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lopusBest ResponseYou've already chosen the best response.0
http://www.khanacademy.org/math/probability/v/probabilityusingcombinations
 one year ago

wioBest 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.
 one year ago

wioBest 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?
 one year ago

cinarBest ResponseYou've already chosen the best response.0
listen I'll be right back ok, give me couple minutes, I need to do something..
 one year ago

wioBest 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
 one year ago

wioBest 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
 one year ago

cinarBest ResponseYou've already chosen the best response.0
you'are right, I made a typo..
 one year ago

wioBest ResponseYou've already chosen the best response.1
So we have \((10+26+26)^{10}  ((10+26)^{10} + (10+26)^{10} + (26+26)^{10}  (10^{10} + 26^{10} + 26^{10})) \)
 one year ago

wioBest ResponseYou've already chosen the best response.1
Do you understand how I did that?
 one year ago

cinarBest ResponseYou've already chosen the best response.0
it is like union of three set..
 one year ago

wioBest ResponseYou've already chosen the best response.1
Are you doing probability with set theory?
 one year ago

wioBest ResponseYou've already chosen the best response.1
So do you or don't you understand how I got what I got?
 one year ago

cinarBest ResponseYou've already chosen the best response.0
so, AUBUC=A+B+CAIBAICBIC+AIBIC I=intersection
 one year ago

cinarBest ResponseYou've already chosen the best response.0
how about part b, any idea..
 one year ago

wioBest ResponseYou've already chosen the best response.1
Do you have any ideas where to start?
 one year ago

cinarBest ResponseYou've already chosen the best response.0
preceded by a letter means a1 or 1a
 one year ago

wioBest ResponseYou've already chosen the best response.1
I'm pretty sure it means a1
 one year ago

cinarBest ResponseYou've already chosen the best response.0
hmm, then password cannot be begun by a number..
 one year ago

wioBest 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?
 one year ago

cinarBest ResponseYou've already chosen the best response.0
and also no number is successive
 one year ago

wioBest 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.
 one year ago

wioBest ResponseYou've already chosen the best response.1
so 1a23456789 would be ok too.
 one year ago

cinarBest ResponseYou've already chosen the best response.0
that' why I am confused..
 one year ago

cinarBest ResponseYou've already chosen the best response.0
listen I need some break to eat something..
 one year ago

wioBest 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.
 one year ago

wioBest 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
 one year ago

wioBest 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)
 one year ago
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