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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. \(\frac{1530}{100,000}\)
[Reveal] Spoiler: OA

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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 5-digit 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: M26-19 [#permalink]

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New post 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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New post 20 Mar 2015, 05:50
sairajeshg wrote:
In the above problem we can choose digit 6 out of 10, how come it is 6 out of 5.


Where do we have 6 out of 5? How it can be choosing 6 out of 5? We have 5C3, which is the number of ways to choose which 3 digits (out of 5) will be 6's.
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New post 05 May 2016, 14:21
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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New post 10 May 2016, 20:08
I stand corrected. Thanks.

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

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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 = 10-1 = 9..
so possiblity = 9*9*5C3 = 810.. as explained by others..

the way you are taking, you can have a 5-digit code with all 6 - 66666 or 4 6s - 66661 or 16666 etc..

but Q asks us EXACTLy 3...

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New post 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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New post 05 Jun 2016, 04:52
glochou wrote:
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?


5C3 is the number of ways to choose which 3 digits will be 6's out of 5 digits we have. The remaining 2 places will be taken by non-sixes (9*9 combination).
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Update (21-Jun-16): 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 0-9 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,
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Last edited by Linhbiz on 20 Jun 2016, 18:18, edited 1 time in total.

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Re: M26-19 [#permalink]

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New post 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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New post 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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New post 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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New post 05 Sep 2016, 23:41
Hi Bunnel,

Kindly reply.

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New post 12 Oct 2016, 06:58
I think this is a high-quality question.

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New post 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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New post 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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New post 20 Jan 2017, 06:13
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 36366


Total = 30 Total ways of arranging the digits (Case 1 + Case 2)

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New post 10 Mar 2017, 19:38
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 36366


Total = 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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New post 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 36366


Total = 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/baker-s-dozen-128782-20.html

Hope it helps.
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Re: M26-19   [#permalink] 11 Mar 2017, 04:04

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