A password is made from 6 characters , 4 of them are digits (0-9) and 2 are letters a-z (26 letters only lower case) each digit and letter can appear once at most.
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Someone choose a random password he remembers the letters and digits that make his password however he does not remember their order , what is the probability of getting it right in a single try?
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He remembers that the letters were used as the first 2 characters but does not remember the letters and digits what is the probability of getting it right in a single try?
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what is the probability of the password having the letter a and at least one of the digits 1,2?
I will start with "2" because I might have got it right ( not sure if it is right or just a coincidence ) if the first 2 characters are letters then we have $26*25*10*9*8*7$ and since we want him to get it right from a single try we will do $\frac 1 {26*25*10*9*8*7}$ $=$ $3.05*10^{-7}$ which is right according to the book but I am not sure if it is a coincidence.
for "1" since he knows the digits and letters he used we only need to sort them so my thought would be $6!$ and divide by $26*25*10*9*8*7$ but it is not right so I tried also sorting the digits first $4!$ and then the letters so we get $4!*2!$ and again divide by $26*25*10*9*8*7$ but again it is not right , then tried picking the digits first ${6 \choose 4}$ and multiply by the 2 characters left ${2 \choose 2}$ and divide by $26*25*10*9*8*7$ which is also not right.. the answer should be $1.4*10^{-3}$
And for "3" I thought about using Complement so my try was $1-$ $\frac {{6 \choose 4}8*2} {26*25*10*9*8*7}$ but I am not even close as the answer should be $0.051$
Thank you for any help , tips and hints all is appreciated!
Best Answer
Your answer and reasoning for problem 2 are both correct.
Since he knows the letters and digits used in the password, the only thing he needs to guess is the order in which they appear. There are $6!$ possible orders in which six distinct characters could appear, of which only one is correct. Hence, the probability that he guesses correctly the first time is $$\frac{1}{6!} = \frac{1}{720}$$
There are $\binom{26}{2}$ ways to select two distinct letters, $\binom{10}{4}$ ways to select four distinct digits, and $6!$ ways to arrange six distinct characters, so there are $$\binom{26}{2}\binom{10}{4}6!$$ possible passwords.
We will count the favorable cases using the Inclusion-Exclusion Principle.
The number of passwords which contain the letter a and the digit 1 is $$\binom{1}{1}\binom{25}{1}\binom{1}{1}\binom{9}{3}6! = \binom{25}{1}\binom{9}{3}6!$$ since we must choose the letter a, one of the other $25$ letters, the digit 1, and three of the other nine digits, then arrange the six distinct characters.
By symmetry, the number of passwords which contain the letter a and the digit 2 is also $$\binom{25}{1}\binom{9}{3}6!$$
However, if we add these two cases, we will have counted those cases in which the password contains the letter a and both the digits 1, 2 twice, once when we counted passwords containing the letter a and the digit 1 and once when we counted passwords containing the letter a and the digit 2. We only want to count them once, so we must subtract them from the total.
The number of passwords which contain the letter a and both the digits 1 and 2 is $$\binom{1}{1}\binom{25}{1}\binom{2}{2}\binom{8}{2}6! = \binom{25}{1}\binom{8}{2}6!$$ since we must select the letter a, one of the other $25$ letters, both the digits 1 and 2, and two of the other eight digits, then arrange the six distinct characters.
By the Inclusion-Exclusion Principle, the number of favorable passwords is $$\binom{25}{1}\binom{9}{3}6! + \binom{25}{1}\binom{9}{3}6! - \binom{25}{1}\binom{8}{2}6!$$ so the probability that an admissible password will contain the letter a and at least one of the digits 1, 2 is $$\frac{2\dbinom{25}{1}\dbinom{9}{3}6! - \dbinom{25}{1}\dbinom{8}{2}6!}{\dbinom{26}{2}\dbinom{10}{4}6!} = \frac{2\dbinom{25}{1}\dbinom{9}{3} - \dbinom{25}{1}\dbinom{8}{2}}{\dbinom{26}{2}\dbinom{10}{4}}$$ Note that the answer depends only on which characters appear in the password, not the order in which they appear.