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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. \(\frac{1530}{100,000}\)

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.

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.
_________________

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.

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..

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?

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).
_________________

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, HL
_________________

[4.33] In the end, what would you gain from everlasting remembrance? Absolutely nothing. So what is left worth living for? This alone: justice in thought, goodness in action, speech that cannot deceive, and a disposition glad of whatever comes, welcoming it as necessary, as familiar, as flowing from the same source and fountain as yourself. (Marcus Aurelius)

Last edited by Linhbiz on 20 Jun 2016, 18:18, edited 1 time in total.

\(9*8*5!/3!* 2!\) should be 720 right? --> \(9*8*5!/(3!*2!) = 72*4*5/2 = 720\)

Then you'll get the final ans: \(810/10^5\)
_________________

[4.33] In the end, what would you gain from everlasting remembrance? Absolutely nothing. So what is left worth living for? This alone: justice in thought, goodness in action, speech that cannot deceive, and a disposition glad of whatever comes, welcoming it as necessary, as familiar, as flowing from the same source and fountain as yourself. (Marcus Aurelius)

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

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.

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.

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.

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.

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