probability 5 digit ID

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canuckclint: Master | Next Rank: 500 Posts; Posts: 154; Joined: Tue Aug 26, 2008 12:59 pm; Location: Canada; Thanked: 4 times

by canuckclint » Thu Sep 04, 2008 8:11 pm

In a 5 digit ID number, what is the probability of 3 digits are the digit 2?

This one's harder: Probability that 3 digits are the same.

Have fun!

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tishan: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Tue Aug 26, 2008 9:15 pm; Location: INDIA

by tishan » Thu Sep 04, 2008 9:05 pm

hi!!
is the answer 1/30?

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P_mashru: Senior | Next Rank: 100 Posts; Posts: 43; Joined: Wed Jun 11, 2008 5:03 am; Location: Mumbai; GMAT Score:530

by P_mashru » Thu Sep 04, 2008 9:10 pm

1 1 1 10 10

Possibilities of having digit 2 in 3 places out of 5 are 100
Total posibilities - 10^5

Probability - 10^2 / 10^5 = 1/1000

What is OA?

u have not even mentioned options frined

I am I

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parallel_chase: Legendary Member; Posts: 1153; Joined: Wed Jun 20, 2007 6:21 amFollowed by: 2 members; Thanked: 146 times

Re: probability 5 digit ID

by parallel_chase » Fri Sep 05, 2008 4:13 am

canuckclint wrote:In a 5 digit ID number, what is the probability of 3 digits are the digit 2?

This one's harder: Probability that 3 digits are the same.

Have fun!

Probability that exactly 3 digits are 2 = (1*1*1*9*9)/10^5 = 81/10^5

Probability that 3 digits are same = 10*1*1*9*9/10^5 = 81/10^4

Last edited by parallel_chase on Fri Sep 05, 2008 4:40 am, edited 1 time in total.

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4meonly: Legendary Member; Posts: 891; Joined: Sat Aug 16, 2008 4:21 amFollowed by: 1 member; Thanked: 27 times; GMAT Score:660(

Re: probability 5 digit ID

by 4meonly » Fri Sep 05, 2008 4:25 am

parallel_chase wrote: Probability that 3 digits are 2 = (1*1*1*10*10)/10^5 = 1/1000

I think that we should multiply this by 5*4*3 because these three 2's can be on different positions
answer will be 60/1000 = 6/100 = 0,06

because

canuckclint wrote:In a 5 digit ID number, what is the probability of 3 digits are the digit 2?

means that three 2's are not certainly on the first positions

What do you think?

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parallel_chase: Legendary Member; Posts: 1153; Joined: Wed Jun 20, 2007 6:21 amFollowed by: 2 members; Thanked: 146 times

Re: probability 5 digit ID

by parallel_chase » Fri Sep 05, 2008 4:45 am

4meonly wrote:
parallel_chase wrote: Probability that 3 digits are 2 = (1*1*1*10*10)/10^5 = 1/1000
I think that we should multiply this by 5*4*3 because these three 2's can be on different positions
answer will be 60/1000 = 6/100 = 0,06
What do you think?

I have edited my post above.

we have 10 letters (1,2,3,4,5,6,7,8,9,0)

Probability that exactly 3 digits is 2

for first digit we can have only one option i.e. 2
for second digit we can have only one option i.e. 2
for third digit we can have only one option i.e. 2
for fourth digit we can have all the letters except 2 i.e. 9 ways
for fifth digit we can have all the letters except 2 i.e. 9 ways

1*1*1*9*9 = 81

total options = 10^5

probability = 81/10^5

Hope its clear.

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4meonly: Legendary Member; Posts: 891; Joined: Sat Aug 16, 2008 4:21 amFollowed by: 1 member; Thanked: 27 times; GMAT Score:660(

by 4meonly » Fri Sep 05, 2008 5:00 am

Yes,
you made reasonable corrections about 9's instead of 10's

But the question is should we multiply 81/10^5 by 5*4*3 because these three 2's can be on different positions
First 2 can be placed in 5 positions
Second 2 can be placed in 4 positions
Third 2 can be placed in 3 positions

81*60/10^5 = 81*6/10^4 = 486/10000 = 0,0486

what about this comment?
5*4*3

Last edited by 4meonly on Fri Sep 05, 2008 5:32 am, edited 2 times in total.

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sudhir3127: Legendary Member; Posts: 829; Joined: Mon Jul 07, 2008 10:09 pmFollowed by: 3 members; Location: INDIA; Thanked: 84 times

by sudhir3127 » Fri Sep 05, 2008 5:23 am

4meonly wrote:Yes,
you made reasonable corrections about 9's instead of 10's

But the question is should w multiply 81/10^5 by 5*4*3 because these three 2's can be on different positions
First 2 can be placed in 5 positions
Second 2 can be placed in 4 positions
Third 2 can be placed in 3 positions

81*60/10^5 = 81*6/10^4 = 486/10000 = 0,0486

what about this comment?
5*4*3

i go with

120*9^2/10^5

do let us know the OA

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4meonly: Legendary Member; Posts: 891; Joined: Sat Aug 16, 2008 4:21 amFollowed by: 1 member; Thanked: 27 times; GMAT Score:660(

by 4meonly » Fri Sep 05, 2008 5:34 am

120 means 5! ?
explain, please

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sudhir3127: Legendary Member; Posts: 829; Joined: Mon Jul 07, 2008 10:09 pmFollowed by: 3 members; Location: INDIA; Thanked: 84 times

by sudhir3127 » Fri Sep 05, 2008 5:38 am

4meonly wrote:120 means 5! ?
explain, please

arrranging first 2 in any of the 5 places in 5 ways
arrranging second 2 in any of the 4 places in 4 ways
arrranging third 2 in any of the 3 places in 3 ways

now arranging any of the remaining 9 numbers is 2 places using the same logic as above is
2*9^2 ways

thus total is

5*4*3*2*9^2/10^5

hope this is clear...

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Fri Sep 05, 2008 9:15 am

Let's begin with an easier question:

How many 5-digit id numbers can be made if the first three digits are 2, and the last two digits are not 2?

We then have 1 choice for each of the first three digits, and 9 choices for the last 2:

1*1*1*9*9 = 81

If we want to count how many 5-digit id numbers can be made with exactly three 2s and two other digits, we need to multiply the above by the number of ways we can choose three slots out of five to place the digit 2. We would have 5C3 choices for where to put the three 2s, so we should multiply by 5C3 = 10. Note that you shouldn't multiply by 5*4*3, because then you're assuming the order of the 2s matters, and if you shuffle the 2s around within a given id, the number remains the same.

10*9^2 = 810

If we want to count how many ids can be made with exactly three of one digit, and the other two different, we can multiply the above by 10, since there will be 810 ways to have three 0s, 810 ways to have three 1s, etc. So we would have 8100 such ids in total.

If it's a probability question, in each case we would need to divide by the total number of possible ids, which is 10^5.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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schumi_gmat: Master | Next Rank: 500 Posts; Posts: 377; Joined: Wed Sep 03, 2008 9:30 amFollowed by: 2 members; Thanked: 15 times

by schumi_gmat » Fri Sep 05, 2008 11:48 am

I Think Ian we have to multiply by 5!. My solution assuming that first didgit can be 0 in an ID.

no of ways to select 5 digit number is 10^5

no of ways to select digit with 3 2's is 9*9 = 81

the 2's and other two digits can be arranged among each other in 5! ways.

Hence the digits with 3 2's = 81 * 120

Probability = 81*120/ 10^5

What is an OA?

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Fri Sep 05, 2008 2:25 pm

schumi_gmat wrote:I Think Ian we have to multiply by 5!. My solution assuming that first didgit can be 0 in an ID.

no of ways to select 5 digit number is 10^5

no of ways to select digit with 3 2's is 9*9 = 81

the 2's and other two digits can be arranged among each other in 5! ways.

Hence the digits with 3 2's = 81 * 120

Probability = 81*120/ 10^5

What is an OA?

As Ian pointed out, if you use simple factiorials, you're counting identical arrangements multiple times.

For example, let's look at the two codes:

22214
and
22214

Notice anything about those two codes? Indeed, they're identical. However, using your method, we'd have counted them as two separate items.

If we think of the three "2"s as "2a" "2b" and "2c", the two arrangements we see above could be:

(2a)(2b)(2c)14
and
(2b)(2a)(2c)14

or a number of other permutations of our 2s.

Again, as Ian stated (he's usually right, you know!), we really want to use the combinations formula, because what we're really counting is the number of different subsets of 3 items we can make out of a total of 5 items: 5C3

nCk = n!/k!(n-k)! = 5!/3!2! = 5*4*3*2/3*2*2 = 20/2 = 10

So, the chance of getting exactly 3 2s in a 5 digit code (assuming the first digit can be 0) is:

5C3 * 1^3 * 9^2 / 10^5 = 810/100000 = 81/10000

and to expand it to any 3 identical digits we just multiply by 10 (since there are 10 different digits that could be triplicated) to get 81/1000.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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schumi_gmat: Master | Next Rank: 500 Posts; Posts: 377; Joined: Wed Sep 03, 2008 9:30 amFollowed by: 2 members; Thanked: 15 times

by schumi_gmat » Fri Sep 05, 2008 9:07 pm

Thanks for pointing out my mistake. But then i thought about it and have just one doubt.

We know that no of ways to select 3 2's is 81

Now we have to arrange the 5 digits with 3 2's, so it will be done in 5P5/3! and this is according to the resources on beat the gmat

https://regentsprep.org/regents/math/permut/LpermRep.htm

I am confused between what is correct because i think both ur solution and this link are trying to solve a similar problem. Can you please explain?
Thanks

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schumi_gmat: Master | Next Rank: 500 Posts; Posts: 377; Joined: Wed Sep 03, 2008 9:30 amFollowed by: 2 members; Thanked: 15 times

by schumi_gmat » Fri Sep 05, 2008 9:26 pm

I think i got my answer -

Here are the 5P5/3! = 20 combinations

2 2 2 x y
2 2 x 2 y
2 2 x y 2
2 x 2 y 2
2 x y 2 2
x 2 y 2 2
x y 2 2 2
y x 2 2 2
y 2 x 2 2
2 y x 2 2
2 y 2 x 2
2 2 y x 2
2 2 2 y x
x 2 2 2 y
y 2 2 2 x
2 x 2 2 y
2 y 2 2 x
x 2 2 y 2
y 2 2 x 2

Let me know if this is correct way of doing it. Hence now the answer is 81 * 20/ 10^5

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