anonymous
  • anonymous
password question, need some help..
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
1 Attachment
lopus
  • lopus
http://www.khanacademy.org/math/probability/v/probability-using-combinations
anonymous
  • anonymous
a) Let's try first by calculating the total number of passwords without requirements, and then subtracting the total number of passwords which DON'T meet the requirements.

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anonymous
  • anonymous
Cinar, can you do this?
anonymous
  • anonymous
62^10
anonymous
  • anonymous
that is total only..
anonymous
  • anonymous
Now how many don't meet the 'at least one digit' rule, how many don't meet the 'at least one lower case letter', and how many don't meet the 'at least one upper case letter' requirement?
anonymous
  • anonymous
10^52+10^36+10^36
anonymous
  • anonymous
listen I'll be right back ok, give me couple minutes, I need to do something..
anonymous
  • anonymous
\( (26 +26)^{10}\) Have no digit \( (26+10)^{10} \) Have no lower case \((26+10)^{10} \) Have no upper case
anonymous
  • anonymous
We still double count the ones who break two or more rules, so we have to get rid of them. \(10^{10} \) break both upper and lower case rule \(26^{10} \) break both upper case and digit rule \(26^{10} \) break both lower case and digit rule \(0\) break all three rules, because it must contain at least a digit or letter
anonymous
  • anonymous
you'are right, I made a typo..
anonymous
  • anonymous
So we have \((10+26+26)^{10} - ((10+26)^{10} + (10+26)^{10} + (26+26)^{10} - (10^{10} + 26^{10} + 26^{10})) \)
anonymous
  • anonymous
Do you understand how I did that?
anonymous
  • anonymous
hold on..
anonymous
  • anonymous
it is like union of three set..
anonymous
  • anonymous
Yeah
anonymous
  • anonymous
Are you doing probability with set theory?
anonymous
  • anonymous
Discreet math?
anonymous
  • anonymous
Discreet math
anonymous
  • anonymous
So do you or don't you understand how I got what I got?
anonymous
  • anonymous
so, AUBUC=A+B+C-AIB-AIC-BIC+AIBIC I=intersection
anonymous
  • anonymous
Yeah
anonymous
  • anonymous
got it..
anonymous
  • anonymous
how about part b, any idea..
anonymous
  • anonymous
Do you have any ideas where to start?
anonymous
  • anonymous
let see
anonymous
  • anonymous
preceded by a letter means a1 or 1a
anonymous
  • anonymous
I am confused..
anonymous
  • anonymous
I'm pretty sure it means a1
anonymous
  • anonymous
hmm, then password cannot be begun by a number..
anonymous
  • anonymous
Well, one thing to ask is, must all digits be preceded by a number, or just at least one digit?
anonymous
  • anonymous
and also no number is successive
anonymous
  • anonymous
22 or 555 not allowed
anonymous
  • anonymous
a2ddd3y8jj8i898 is ok
anonymous
  • anonymous
I think a123456789 is also okay, because it has one digit immediately preceded by a letter. They didn't say ALL must follow the rule, just one.
anonymous
  • anonymous
so 1a23456789 would be ok too.
anonymous
  • anonymous
that' why I am confused..
anonymous
  • anonymous
listen I need some break to eat something..
anonymous
  • anonymous
I am back..
anonymous
  • anonymous
Ok. So one thing we can take into consideration is that for a password to have a digit and be invalid, it would have to have to be 1... 11.... 111.... basically a bunch of digits in order with no letters in between.
anonymous
  • anonymous
So one way is 1 digit and then 9 letters, with at least one capital and one lowercase
anonymous
  • anonymous
So find how many ways we can have a password that is valid in (a) and not valid in (b) and subtract that from (a)

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