10 3 place combination which can take 2.

if first 3 places have taken 2 as 2 2 2 _ _ then remaining places can take values between 0 -9 except 2. i.e 9.

therefore all possible such IDs 10*9*9 = 810

Overall possible combinations for 5 digit ID = $$10 ^5$$

Therefore $$\frac{810}{10^5} * 100 = 0.81%.$$ Ans A
_________________

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".

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15448
Followers: 649

Kudos [?]: 209 [0], given: 0

Re: In a 5 digit ID number, what is the probability of exactly t [#permalink]

### Show Tags

09 Aug 2015, 01:15
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.
_________________
Intern
Joined: 17 May 2015
Posts: 39
Followers: 1

Kudos [?]: 5 [0], given: 155

Re: In a 5 digit ID number, what is the probability of exactly t [#permalink]

### Show Tags

05 May 2016, 06:08
Why have you done 5C3 and not 5!/3!. In my opinion the order matters so the 5!/3! should remove the three 2s arrangeemnt out of the equation

chetan2u wrote:
hi i think it does not req such long calculations my ans A...
there are 5 digits each place can have any of ten digits so total posb=10^5...
now three places only two can be there and rest two can have any of remaining nine digits...
so posb=1*1*1*9*9*5c3..
prob=.81%... although i might be still missing something.. ill try again
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7374
Location: Pune, India
Followers: 2288

Kudos [?]: 15113 [0], given: 224

Re: In a 5 digit ID number, what is the probability of exactly t [#permalink]

### Show Tags

09 May 2016, 21:15
Expert's post
1
This post was
BOOKMARKED
Why have you done 5C3 and not 5!/3!. In my opinion the order matters so the 5!/3! should remove the three 2s arrangeemnt out of the equation

chetan2u wrote:
hi i think it does not req such long calculations my ans A...
there are 5 digits each place can have any of ten digits so total posb=10^5...
now three places only two can be there and rest two can have any of remaining nine digits...
so posb=1*1*1*9*9*5c3..
prob=.81%... although i might be still missing something.. ill try again

Responding to a pm:

Quote:
I am confused on the arrangement part of it. 3 2s and 2 not2s cannot result in 5C3 or 5!/2!3! . My answer is 5!/3! since i am considering this as a permutation of 5 elements with 3 identical and other 2 may or may be not identical. How do I solve this question with arrangement with repetition standpoint.

5C3 means you are selecting 3 out of 5.
Here, you are selecting 3 spots out of the available 5 spots (1st digit, 2nd digit, 3rd digit, 4th digit, 5th digit). Now you have all such arrangements:

____ 2 ____ 2 2

2 ____ ____ 22

2 ____ 2 ____ 2

... etc

In the two blanks, you can have any of them 9 remaining digits. You will get all ids of the from
23224
23223
32522
etc
i.e. all ids where the remaining two digits are the same or are distinct.
So favourable cases = 5C3 * 9 * 9

In case, you want to specifically arrange, you will need to take two cases:

Case 1: Three 2s and two distinct other digits
No of ways = 9C2 * 5!/3! = 720

Case 2: Three 2s and two another digit
No of ways = 9C1 * 5!/3!*2! = 90

Total no of favourable ways = 720 + 90 = 810

Required probability = 810/100000 = .81%
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Manager
Status: One Last Shot !!!
Joined: 04 May 2014
Posts: 245
Location: India
Concentration: Marketing, Social Entrepreneurship
GMAT 1: 630 Q44 V32
GMAT 2: 680 Q47 V35
Followers: 4

Kudos [?]: 87 [0], given: 141

Re: In a 5 digit ID number, what is the probability of exactly t [#permalink]

### Show Tags

09 May 2016, 22:19
1
This post was
BOOKMARKED
Step 1:
Choose 3 places where 2 can go
i.e. ways to choose 3 out of 5 places
5C3 = 10

Step 2:
Probability of placing 2's in those 3 places
$$\frac{1}{10}*\frac{1}{10}*\frac{1}{10}$$
(Prob of choosing one digit out of 10 is $$\frac{1}{10}$$)

Step 3:
Probability of NOT placing 2's in the remaining 2 places
$$\frac{9}{10}*\frac{9}{10}$$
(Prob of NOT choosing one digit out of 10 is $$1-\frac{1}{10}$$)

step 1 AND step 2 AND step 3

$$10 * (\frac{1}{10})^3 * (\frac{9}{10})^2$$
=> $$\frac{0.81}{100}$$
=> 0.81%

Option A

Btw, this question is a classic implementation of Binomial theorem in probability.
In general, the questions like 'find out the probability of EXACTLY n successes out of m trials', could be easily solved using Binomial theorem.
Here's the formula:
Probability of 'r' success in 'n' trials where 'p' is the probability of 1 success is given by
$$nCr * p^r * (1-p)^{n-r}$$
_________________

One Kudos for an everlasting piece of knowledge is not a bad deal at all...

------------------------------------------------------------------------------------------------------------------------
Twenty years from now you will be more disappointed by the things you didn't do than by the ones you did do. So throw off the bowlines. Sail away from the safe harbor. Catch the trade winds in your sails. Explore. Dream. Discover.
-Mark Twain

Intern
Joined: 28 Dec 2015
Posts: 42
Followers: 2

Kudos [?]: 2 [0], given: 62

Re: In a 5 digit ID number, what is the probability of exactly t [#permalink]

### Show Tags

18 Jul 2016, 01:35
Bunuel wrote:
modirashmi wrote:
5 digit no can be arranged in 9*10*10*10*10 ways ( if repetition allowed )

3 2digit no can be inserted in _ _ _ _ _ 5 digit ID no in 2 ways

(i) ID starting with 2 so thr are 5 ways in whch 3-2's can be placed in 5 slots
222--,22--2,2--22,2-2-2,2-22-
remaining 2 digits can be filled in 9ways ( 0-9 digits excluding 2) each, and they can be arranged among themselves in 2 ways
so total 9*9*5*2 ways

(ii) ID not starting with 2 ,, thr are 4 ways -222-,-22-2,-2-22-,--222
1st digit can be filled by any digit 1-9 excluding 2, so 8 ways
5th slot can be filled by 9 ways
n both themselves can be arranged among themselves in 2 ways
so total 8*9*4*2

that brings us to (9*9*5*2 + 8*9*4*2)/9*10*10*10*10
= 9*2(45+32)/9*10*10*10*10
= 2*77/10000 = 1.52%

I hope the logic works out to be correct

First: there is some problems in calculations, then the answer doesn't match with the choices provided.

Hi bunuel,

we have different combinations of the number 222_ _ using 5C3 but don't we have to look after the arrangement of the remaining last two digits??I am finding it confusing
BSchool Forum Moderator
Status: Aiming 800 Q51 V51
Joined: 18 Jul 2015
Posts: 1667
Location: India
GMAT 1: 670 Q50 V32
GMAT 2: 700 Q50 V34
GPA: 3.65
WE: Brand Management (Health Care)
Followers: 39

Kudos [?]: 399 [0], given: 51

In a 5 digit ID number, what is the probability of exactly t [#permalink]

### Show Tags

21 Jul 2016, 06:17
Ashishsteag wrote:
Bunuel wrote:
modirashmi wrote:
5 digit no can be arranged in 9*10*10*10*10 ways ( if repetition allowed )

3 2digit no can be inserted in _ _ _ _ _ 5 digit ID no in 2 ways

(i) ID starting with 2 so thr are 5 ways in whch 3-2's can be placed in 5 slots
222--,22--2,2--22,2-2-2,2-22-
remaining 2 digits can be filled in 9ways ( 0-9 digits excluding 2) each, and they can be arranged among themselves in 2 ways
so total 9*9*5*2 ways

(ii) ID not starting with 2 ,, thr are 4 ways -222-,-22-2,-2-22-,--222
1st digit can be filled by any digit 1-9 excluding 2, so 8 ways
5th slot can be filled by 9 ways
n both themselves can be arranged among themselves in 2 ways
so total 8*9*4*2

that brings us to (9*9*5*2 + 8*9*4*2)/9*10*10*10*10
= 9*2(45+32)/9*10*10*10*10
= 2*77/10000 = 1.52%

I hope the logic works out to be correct

First: there is some problems in calculations, then the answer doesn't match with the choices provided.

Hi bunuel,

we have different combinations of the number 222_ _ using 5C3 but don't we have to look after the arrangement of the remaining last two digits??I am finding it confusing

5C3 means 5 digits in which 3 are of one kind and 2 are of another kind. => we are already considering the arrangement of the other two as well.

If you are still confused look at the solution given by Karishma above. She has mentioned two distinct cases you can split your question into.
_________________

Good Luck

Manager
Joined: 12 Nov 2016
Posts: 195
Followers: 0

Kudos [?]: 7 [0], given: 135

Re: In a 5 digit ID number, what is the probability of exactly t [#permalink]

### Show Tags

18 Apr 2017, 19:48
What I don't understand about this problem is the possibility of ten digits for the first digit of this five digit number? Doesn't the first digit technically have to be from 1-9 in order to constitute a five digit number?
Math Expert
Joined: 02 Sep 2009
Posts: 38889
Followers: 7735

Kudos [?]: 106146 [0], given: 11607

Re: In a 5 digit ID number, what is the probability of exactly t [#permalink]

### Show Tags

18 Apr 2017, 21:01
Nunuboy1994 wrote:
What I don't understand about this problem is the possibility of ten digits for the first digit of this five digit number? Doesn't the first digit technically have to be from 1-9 in order to constitute a five digit number?

5-digit ID number can start with 0. For example, you can have an ID number 01234.
_________________
Intern
Joined: 19 Dec 2013
Posts: 13
Followers: 0

Kudos [?]: 1 [0], given: 4

Re: In a 5 digit ID number, what is the probability of exactly t [#permalink]

### Show Tags

18 Apr 2017, 22:02
rkassal wrote:
1x1x1x9x9 = 81. This much is understood. I also understand that there are 5C3 ways in which the number 2 can be distributed among the 5 places. But what happens when the other 2 places are filled by the same number? For example, if the ID has the digits 22255 then these can form 5!/3!x2! numbers i.e. 10 numbers. But if the ID has 22237, for example, then there are 5!/3! = 20 combinations. Therefore, we cannot simply perform the multiplication of 81 with 5C3.
In fact, we have to treat these cases separately as follows:
1x1x1x9x8 x 5!/3! = 1440 (here the other two numbers are different)
1x1x1x9x1 x 5!/3!x2!= 90 (here the other two numbers are the same)
Therefore, the total possibilities are 1440+90=1530.
Hence the prob. is 1530/90,000= .017%

Hi rkassal

I think that you are conceptually correct but the only issue with your solution is that you haven't took the correct total outcome which should come as 10^5 instead of 9*10^4
So now when you calculate while putting these two figure you will arrive to correct solution that is option - C (1530/10000) as 1.53%

Sent from my A0001 using GMAT Club Forum mobile app
Intern
Joined: 19 Mar 2015
Posts: 15
Location: India
Followers: 0

Kudos [?]: 2 [0], given: 1

In a 5 digit ID number, what is the probability of exactly t [#permalink]

### Show Tags

19 Apr 2017, 23:34
In a 5 digit ID number, what is the probability of exactly three digits are the digit 2?

(A) 0.81%
(B) 1%
(C) 1.53%
(D) 0.081%
(E) 1.44%

sollution) Number of ways three 2s can be arranged in 5 slots =>5!/3!=10
Number of ways three 2s can be arranged in 4 slots => 4!/3!=4
Number of ways by which the 5 digit can be formed without having 2 at 10000's position =>10-4=6
so number of 5 digit codes can be formed with exactly 2 digit => 6*9*9+4*8*9=774
total number of 5 digits => 9*10*10*10*10= 90000
probability of exactly three digits are the digit 2=> 774/90000 = .0086
As the solutions are in percentage the answer = .86%

sollution) Number of ways three 2s can be arranged in 5 slots =>5!/3!=10
so number of 5 digit codes can be formed with exactly 2 digit => 10*9*9=810
total number of 5 digits => 10*10*10*10*10= 100000
probability of exactly three digits are the digit 2=>810/100000=.0081
As the solutions are in percentage the answer = .81%
In a 5 digit ID number, what is the probability of exactly t   [#permalink] 19 Apr 2017, 23:34

Go to page   Previous    1   2   [ 32 posts ]

Similar topics Replies Last post
Similar
Topics:
19 What is the probability of rolling the same number exactly three times 11 18 Apr 2017, 23:52
23 What is the probability of getting exactly three heads on 5 03 Sep 2016, 08:01
22 What is the probability that a three digit number is divis 8 21 Apr 2017, 00:08
3 What is the Probability of getting exactly 2 sixes on three 3 21 Oct 2014, 15:41
6 What is the probability of creating a three digit number wit 10 24 Mar 2015, 06:40
Display posts from previous: Sort by