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lopus Group TitleBest ResponseYou've already chosen the best response.0
http://www.khanacademy.org/math/probability/v/probabilityusingcombinations
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Cinar, can you do this?
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
that is total only..
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
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?
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
10^52+10^36+10^36
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
listen I'll be right back ok, give me couple minutes, I need to do something..
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
\( (26 +26)^{10}\) Have no digit \( (26+10)^{10} \) Have no lower case \((26+10)^{10} \) Have no upper case
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
you'are right, I made a typo..
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
So we have \((10+26+26)^{10}  ((10+26)^{10} + (10+26)^{10} + (26+26)^{10}  (10^{10} + 26^{10} + 26^{10})) \)
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Do you understand how I did that?
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
it is like union of three set..
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Are you doing probability with set theory?
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
Discreet math
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
So do you or don't you understand how I got what I got?
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
so, AUBUC=A+B+CAIBAICBIC+AIBIC I=intersection
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
how about part b, any idea..
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Do you have any ideas where to start?
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
preceded by a letter means a1 or 1a
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
I am confused..
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
I'm pretty sure it means a1
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
hmm, then password cannot be begun by a number..
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Well, one thing to ask is, must all digits be preceded by a number, or just at least one digit?
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
and also no number is successive
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
22 or 555 not allowed
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
a2ddd3y8jj8i898 is ok
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
so 1a23456789 would be ok too.
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
that' why I am confused..
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
listen I need some break to eat something..
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
So one way is 1 digit and then 9 letters, with at least one capital and one lowercase
 2 years ago

wio Group TitleBest ResponseYou've already chosen the best response.1
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)
 2 years ago
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