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ID plates-- proability question

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Re: ID plates-- proability question

by parallel_chase » Sun Dec 28, 2008 3:55 pm
nh8404052006 wrote:A warehouse contains 10,000 gadgets, and each gadget is given an unique ID ranging from 1 to 10,000. What is the probability that at least one number 8 appears on thsoe ID plates?
1 digit ID

no. of ways 8 cannot appear = 9

2 digit ID
no. of ways 8 cannot appear = 9*9 = 81

3 digit ID
no. of ways 8 cannot appear = 9*9*9 = 729

4 digit ID
no. of ways 8 cannot appear = 9*9*9*9 = 6561

5 digit ID
no. of ways 8 cannot appear = 1*1*1*1 = 1 [this is because we have only 1 five digit number i.e. 10000]


9+81+729+6561+1 = 7381


Probability of not having 8 in any of the IDs = 7381/10000


Probability of having at least one 8 in the IDs = 1- 7381/10000 =
2619/10000

I am not sure if I have done this correctly. whats the OA?
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by nh8404052006 » Sun Dec 28, 2008 6:26 pm
OA : 1 - (9/10)^4

Please note the OA may be wrong.
Last edited by nh8404052006 on Sun Dec 28, 2008 8:09 pm, edited 1 time in total.
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by cramya » Sun Dec 28, 2008 7:52 pm
3 digit ID
no. of ways 8 cannot appear = 9*9*9 = 729
Since 1 and 8 cannot apear in this shold this be

8*9*9


Anyways I am getting nowghere close to OA also.

Atleast one 8

1 digit no

1

2 digit no

1 10 + 9 1 = 19

3 digit no

1 10 10 + 9 1 10 + 9 10 1 = 280

4 digit no

1 10 10 10 + 9 1 10 10 + 9 10 1 10 + 9 10 10 1 = 3700

P = 4000/10000 = 2/5

I missed something also!
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by mental » Mon Dec 29, 2008 12:02 am
I am getting 1 - (9/10)^4

There has to be a smaller method:

If the num is single digit: it can be any thing from 1 ~ 9 including 8
if we exclude 8, it will be 8 choices

double digit: it will be from 10~99. Tens digit 1~9, units 0~9.
we exclude digit 8, it will be 8*9 choices

triple digit: it will be 100~999
we exclude digit 8, it will be 8*9*9 choices

four digits: it will be 1000~9999
we exclude digit 8, it will be 8*9*9*9 choices

Five digits: only 1 choice: 10000

Total choices without digit 8 = 8 + 8*9 + 8*9*9 + 8*9*9*9 + 1
rearrange = 1 + 8 + 8*9 + 8*9*9 + 8*9*9*9
= 9 + 8*9 + 8*9*9 + 8*9*9*9
= 9(1 + 8 + 8*9 + 8*9*9) = 9(9 + 8*9 + 8*9*9)
= 9*9(1 + 8 + 8*9) = 9*9(9 + 8*9) = 9*9*9*9 = 9^4

total choices of numbers inclusive of digit 8 = 1~10000 = 10000 = 10^4

ways a number can be chosen without 8 = 9^4/10^4 = (9/10)^4

with atleast one 8 = 1 - (9/10)^4
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by cramya » Mon Dec 29, 2008 12:06 am
Mental,
Good solution.

I should have taken the 1- p(none) approach also and for some reason started of with finding the 8 numbers.

Regards,
Cramya
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by nh8404052006 » Mon Dec 29, 2008 2:34 am
Mental, I have just only one word: FASINATING!
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Re: ID plates-- proability question

by Stuart@KaplanGMAT » Mon Dec 29, 2008 2:51 am
nh8404052006 wrote:A warehouse contains 10,000 gadgets, and each gadget is given an unique ID ranging from 1 to 10,000. What is the probability that at least one number 8 appears on thsoe ID plates?
What's the source of this question? In future, please always cite the source and provide answer choices. If it's not a GMAT style multiple choice question, then please let us know that too, since there's an excellent chance that the question is irrelevant to the GMAT.

This particular question would never appear as written, since the probability that at least one number 8 appears on those ID plates is 100%, since all 10000 have to have unique ID numbers. I'm guessing the question that you want us to address is:

"If a plate is chosen at random, what's the probability that it contains at least one 8?"

The best way to approach is definitely with the "one minus" method, using our old buddy equation:

Prob Want = 1 - (Prob don't want)

Here, we don't want a plate with no 8s.

Pretending that 10000 is just 0000 (since there's no "0000" it doesn't hurt us to do so), we see that there are 4 digits in each code (i.e. 0000 to 9999... we're looking at "1" as "0001", and so on; it's safe to add "0"s since we don't want those anyway). The different ways to get no "8"s is:

9*9*9*9 = 9^4

Since the total number of possibilities is 10^4, the probability of getting at least one 8 is, indeed:

1 - 9^4/10^4 = 1- (9/10)^4
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by mental » Mon Dec 29, 2008 2:58 am
Thanks Stuart for the crisp reply

I was thinking about that and said in my post:
There has to be a smaller method:
your answer is simple and removed certain doubts
Pretending that 10000 is just 0000 (since there's no "0000" it doesn't hurt us to do so), we see that there are 4 digits in each code (i.e. 0000 to 9999... we're looking at "1" as "0001", and so on; it's safe to add "0"s since we don't want those anyway). The different ways to get no "8"s is:
9*9*9*9 = 9^4
I was unable to accommodate for 10000.
THANKS
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by nh8404052006 » Mon Dec 29, 2008 10:14 pm
Hi, Stuart

Thank. your answer is very concise and brilliant

Source: probability_note.doc

The question is not a multiple choice question.
The author of the document just writes down the question and the answer without any other details.
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