ID plates-- proability question
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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?
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1 digit IDnh8404052006 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?
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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OA : 1 - (9/10)^4
Please note the OA may be wrong.
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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Since 1 and 8 cannot apear in this shold this be3 digit ID
no. of ways 8 cannot appear = 9*9*9 = 729
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!
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
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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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.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?
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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Thanks Stuart for the crisp reply
I was thinking about that and said in my post:
THANKS
I was thinking about that and said in my post:
your answer is simple and removed certain doubtsThere has to be a smaller method:
I was unable to accommodate for 10000.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
THANKS
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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.
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.