M26-19

Question

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}

Bunuel · Accepted Answer

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}

The total number of 5-digit codes is  10^5 . It's important to note that it's not  9*10^4 , as the first digit  can  be zero in a password.
 
The number of passwords with three digits as 6 can be calculated as  9*9*C^3_5 = 810 . We have 9 choices for each of the two remaining digits (not 6), resulting in  9*9 . The term  C^3_5  represents the number of ways to choose the positions for the 6s among the five digits (essentially deciding which three out of ***** will be 6s).
 
The probability is therefore,  P=\frac{favorable}{total}=\frac{810}{10^5} .

Answer: B

wmichaelxie · Answer

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.

chetan2u · Answer

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...

glochou · Answer

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?

Bunuel · Answer

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).

fuzzywuzzy · Answer

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

srikanth9502 · Answer

Guys,

After spending sometime on the question, I could understand why the answer is 810 but not 1530. Here is my explanation.

Number of ways in which 3 places are occupied by the digit 6 and the remaining 2 places are occupied by two different digits , say 1 and 2

Now, let us see what happens when we calculate 9*8*5!/(3!) = 1440 ways

-> 9*8 implies we are considering all the arrangements pertaining to two digits. As we took digits 1 and 2 as our examples, we happen to consider the arrangements  1   2  and  2   1  separately

Now let us consider the arrangement  1   2  followed by three 6's. when you multiply this arrangement of  1   2   6   6   6  with 5!/(3!), you consider the cases including the ones in which 2 comes before 1 . For example  2   1   6   6   6  is also considered as 5!/(3!) includes all the possible arrangements.

But as explained earlier, since 9*8 considers the arrangement  2   1  as distinct from  1   2  , you therefore multiply the arrangement  2   1   6   6   6  with 5!/(3!) separately, resulting in the double counting since all the arrangements pertaining to three 6's and 1,2 are already counted.

Therefore you should divide the computation 9*8*5!/(3!) by 2! inorder to avoid the double counting.

So the answer is 720+ 90 (Number of ways in which two other digits are same i.e 9*1*5!/(3!*2!)) = 810

BrentGMATPrepNow · Answer

We're going to apply the MISSISSIPPI rule at the end of my solution, so here's what it says:
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are  n  objects where  A  of them are alike, another  B  of them are alike, another  C  of them are alike, and so on, then the total number of possible arrangements =  n !/ A !)( B !)( C !)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows: 
There are  11  letters in total 
There are  4  identical I's
There are  4  identical S's
There are  2  identical P's
So, the total number of possible arrangements =  11 !/ 4 !)( 4 !)( 2 !)]

-----NOW ONTO THE QUESTION------------------------------

We'll consider two cases: 
case i: We have three 6's and two DIFFERENT digits (e.g., 66612)
case ii: We have three 6's and two IDENTICAL digits (e.g., 66677)

case i: We have three 6's and two DIFFERENT digits (e.g., 66612)  
We already have three 6's. So, we must select 2 different digits from (0,1,2,3,4,5,7,8 and 9)
We can do this in 9C2 ways (= 36  ways)
Now that we've selected our 5 digits, we must ARRANGE them, which means we can use the MISSISSIPPI rule. 
We can arrange 3 identical digits and 2 different digits in 5!/3! ways =  20  ways
So, we can have three 6's and two DIFFERENT digits in ( 36 )( 20 ) ways (=  720  ways)

case ii: We have three 6's and two IDENTICAL digits (e.g., 66677)  
We already have three 6's. So, we must select 1 digit (which we'll duplicate)
Since we're selecting 1 digit from (0,1,2,3,4,5,7,8,9) we can do so in  9  ways
Now that we've selected our 5 digits, we must ARRANGE them, which means we can use the MISSISSIPPI rule. 
We can arrange 3 identical 6's and 2 other identical digits in 5!/3!2! ways =  10  ways
So, we can have three 6's and two IDENTICAL digits in ( 9 )( 10 ) ways (=  90  ways)

So, TOTAL number of ways to have three 6's =  720  +  90  =  810

Since there are 100,000 possible 5-digit codes, P(having exactly three 6's) = 810/100,000

Answer: B

Cheers,
Brent

VeritasPrepErika · Answer

Another (slightly more painful) option here if you're struggling with the repeated terms:

First,  carefully  work out all of the possible arrangements of the three 6's:

6 6 6 __ __
6 6 __ 6 __
6 6 __ __ 6
6 __ 6 6 __
6 __ 6 __ 6
6 __ __ 6 6
__ 6 6 6 __
__ 6 6 __ 6
__ 6 __ 6 6
__ __ 6 6 6

So there are 10 ways that we can arrange the 6's.

Then, we deal with the other two blanks. For each of the 10 options above, the two remaining blanks can be any digit other than 6, giving 9 options for each blank. This means that for each of the 10 options above, we have 9*9=81 total arrangements of digits for the two remaining blanks. Combining these two pieces of information, 10 arrangements of the three 6's * 81 arrangements of the remaining two blanks each = 810 favorable arrangements.

Since there are 10^5 arrangements total, the answer is B.

When you have one severely limiting factor (like needing three 6's), it can sometimes be relatively quick to brute force just that piece, then use simple combinatorics principles (like the slot method used here) for the rest. However, if you use this method, be extremely careful not to miss any options. Here, I stayed organized by locking the first two 6's and finding all of the options to place the third 6. Then, keeping the first 6 locked, I moved the second 6 by one space, locked it, and found all of the options to place the third 6. I repeated this process for all options for the second 6 before moving the first 6 one space, locking it, and repeating the process from the beginning.

SDW2 · Answer

Hello,
Can someone explain how does the duplication of 2 remaining digits ("2" in the adjacent example) is taken care of in this method? like how does one of these two 66622 and 66622 is removed using the method shown above?
 Bunuel   VeritasKarishma   IanStewart

KarishmaB · Answer

Not sure what your doubt is. Taking both possible scenarios.

Problem 1: Is 66622 allowed?
There is no issue with duplication of the other number. 
66622 is an acceptable 5 digit code.

5C3 picks 3 of the 5 places to give to 6. 
So the 3 spots picked could be first, second and third. 666 __ __
Now the other 2 spots can be filled in 9*9 ways (no 6 on either). We could put 2 in both the spots. 66622
This is acceptable.

Problem 2: Does 66622 appear twice?
There is no duplication of 66622. 
9*9 is the way you can fill __ __ (two distinct places using any one of 9 digits)
So 22 will occur only once.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

saransh2797 · Answer

Hi Karishma, I think 22 will appear twice. Since 9x9 means this- 6662 2 and 666 2 2. I have bolded the 2 for your ease. That's why previous person asked how do we account for this duplicacy. Thanks

phmahi1997 · Answer

I like the solution - it’s helpful.

gagan03 · Answer

Why do we use 5C3 in this case, even though it is possible that out of 5 digits, 3 are 6 and the other 2 are different? In such a case, the number of permutations should be 5!/3! instead.

Bunuel · Answer

It's not about permutation.  5C3 is the number of ways to choose which 3 digits will be 6's out of 5 digits we have (so which three * out of *****); it's not the arrangement of the five digits. The remaining 2 places will be taken by non-sixes (9*9 combinations).

jacopoverdolini · Answer

Hi  Bunuel  my approach to this was a little different and I don't know if it's correct in general or it works only for this particular question.

Since we want to know the probability of exactly three 6 and two non six, i set it up like this:

Y Y Y N N

with Y and N being yes and no answer to "is this number a 6?"

then we need to find the probability of Yes and Probability of No, clearly 1/10 *1/10 *1/10 *9/10 *9/10
yielding 81/100000.
Then this three Yes and two No can be arranged in 5!/3!2! = 10 ways, multiplying 81 by 10 we get 810/100000.
Choice: B

braddouglas10 · Answer

Bunuel

Do you have any similar problems you can please send?

Bunuel · Answer

Check  Constructing Numbers, Codes and Passwords Questions  in our  Special Questions Directory .

Hope it helps.

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