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Re M2619 [#permalink]
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16 Sep 2014, 01:25
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Official Solution:A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digits 6? A. \(\frac{860}{90,000}\) B. \(\frac{810}{100,000}\) C. \(\frac{858}{100,000}\) D. \(\frac{860}{100,000}\) E. \(\frac{1530}{100,000}\) Total # of 5digit codes is \(10^5\), notice that it's not \(9*10^4\), since in a code we can have zero as the first digit. # of passwords with three digits 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have \(9*9\) and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have. \(P=\frac{favorable}{total}=\frac{810}{10^5}\). Answer: B
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Re: M2619 [#permalink]
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19 Mar 2015, 17:27
In the above problem we can choose digit 6 out of 10, how come it is 6 out of 5.



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Re: M2619 [#permalink]
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20 Mar 2015, 05:50



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Pretty sure it should be 1/100 since the # of passwords with three digits 6 should be 10∗10∗C35=10*10*10=1000
Each out of two other digits (not 6) should have 10 choices if you are considering 10 possible digits when calculating the total possible outcomes for the denominator (10,000) in the first step.



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I stand corrected. Thanks.
Last edited by wmichaelxie on 10 May 2016, 20:58, edited 1 time in total.



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Re: M2619 [#permalink]
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10 May 2016, 20:55
wmichaelxie wrote: Pretty sure it should be 1/100 since the # of passwords with three digits 6 should be 10∗10∗C35=10*10*10=1000
Each out of two other digits (not 6) should have 10 choices if you are considering 10 possible digits when calculating the total possible outcomes for the denominator (10,000) in the first step. Hi you are wrong here.. If you are considering other digits than 6... total digits = 101 = 9.. so possiblity = 9*9*5C3 = 810.. as explained by others.. the way you are taking, you can have a 5digit code with all 6  66666 or 4 6s  66661 or 16666 etc.. but Q asks us EXACTLy 3... Quote: Bunuel, give me kudos bro NOW who deserves kudos
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Re: M2619 [#permalink]
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03 Jun 2016, 10:16
I understand that 5C3 is how many ways that the digit '6' is allocated in each of the five slots, but what about the other two digits? Where do you account for the number of ways that those two digits can be allocated in each of the five slots?



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Re: M2619 [#permalink]
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05 Jun 2016, 04:52



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Update (21Jun16): it's ok now, I know why I was wrong, xx is double counted so need to divide by 2 again > probability = \(810/10^5\)Hi Bunuel, I am always not so good at Combination & Probability, it's sometimes hard to follow your explanation. (Edited to fit with my understanding)I came out with this kind of different understanding about the question. Basically, the number of desired outcomes is the # arrangement of: 666xy  where xx is any number from 09 except 6. Since we have 2 repetition: 3 of 6, for each and every case of xy (9*9=81 cases) So the # of desired outcomes = \((5!/3!)*9*9 = 1620\) > different result. Could you kindly help to point out my incorrectness? Thanks, HL
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Last edited by Linhbiz on 20 Jun 2016, 18:18, edited 1 time in total.



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Re: M2619 [#permalink]
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19 Jun 2016, 03:37
no of favourable cases:
1. 3 digits are 6 and rest two are different digits: 9*8*5!/3!* 2!= 1440 (6 6 6 9 8)
2. 3 digits are 6 and rest two digits are same: 9* 5!/3!* 2!=90( 6 6 6 8 8)
total 1440+90= 1530
probabiltiy =1530/10^5
plz crct me.



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Re: M2619 [#permalink]
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20 Jun 2016, 18:26
vishnu440 wrote: no of favourable cases:
1. 3 digits are 6 and rest two are different digits: 9*8*5!/3!* 2!= 1440 (6 6 6 9 8)
2. 3 digits are 6 and rest two digits are same: 9* 5!/3!* 2!=90( 6 6 6 8 8)
total 1440+90= 1530
probabiltiy =1530/10^5
plz crct me. vishnu440: The way you're thinking is correct On calculation of #1 though: \(9*8*5!/3!* 2!\) should be 720 right? > \(9*8*5!/(3!*2!) = 72*4*5/2 = 720\) Then you'll get the final ans: \(810/10^5\)
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Re: M2619 [#permalink]
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27 Aug 2016, 06:55
Hi Bunuel, do let us know why you haven't accounted for the arrangement of the remaining digits in the solution. ie your solution will count 63566 and 65366 as one case only in the numerator, while it will count both cases in the denominator. As someone above pointed out, I calculate 1530/10^6 as well. Kindly elaborate as to what we're missing. Regards,



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Re: M2619 [#permalink]
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05 Sep 2016, 23:41
Hi Bunnel,
Kindly reply.



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Re M2619 [#permalink]
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12 Oct 2016, 06:58
I think this is a highquality question.



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Re: M2619 [#permalink]
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05 Nov 2016, 23:00
Hi guys,
Alternate solution that worked for me  let me know what you think!
1. The probability of choosing a 6 is (1/10) 2. There are 5 digits, so the combined probability is (1/10)*(1/10)*(1/10)*(9/10)*(9/10) = 81/100,000 3. There are 5C3 ways to organize the 3 6's you picked
The answer is 10*(81/100,000) = 810/100,000



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Re: M2619 [#permalink]
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02 Jan 2017, 23:07
If the remaining two digits are different, then possible outcomes =9*8*5!/3! (3! for the three 6's)
if the remaining two digits are same, then possible outcomes =9*5!/(3!*2!) (3! for the three 6's and 2! for the remaining two equal digits)
Total Possible outcomes=9*8*5!/3!+9*5!/(3!*2!) = 1530.
Kindly correct the errors i have made.



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Sumanth8492 wrote: If the remaining two digits are different, then possible outcomes =9*8*5!/3! (3! for the three 6's)
if the remaining two digits are same, then possible outcomes =9*5!/(3!*2!) (3! for the three 6's and 2! for the remaining two equal digits)
Total Possible outcomes=9*8*5!/3!+9*5!/(3!*2!) = 1530.
Kindly correct the errors i have made. +++ I got the same answer: Case#1:  666_ _ (the last two digits could be 00,11,22,33,44,55,77,88,99): No. of ways to arrange these = 5!/3!2! = 10 Probability of 3 6's = (1/10)(1/10)(1/10)(9/10)(1/10) = 9/100000 // 9 ways to select the first digit (other than 6) and only one way to select the same number as the previous number that we have selected.No. of arrangement = 9/10^5 * 10 = 90/10^5 Case#2:  666_ _ (the last two digits could be 01,10,12,23,etc): No. of ways to arrange these = 5!/3!1!1! = 20 Probability of 3 6's = (1/10)(1/10)(1/10)(9/10)(8/10) = 72/100000 // 9 ways to select the first digit (other than 6) and 8 ways to select the last digit that is not 6 and not the previous number already selected.No. of arrangement = 72/10^5 * 20 = 1440/10^5 Case1 + Case2 = Total probability = (90+1440)/10^5 = 1530/10^5 Experts, am I counting anything twice between Case 1 and 2? I do get 81 possibilities (9+72 = 81), but I get different number of possible arrangements. I get total 30 arrangements whereas in the official answer I see only 10. Am I missing something. Please guide me. I also tried with only 3 numbers (2,3,6) instead of 10 (0,1,2,3,4,5,6,7,8,9) and got similar arrangement possibilities i.e 30. B = 6 A = 2 C = 3 1 BBBAC 66623 2 BBBCA 66632 3 BBACB 66236 4 BBCAB 66326 5 BBABC 66263 6 BBCBA 66362 7 BACBB 62366 8 BABBC 62663 9 BABCB 62636 10 BCABB 63266 11 BCBBA 63662 12 BCBAB 63626 13 ACBBB 23666 14 ABBBC 26663 15 ABCBB 26366 16 ABBCB 26636 17 CABBB 32666 18 CBBBA 36662 19 CBABB 36266 20 CBBAB 36626 1 BBBAA 66622 2 AABBB 22666 3 ABBBA 26662 4 ABBAB 26626 5 ABABB 26266 6 BBBCC 66633 7 CCBBB 33666 8 CBBBC 36663 9 CBBCB 36636 10 CBCBB 36366Total = 30 Total ways of arranging the digits (Case 1 + Case 2)



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nn2015 wrote: Sumanth8492 wrote: If the remaining two digits are different, then possible outcomes =9*8*5!/3! (3! for the three 6's)
if the remaining two digits are same, then possible outcomes =9*5!/(3!*2!) (3! for the three 6's and 2! for the remaining two equal digits)
Total Possible outcomes=9*8*5!/3!+9*5!/(3!*2!) = 1530.
Kindly correct the errors i have made. +++ I got the same answer: Case#1:  666_ _ (the last two digits could be 00,11,22,33,44,55,77,88,99): No. of ways to arrange these = 5!/3!2! = 10 Probability of 3 6's = (1/10)(1/10)(1/10)(9/10)(1/10) = 9/100000 // 9 ways to select the first digit (other than 6) and only one way to select the same number as the previous number that we have selected.No. of arrangement = 9/10^5 * 10 = 90/10^5 Case#2:  666_ _ (the last two digits could be 01,10,12,23,etc): No. of ways to arrange these = 5!/3!1!1! = 20 Probability of 3 6's = (1/10)(1/10)(1/10)(9/10)(8/10) = 72/100000 // 9 ways to select the first digit (other than 6) and 8 ways to select the last digit that is not 6 and not the previous number already selected.No. of arrangement = 72/10^5 * 20 = 1440/10^5 Case1 + Case2 = Total probability = (90+1440)/10^5 = 1530/10^5 Experts, am I counting anything twice between Case 1 and 2? I do get 81 possibilities (9+72 = 81), but I get different number of possible arrangements. I get total 30 arrangements whereas in the official answer I see only 10. Am I missing something. Please guide me. I also tried with only 3 numbers (2,3,6) instead of 10 (0,1,2,3,4,5,6,7,8,9) and got similar arrangement possibilities i.e 30. B = 6 A = 2 C = 3 1 BBBAC 66623 2 BBBCA 66632 3 BBACB 66236 4 BBCAB 66326 5 BBABC 66263 6 BBCBA 66362 7 BACBB 62366 8 BABBC 62663 9 BABCB 62636 10 BCABB 63266 11 BCBBA 63662 12 BCBAB 63626 13 ACBBB 23666 14 ABBBC 26663 15 ABCBB 26366 16 ABBCB 26636 17 CABBB 32666 18 CBBBA 36662 19 CBABB 36266 20 CBBAB 36626 1 BBBAA 66622 2 AABBB 22666 3 ABBBA 26662 4 ABBAB 26626 5 ABABB 26266 6 BBBCC 66633 7 CCBBB 33666 8 CBBBC 36663 9 CBBCB 36636 10 CBCBB 36366Total = 30 Total ways of arranging the digits (Case 1 + Case 2) I also got 1530 for the exact same reason as above. It's boiling down to the concept of either 9*9*(5c3) or 9*8*(5!/3!)+9*[5!/(3!2!)] and I cannot understand why you would pick the former. Using 5c3 gives the number of different ways to get a subset of 3 digits from a set of 5, but the latter provides a method for determining the number of possibilities to arrange 5 digits, of which 3 are equivalent, and the number of possibilities to arrange 5 digits, of which 3 are equivalent to each other and the other are equivalent to each other. I'm hoping someone can correct me because I'm convinced the answer is 1530/10^5.



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Re: M2619 [#permalink]
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11 Mar 2017, 04:04
brooklyndude wrote: nn2015 wrote: Sumanth8492 wrote: If the remaining two digits are different, then possible outcomes =9*8*5!/3! (3! for the three 6's)
if the remaining two digits are same, then possible outcomes =9*5!/(3!*2!) (3! for the three 6's and 2! for the remaining two equal digits)
Total Possible outcomes=9*8*5!/3!+9*5!/(3!*2!) = 1530.
Kindly correct the errors i have made. +++ I got the same answer: Case#1:  666_ _ (the last two digits could be 00,11,22,33,44,55,77,88,99): No. of ways to arrange these = 5!/3!2! = 10 Probability of 3 6's = (1/10)(1/10)(1/10)(9/10)(1/10) = 9/100000 // 9 ways to select the first digit (other than 6) and only one way to select the same number as the previous number that we have selected.No. of arrangement = 9/10^5 * 10 = 90/10^5 Case#2:  666_ _ (the last two digits could be 01,10,12,23,etc): No. of ways to arrange these = 5!/3!1!1! = 20 Probability of 3 6's = (1/10)(1/10)(1/10)(9/10)(8/10) = 72/100000 // 9 ways to select the first digit (other than 6) and 8 ways to select the last digit that is not 6 and not the previous number already selected.No. of arrangement = 72/10^5 * 20 = 1440/10^5 Case1 + Case2 = Total probability = (90+1440)/10^5 = 1530/10^5 Experts, am I counting anything twice between Case 1 and 2? I do get 81 possibilities (9+72 = 81), but I get different number of possible arrangements. I get total 30 arrangements whereas in the official answer I see only 10. Am I missing something. Please guide me. I also tried with only 3 numbers (2,3,6) instead of 10 (0,1,2,3,4,5,6,7,8,9) and got similar arrangement possibilities i.e 30. B = 6 A = 2 C = 3 1 BBBAC 66623 2 BBBCA 66632 3 BBACB 66236 4 BBCAB 66326 5 BBABC 66263 6 BBCBA 66362 7 BACBB 62366 8 BABBC 62663 9 BABCB 62636 10 BCABB 63266 11 BCBBA 63662 12 BCBAB 63626 13 ACBBB 23666 14 ABBBC 26663 15 ABCBB 26366 16 ABBCB 26636 17 CABBB 32666 18 CBBBA 36662 19 CBABB 36266 20 CBBAB 36626 1 BBBAA 66622 2 AABBB 22666 3 ABBBA 26662 4 ABBAB 26626 5 ABABB 26266 6 BBBCC 66633 7 CCBBB 33666 8 CBBBC 36663 9 CBBCB 36636 10 CBCBB 36366Total = 30 Total ways of arranging the digits (Case 1 + Case 2) I also got 1530 for the exact same reason as above. It's boiling down to the concept of either 9*9*(5c3) or 9*8*(5!/3!)+9*[5!/(3!2!)] and I cannot understand why you would pick the former. Using 5c3 gives the number of different ways to get a subset of 3 digits from a set of 5, but the latter provides a method for determining the number of possibilities to arrange 5 digits, of which 3 are equivalent, and the number of possibilities to arrange 5 digits, of which 3 are equivalent to each other and the other are equivalent to each other. I'm hoping someone can correct me because I'm convinced the answer is 1530/10^5. First of all this is a hard question. Next, let me assure you that the answer is 100% correct. I tried to explain the reasoning above. You can check alternative solutions here: https://gmatclub.com/forum/bakersdozen12878220.htmlHope it helps.
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