How Many Possible Passwords Meet Specific Criteria in a Bank Security System?

[adcruze]

Hello, I'm a science student, trying to solve question regarding probability...can anyone help me...thanks!

A bank security system uses password that consist of five letters followed by a single digit.
1. How many passwords are possible?
2.How many passwords consists of three A's and two B's and end in an even digit?
3.If you forget your password but remember that it has the characteristics described in part (2), what is the probability that you will guess the password correctly on the first attempt?

 

[berkeman]

Welcome to the PF, adcruze. I've moved your post into a Homework Help forum -- the Tutorials forum that you initially posted in is for tutorials that explain how to do something, not the original question.

Now to your questions. Could you please show us some of your own work, so that we can provide tutorial help? Like on question -1-, what do you think the relevant equation is that you will use to solve it?

 

[EugP]

I will give you a small hint.
Think of each letter in the password as a place holder. So in your case you have five place holders:

[1][2][3][4][5]

Now you know that each place holder has to have a letter. Assuming the passowrd is in english the answer is 26. So there can be one of 26 letters in each placeholder.

You should be able to take it from here.

Good Luck!

 

[disregardthat]

Sorry... I thought this thread was old, but it wasn't :S

If the thread starter isn't answering may I ask for the answers?

 

[berkeman]

Sorry... I thought this thread was old, but it wasn't :S

If the thread starter isn't answering may I ask for the answers?

No worries. If you want to post a similar problem and offer your answer, we can let you know if you understand the principle or not.

 

[disregardthat]

Ok, let's say a password consists of three letters followed by two digits. Assume the english number of letters.
How many different passwords can you produce?
Would the answer be: 26^3*10^2?
it consists of two a's and two b's and the two digits are uneven.
would the answer be: (3*2)(3)(5^2) = 450
and the probability for choosing the right password if you know that it has the same characteristics as in the previous question is 450/(26^3*10^2)=45/(10*26^3)?

 

[berkeman]

Ok, let's say a password consists of three letters followed by two digits. Assume the english number of letters.
How many different passwords can you produce?
Would the answer be: 26^3*10^2?

Yes.

it consists of two a's and two b's and the two digits are uneven.
would the answer be: (3*2)(3)(5^2) = 450

I'm not sure I understand what this problem statement means. What is the 2 a's and 2 b's part?

 

[EugP]

Ok, let's say a password consists of three letters followed by two digits. Assume the english number of letters.
How many different passwords can you produce?
Would the answer be: 26^3*10^2?
it consists of two a's and two b's and the two digits are uneven.
would the answer be: (3*2)(3)(5^2) = 450
and the probability for choosing the right password if you know that it has the same characteristics as in the previous question is 450/(26^3*10^2)=45/(10*26^3)?

If your password has 3 letters and 2 digits, how can you have 2 a's and 2 b's?

 

[berkeman]

I think they're separate questions, but I'm confused as well.

 

[disregardthat]

Sorry, i meant two a's and one b.
And as it referred to the thread starters second question, i meant: "how many passwords can be produced with these characteristics?" I will be more clear next time.

 

[berkeman]

Sorry, i meant two a's and one b.

So the question and answer now are:

it consists of two a's and one b, and the two digits are uneven.
would the answer be: (3*2)(3)(5^2) = 450

So you mean a password with three letters and two digits, where the 3 letters are the different combinations of 2 a's and one b, and the two digits are only odd?

Well, the two odd digits matches the 5^2 part of your answer. For the combinations of 2 a's and 1 b:

aab
aba
baa

Seems like the 1 b can only be in 3 places... how did you get the (3*2)(3) part? Sorry if I'm still not understanding something.

 

[disregardthat]

Yeah, i messed up, i see this now.. OF course x|
well, how about 3*5^2

 

[berkeman]

Works for me!