MHB What is the probability of a password containing exactly 1 or 2 integers?

[iamlorde]

Greetings! Can someone offer help?

A computer system uses passwords that contain exactly
eight characters, and each character is one of the 26 lowercase
letters (a–z) or 26 uppercase letters (A–Z) or 10 integers (0–9).
Assume all passwords are equally likely. Let A and B denote
the events that consist of passwords with only letters or only
integers, respectively. Determine the following probability:

P (Password contains exactly 1 or 2 integers)

I proceeded like so:

$$

\frac{{52}^{7}\cdot10}{{62}^{8}}\ + \frac{{52}^{6}\cdot{10}^{2}}{{62}^{8}}\ = 0.0561

$$

The answer is supposed to be: 0.630, but I have no idea how to reach it.

 

[Jameson]

Hi iamlorde,

Welcome to MHB! :)

Is this the whole problem? I ask because the events A and B are introduced but not used in the question you pose.

Assuming that you can reuse any character or letter if need be, which is consistent with how real passwords work, then I agree with your answer and your method.

 

[iamlorde]

Jameson, thank you for your reply. I hadn't given the rest because I had answered them already. Here is the entirety.

A computer system uses passwords that contain exactly
eight characters, and each character is one of the 26 lowercase
letters (a–z) or 26 uppercase letters (A–Z) or 10 integers (0–9).
Assume all passwords are equally likely. Let A and B denote
the events that consist of passwords with only letters or only
integers, respectively. Determine the following probabilities:
(a) P(A∪ B) (b) P(A′∪ B)
(c) P (Password contains exactly 1 or 2 integers)

(a) $$
P(A \cup B) =

\frac{{52}^{8}}{{62}^{8}}\ + \frac{{10}^{8}}{{62}^{8}}\ = 0.2448
$$

(b) $$
P(A' \cup B) = \frac{{62}^{8} - {52}^{8}}{{62}^{8}}\ = 0.7552
$$

 

[iamlorde]

The following is my visualization of the problem (c).

At this point I'm acknowledging that set (Password contains exactly 1 or 2 integers) is exclusive from set A & B.

Yet I do not know how to quantitatively approach this problem.

Thank you.

 

[Fallen Angel]

Hi,

I think you are missing the position where the integers appear, count it and then the result is 0.6302

 

[iamlorde]

Hi,

I think you are missing the position where the integers appear, count it and then the result is 0.6302

I'm sorry but I still can't get the answer.
My logic is as follow:

P[(passwords with exactly 1 integer) OR (passwords with exactly 2 integer)]

Therefore:

[A-Za-z]*[A-Za-z]*[A-Za-z]*[A-Za-z]*[A-Za-z]*[A-Za-z]*[A-Za-z]*[0-9]
or
[52]*[52]*[52]*[52]*[52]*[52]*[52]*[10] so: 52^7 * 10 over 62^8

then:

[A-Za-z]*[A-Za-z]*[A-Za-z]*[A-Za-z]*[A-Za-z]*[A-Za-z]*[0-9]*[0-9]
or
[52]*[52]*[52]*[52]*[52]*[52]*[10]*[10] so: 52^6 * 10^2 over 62^8Then adding both results, would theoretically give the answer. But my logic is flawed and I can't seem to understand why.

 

[Fallen Angel]

Hi,

The point of my previous answer is that the integer doesn't need to be in last position.You got that there are $\dfrac{52^7\times 10}{62^8}$ passwords with the last character being an integer, so in order to count passwords with exactly one integer we have to multiply it by $8$.

Now for counting passwords with exactly two integers we got $\dfrac{52^7\times 10^2}{62^8}$ where integers are in positions one and two. Now multyplying by $\dfrac{8\times 7}{2}$ we got all possible possitions for the two integers, adding it we reach that $P(...)=0.6302$.

Note:There is a two in the denominator because having integers in position one and two is the same as having integers in positions two and one.