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3 digits

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by shahdevine » Sat Aug 29, 2009 6:52 pm
What is the probability that someone writing a 3-digit number will write one that is divisible by ten and has no digit used more than once?

a) 9/125
b) 2/25
c) 1/10
d) 18/25
e) 4/5

oa after some discussion
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by PussInBoots » Sat Aug 29, 2009 7:38 pm
Total 3-digit numbers: 9 * 10 * 10 = 900

3-digit numbers that satisfy the problem: 9 * 8 * 1 = 72

72 / 900 = 2/25
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by shahdevine » Sat Aug 29, 2009 8:03 pm
PussInBoots wrote:Total 3-digit numbers: 9 * 10 * 10 = 900

3-digit numbers that satisfy the problem: 9 * 8 * 1 = 72

72 / 900 = 2/25
thx...here are some stupid questions:

1) why is total of 3 digit numbers 9x10x10? is it because you have to discount the possibility of farthest left digit column having zeroes?

2) how did you find the 3-digit numbers that satisfy the problem? I get lost in the language of "has no digit used more than once". Could you explain what that exactly means and give permutation combination translation?

3) it seems probability problems are actually combination problems divided into two parts via numerator/denominator? is this the most consistent strategy to get all these types of problems?

thanks.
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by bharathh » Sat Aug 29, 2009 8:09 pm
I get 1/10

Here's what I did.

There are 3 digits with no repetitions... So max numbers possible = 10.9.8

The number has to be divisible by 10. So the last digit should always be 0. Also 0 cannot be used for the first or the second digit ... So there are 9.8 possibilities.

Probability = 9.8/10.9.8 = 1/10
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by shahdevine » Sat Aug 29, 2009 8:19 pm
oa is b
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by bharathh » Sat Aug 29, 2009 8:25 pm
Ah k... I see my mistake now.

I calculated total number of 3 digit numbers that have no numbers repeated... and not total # of 3 digit numbers.

In fact you don't even have to calculate the total number of possibilties... from 100 to 999 inclusive there are 900 numbers.

The second part is correct though.
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