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In a 5 digit ID number, what is the probability of exactly t [#permalink]
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30 Dec 2009, 16:08
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Re: ID number. [#permalink]
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31 Dec 2009, 02:28
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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 32's can be placed in 5 slots 222,222,222,222,222 remaining 2 digits can be filled in 9ways ( 09 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 222,222,222,222 1st digit can be filled by any digit 19 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



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Re: ID number. [#permalink]
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31 Dec 2009, 07:31
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 32's can be placed in 5 slots 222,222,222,222,222 remaining 2 digits can be filled in 9ways ( 09 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 222,222,222,222 1st digit can be filled by any digit 19 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.
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Re: ID number. [#permalink]
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31 Dec 2009, 12:09
I have been trying this problem for sometime and here is my approach.. Total number of 5 digit numbers = 9999910000+1=90,000 No. of ways of arranging 3 digits as 2 = 5C3=10.
Arrangements are as follows. Here X can be any digit other than 2
1) 222XX 2) 2X22X 3) 2XX22 4) X2X22 5) X22X2 6) X222X 7) 2X2X2 8) XX222 9) X22X2 10)22XX2
Here the first X cannot be digit '0' so no. of arrangements with first digit as 'X' = 8*9=72. Total such arrangements = 72*5=360
no. of arrangements with first digit as '2'=9*9=81 Total such arrangements = 81*5=405
Now there is some double counting where 'XX' are together.I have counted 00,11,33,44,55,66,77,88,99 as different numbers.. So need to subtract these..
In combination XX222 '00' is not present. So 8 combinations In combinations '222XX', '22XX2', '2XX22', there are 3*9=27 combinations.
Hence 765(27+8)= 730 Total probability= (730/90,000)*100 ~ 0.81%. IMO answer is A



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Re: ID number. [#permalink]
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31 Dec 2009, 13:00
acegre wrote: I have been trying this problem for sometime and here is my approach.. Total number of 5 digit numbers = 9999910000+1=90,000 No. of ways of arranging 3 digits as 2 = 5C3=10.
Arrangements are as follows. Here X can be any digit other than 2
1) 222XX 2) 2X22X 3) 2XX22 4) X2X22 5) X22X2 6) X222X 7) 2X2X2 8) XX222 9) X22X2 10)22XX2
Here the first X cannot be digit '0' so no. of arrangements with first digit as 'X' = 8*9=72. Total such arrangements = 72*5=360
no. of arrangements with first digit as '2'=9*9=81 Total such arrangements = 81*5=405
Now there is some double counting where 'XX' are together.I have counted 00,11,33,44,55,66,77,88,99 as different numbers.. So need to subtract these..
In combination XX222 '00' is not present. So 8 combinations In combinations '222XX', '22XX2', '2XX22', there are 3*9=27 combinations.
Hence 765(27+8)= 730 Total probability= (730/90,000)*100 ~ 0.81%. IMO answer is A Couldn't understand this part: Quote: Now there is some double counting where 'XX' are together.I have counted 00,11,33,44,55,66,77,88,99 as different numbers.. So need to subtract these..
In combination XX222 '00' is not present. So 8 combinations In combinations '222XX', '22XX2', '2XX22', there are 3*9=27 combinations. What exactly do you mean by double counting. Request you to explain this in detail.... Probably would be better if you could use some numbers as example and explain. Thanks, JT
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Re: ID number. [#permalink]
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31 Dec 2009, 15:30
Ok let me try with an example..Though not sure how to explain more clearly.. For the combination, say 2XX22, numbers can vary between be 20022 to 29922. Hence total of 2992220022+1=100 numbers. Of these 100 numbers, 19 numbers have more than three 2s. Hence 10019=81. Of these 81 numbers the following numbers are counted twice..Here when the 100th digit and 1000th digit are same, the numbers are counted twice though infact they are the same... 20022,21122,23322,24422,25522,26622,27722,28822,29922 Hence need to subtract 9. This happens again with 22XX2, 9 times as well.. 22002,22112,22332,22442,22552,22662,22772,22882,22992... For XX222, only 8 such combinations exist.. 11222,33222,44222,55222,66222,77222,88222,99222. Here key is to look for unique combinations.. I hope this clarifies your question..And maybe there is a simpler method to solve this



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Re: ID number. [#permalink]
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01 Jan 2010, 03:52
@ acegre: OK I am still a bit lost as to why you do the substraction. Let me tell you my understanding. Might be u can see where am I leading to. Lets revist the question again: In a 5 digit ID number, what is the probability of exactly three digits are the digit 2?Approach: As rightly done by we could have 5C3 ways of constructing such number i.e. 10 ways. 1. 222XX 2. 22XX2 3. 2XX22 4. 2X2X2 5. 22X2X 6. 2X22X 7. XX222 8. X222X 9. X2X22 10.X22X2 Note: Although you came up with 10 ways, your listed 10 ways had an issue. Please check with the combination in ur post. Considering the first 6 ways, which have 2 in as the first start digit (from left): In 222XX, the first X from left can be filled up in 9 different ways (total 10 digits and we exclude digit 2 from this. Hence 9 digits) Similarly the second X from left can also be filled up in 9 different ways. Reason Both X can be same also as there isn't any condition on it. Therefore 222XX can be filled up in 9X9 ways i.e. 81. (I guess u got his number with a different approach which is also correct. I have used the generic approach followed in probability! But no worries as we both are on the same page ultimately) Now comes the substraction which I do not understand the reason for! To make my point clear, I would list the 222XX combinations for you: Code: 22211 22231 22241 22251 22261 22271 22281 22291 22201 22213 22233 22243 22253 22263 22273 22283 22293 22203 22214 22234 22244 22254 22264 22274 22284 22294 22204 22215 22235 22245 22255 22265 22275 22285 22295 22205 22216 22236 22246 22256 22266 22276 22286 22296 22206 22217 22237 22247 22257 22267 22277 22287 22297 22207 22218 22238 22248 22258 22268 22278 22288 22298 22208 22219 22239 22249 22259 22269 22279 22289 22299 22209 22210 22230 22240 22250 22260 22270 22280 22290 22200 In the above table I dont see any number which is repeated or duplicated as per you. Hence this is where I am lost! Anyway.. continuing the above approach... in the first 6 ways, we can have 81 X 6 = 486 numbers! For the remaining 4 ways, we have the following explanation: 7. XX222 8. X222X 9. X2X22 10.X22X2 In the above sequence of numbers, the first X from left can be filled up by 8 different digits. (Out of the 10 digits, we remove 2 and 0 leaving only 8 possible digits) The second X can anyway be filled up by 9 different digits (Leaving only 2 out of the 10 digits) Hence for each way we have 8X9 = 72 numbers. Therefore for 4 ways we have 72X4 = 288 numbers! Summing the 10 ways, we get 486 + 288 = 774 numbers Total number possible = 9 X 10 X 10 X 10 X 10 = 90000 numbers Therefore probability % = (774 / 90000) X 100 = 0.86%. I know the answer isn't among the options and this makes me feel a bit less confident. Do let me know if you identify any loops in my approach! For the additional question stated: Find the probability that any three digits (exactly three) are the same.the approach is as follows: Let us consider the following combinations. The combination would be same as consider for 2. Only difference is that we can replace 2 with Y for now. Hence the combinations are: 1. YYYXX 2. YYXXY 3. YXXYY 4. YXYXY 5. YYXYX 6. YXYYX 7. XXYYY 8. XYYYX 9. XYXYY 10.XYYXY Considering the first 6 ways: YYYXX  We can choose first Y in 9 different ways (Leave out 0 from 10 digits). Once first Y is selected, the other Ys would be the same and hence no selection required. The first X can be selected in 9 different ways (leave out the digit selected for Y from 10 digits). The second X can be selected in 8 ways (leave out the digit selected for Y and for the first X from 10 digits). Therefore we have YYYXX as 9 x 1 x 1 x 9 x 8 = 648 numbers Hence for 6 ways we have 648 x 6 = 3888 numbers Considering the next 4 ways: XXYYY  We can chose the first Y in 10 different ways. As before, once first Y is selected, the other Ys would be the same and hence no selection required. Returning to X, the first X can be filled in 8 ways (leave out 0 and Y digit from 10 digits) and the second X can be selected in 8 ways (leave out the digit selected for Y and for the first X from 10 digits). Therefore we have XXYYY as 8 x 8 x 10 x 1 x 1 = 880 numbers Hence for the 4 ways we have 880 x 4 = 3520 numbers Summing the numbers possible in 10 ways = 3888 + 3520 = 7408 Total Numbers = 90000 (as before) Therefore probability = 7408/90000 = 0.0823 = 8.23% Please feel free to share your analysis... Thanks, JT
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Re: ID number. [#permalink]
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01 Jan 2010, 06:22
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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: ID number. [#permalink]
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01 Jan 2010, 07:20
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 Chetan, I guess this is wat you r missing. Please check my explanation below: First of all we would need to consider total number of 5 digit IDs as : 9 x 10 x 10 x 10 x 10 = 90000 and not \(10^5\) as we cannot have 0 in the first place, else it would become 4 digit ID. Now as you consider for 3 places to be filled with digit 2, the possible number of combinations = 1 x 1 x 1 x 9 x 9 x 5c3 = 810 number. Again this includes the possibility of having digit 0 as the first or having digit 0 as both first and second. We need to subtract the number falling into this category from 810. Hence we again arrive at the 4 combinations as follows where 0 could be the first or second: XX222 X222X X2X22 X22X2 Its only in them we can have 0 as first. If u notice, we could have 0 as both first and second in the first combination only. For 0 as the first digit, we have the following posb = 1 x 1 x 1 x 1 x 9 x 4 = 36 For 0 as both as first and second digit we have only 1 posb i.e. 00222 or 222. Therefore we need to subtract 36+1 = 37 from 810. This gives us the actuall posb = 81037 = 773 Hence probability % = \(\frac{773}{90000}\) x 100 = 0.858% \(\approx\) 0.86% I know I do not match the options given but am a little lost as I feel the logic to be correct! I hope am on the right track! Thanks, JT
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Re: ID number. [#permalink]
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01 Jan 2010, 08:12
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Bunuel wrote: jeeteshsingh 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 posb=1*1*1*9*9*5c3.. prob=.81%... although i might be still missing something.. ill try again Chetan, I guess this is wat you r missing. Please check my explanation below: First of all we would need to consider total number of 5 digit IDs as : 9 x 10 x 10 x 10 x 10 = 90000 and not \(10^5\) as we cannot have 0 in the first place, else it would become 4 digit ID. Now as you consider for 3 places to be filled with digit 2, the possible number of combinations = 1 x 1 x 1 x 9 x 9 x 5c3 = 810 number. Again this includes the possibility of having digit 0 as the first or having digit 0 as both first and second. We need to subtract the number falling into this category from 810. Hence we again arrive at the 4 combinations as follows where 0 could be the first or second: XX222 X222X X2X22 X22X2 Its only in them we can have 0 as first. If u notice, we could have 0 as both first and second in the first combination only. For 0 as the first digit, we have the following posb = 1 x 1 x 1 x 1 x 9 x 4 = 36 For 0 as both as first and second digit we have only 1 posb i.e. 00222 or 222. Therefore we need to subtract 36+1 = 37 from 810. This gives us the actuall posb = 81037 = 773 Hence probability % = \(\frac{773}{90000}\) x 100 = 0.858% \(\approx\) 0.86% I know I do not match the options given but am a little lost as I feel the logic to be correct! I hope am on the right track! Thanks, JT No, that was not the thing chetan2u was missing. First ID number CAN start with 0, no problem with that. It's 5 digit number which can not start with 0. Can't you have ID (credit card) with number 01234? In that case, I guess chetan2u was correct. Total posb = \(10^5\) Posb of having 3 2 digits = 1 x 1 x 1 x 9 x 9 x 5c3 = 810 Therefore probablity % = \(\frac{810}{10^5}\) x 100 = 0.81 % Guess would wait for your solution! Cheers! JT
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Re: ID number. [#permalink]
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01 Jan 2010, 09:54
I was beginning to rethink whether I should restart my Math prep again Thanks for clarifying the same Bunuel! Cheers! JT
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Re: ID number. [#permalink]
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02 Jan 2010, 01:25
Wow..I was making it complicated for no reason.. Thanks for the enlightment! That was a good one.



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Re: ID number. [#permalink]
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07 Jan 2010, 01:25
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My approach was finding total possibilities, and dividing out the number of possibilities with exactly three 2s. Total possibilities = 100000 Two of the digits have 9 possible choices each, and the other three must be 2s, and the number of ways of choosing the three spaces that must have 2s is 5c3 = 10 Desired outcomes = 9*9*1*1*1*10 Probability = 810/100000 = 0.81% Hopefully this is correct!
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Re: ID number. [#permalink]
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08 Jan 2010, 06:20
Uhm, what does ID mean? My approach: there are 2 types of number satisfying: 2abcd (2 of a,b,c,d are digit 2) and Amnpq (A is different from 2; 3 of m,n,p,q are digit 2) 2abcd: 4C2 * 9 * 9 desired outcomes Amnpq: 8* 4C3 *9
Total posibilities: 9*10^4
Probability= (4C2 * 9 * 9 + 8* 4C3 * 9) / (9 * 10^4) =0.0086



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Re: ID number. [#permalink]
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23 Feb 2010, 00:14
Bunuel wrote: Found this question elsewhere and think it was solved incorrectly. Give it a try. Only my solution to follow as no OA.
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%
Additional question: Find the probability that any three digits (exactly three) are the same. Answer to the additional question: favorable outcomes: \(9C1.9C1.5C3.10C1/=8100\) Total possible outcomes: \(10^5\) Probability in %= 8.1%



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Re: ID number. [#permalink]
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23 Feb 2010, 08:43
Bunuel wrote: Found this question elsewhere and think it was solved incorrectly. Give it a try. Only my solution to follow as no OA.
1) In a 5 digit ID number, what is the probability of exactly three digits are the digit 2? 2) Additional question: Find the probability that any three digits (exactly three) are the same. 1) total ways = 10^5 3 digits of 2 and the rest 2 digits are different = 9 * 9 * 5c3 (5c3 to pick 3 places for 2's and the rest possibilities are filled with one of the remaining 9 numbers) probability = 0.81% 2) extension to the above problem  multiply by 10 as there are 10 single digit numbers probability = 8.1%



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Re: ID number. [#permalink]
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25 Feb 2010, 18:26
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%



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Re: ID number. [#permalink]
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26 Feb 2010, 08:35
The question is ambiguous. No where it is mentioned that 00000 could be taken as a 5 digit ID number. Going by actual 5 digit no.s, I think the answer should be 0.86 %
No. of 5 digit numbers = 9*10*10*10*10 = 90,000 Possible no.s with 2 in the ten thousands position = 1*4C2*9*9= 486 Possible no.s with any other no. in the ten thousands position = 8*4C3*9 = 288
Required Probability = (486+288)/90000 = 0.86%



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Re: ID number. [#permalink]
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27 Feb 2010, 19:20
I could swear the answer is .0081% (which I realize isn't among the answers). Using my 5step method:
1) Lay out the number of events (5 digits) _ _ _ _ _
2) Write down one specific example of the desired outome (exactly 3 twos): 2 2 2 (not2) (not2)
3) Label each event with its relevant probability and multiply across: 1/10 1/10 1/10 9/10 9/10 = 81/10^5
4)Determine the number of ways in which we could have the desired outcome: 5 events, 3 desired = 5C3 = 10
5) Multiply the result of step 4 by the result of step 3: 10 x 81/10^5 = 81/10^4 = .0081%



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Re: ID number. [#permalink]
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27 Feb 2010, 21:43
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lennon wrote: I could swear the answer is .0081% (which I realize isn't among the answers). Using my 5step method:
1) Lay out the number of events (5 digits) _ _ _ _ _
2) Write down one specific example of the desired outome (exactly 3 twos): 2 2 2 (not2) (not2)
3) Label each event with its relevant probability and multiply across: 1/10 1/10 1/10 9/10 9/10 = 81/10^5
4)Determine the number of ways in which we could have the desired outcome: 5 events, 3 desired = 5C3 = 10
5) Multiply the result of step 4 by the result of step 3: 10 x 81/10^5 = 81/10^4 = .0081% Lennon  remember, percentage = divided by 100 You have the right answer, but 81/10^4 = 81/100 (or 81%) * 1/100, thus, .81% .0081 is the right answer, but .81 is the PERCENTAGE







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