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

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Re: ID number.  [#permalink]

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New post 04 Mar 2010, 22:24
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1
The answer is simple
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 :done
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Re: ID number.  [#permalink]

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New post 01 Jun 2014, 06:30
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
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Re: In a 5 digit ID number, what is the probability of exactly t  [#permalink]

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New post 05 May 2016, 05:08
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
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Re: In a 5 digit ID number, what is the probability of exactly t  [#permalink]

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New post 09 May 2016, 20:15
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2
akadmin wrote:
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

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

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New post 09 May 2016, 21:19
1
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 \(1-\frac{1}{10}\))

our answer:
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
\(nCr * p^r * (1-p)^{n-r}\)
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Re: In a 5 digit ID number, what is the probability of exactly t  [#permalink]

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New post 18 Jul 2016, 00:35
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
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In a 5 digit ID number, what is the probability of exactly t  [#permalink]

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New post 21 Jul 2016, 05:17
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.
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Re: In a 5 digit ID number, what is the probability of exactly t  [#permalink]

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New post 18 Apr 2017, 18:48
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?
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Re: In a 5 digit ID number, what is the probability of exactly t  [#permalink]

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New post 18 Apr 2017, 20:01
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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.
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Re: In a 5 digit ID number, what is the probability of exactly t  [#permalink]

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New post 18 Apr 2017, 21:02
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%

So the answer is C

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

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New post 19 Apr 2017, 22:34
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%

Assuming that 5 digit doesn't start with '0'

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%

Assuming that 5 digit can start with '0'
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%
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Re: In a 5 digit ID number, what is the probability of exactly t  [#permalink]

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New post 23 Sep 2018, 23:01
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 possibilities=1*1*1*9*9*5c3..
prob=.81%...

chetan2u here arrangements of the digits do not require ?
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Re: In a 5 digit ID number, what is the probability of exactly t  [#permalink]

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New post 24 Sep 2018, 00:17
LoneSurvivor wrote:
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 possibilities=1*1*1*9*9*5c3..
prob=.81%...

chetan2u here arrangements of the digits do not require ?



Hi

Arrangements matters..
So we choose 3 places out of 5 in 5C3 and then we place digit 2 in all 3 places..
Now the remaining 2 places we can place any of two digits so 9*9
Ans 9*9*5C3

Another way..
Case1..
3 are of same type and other two of one type so choose one from 9, so 222aa
They can be arranged in 9C1*5!/(3!2!)=9*10=90
Case 2..
3 are of same type and other 2 from remaining 9 digits but separate, so 222ab
So 9C2*5!/3!=9*4*5*4=720
Combined 90+720=810
Total ways 10*10*10*10*10=10^5

Probability = 810/10^5=0.81/100 or 0.81%
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Re: In a 5 digit ID number, what is the probability of exactly t  [#permalink]

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New post 24 Sep 2018, 01:29
Since it's a code , the number can start with 0.
2 can be selected in 1/10 ways.
Rest can be selected in 9/10 ways
First we select 3 spots for 2 with 5c3.
\(5c3 * (1/10)^3 * (9/10)^2\)
= \(10 * 1/10^3 * 81/10^2\)
= \(81/10^4\)

In percentage , \(81 * 100 / 10^4\) = 0.81 %
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Re: In a 5 digit ID number, what is the probability of exactly t  [#permalink]

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New post 25 Sep 2018, 12:59
You can solve it in 25 second using only one formula:

(5,3) * [(1/10)^3] * [(9/10)^2] = 10 * (1/1000) * (81/100) = 0.0081 = 0.81%
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Re: In a 5 digit ID number, what is the probability of exactly t  [#permalink]

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New post 20 Nov 2018, 05:49
Total number of ways to select any 3 numbers out of 5 is 5C3, i.e, 10 ways.

Total possibilities that exactly 3 digits are '2' is 9*9=81

Hence, Answer = 81/10 = 0.81
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Re: In a 5 digit ID number, what is the probability of exactly t  [#permalink]

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New post 21 Nov 2018, 09:59
This is a very simple question.

Since only three numbers must be 2, the other two numbers must not be 2.

9/10 (the possibility of getting numbers except 2) * 9/10 = 0.81

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

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New post 21 Nov 2018, 10:24
Not sure if this solution is viable upon repetition, but I found [A] to be the correct solution by doing this:

Odds of 3 digits being '2'
1/10 * 1/10 * 1/10 = 1/1000

Odds of other 2 digits NOT being '2'
9/10 * 9/10 = 81/100

Add these two probabilities together and you get 811/1000, or 0.811%
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Re: In a 5 digit ID number, what is the probability of exactly t &nbs [#permalink] 21 Nov 2018, 10:24

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