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Topic: zeroless number (Read 1707 times) |
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Christine
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zeroless number
« on: Aug 19th, 2012, 1:04pm » |
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What is the probability that a n-digit number will not have the digit 0 ? Is it (9/10)^(n-1) ?
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towr
wu::riddles Moderator Uberpuzzler
    
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Re: zeroless number
« Reply #1 on: Aug 19th, 2012, 10:44pm » |
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It is if n > 1 and you don't allow leading zeroes; it's not true for single digit numbers because there the probability is 9/10 instead of 1.
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playful
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Re: zeroless number
« Reply #2 on: Aug 20th, 2012, 2:12pm » |
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Unless I'm missing something, the exponent is n (rather than n-1). 1 digit => p = 0.9 ^ 1 2 digits => p = 0.9 ^ 2
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peoplepower
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Re: zeroless number
« Reply #3 on: Aug 20th, 2012, 4:44pm » |
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on Aug 20th, 2012, 2:12pm, playful wrote:Unless I'm missing something, the exponent is n (rather than n-1). 1 digit => p = 0.9 ^ 1 2 digits => p = 0.9 ^ 2 |
| If we were counting occurrences of any other digit than 0, then you would be correct. But, as towr pointed out, if we do not have leading zeros in our number then the first digit is certainly nonzero and the remaining n-1 digits have the 1/10 probability of being zero. The exception is the convention that the single digit number zero is written 0 rather than being treated as a zero-digit number, which is why when n=1 the probability is 9/10.
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« Last Edit: Aug 20th, 2012, 4:44pm by peoplepower » |
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playful
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Re: zeroless number
« Reply #4 on: Aug 20th, 2012, 5:18pm » |
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Ah... Yes, of course. We won't allow 07 etc. Thanks for waking me up.
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