Question

If you had a 5 digit pin (numbers 0 - 9) with all combinations of digits being equally likely, what is the probability of guessing the pin if you try 3 times with a different pin each time?

Answer 1

I was thinking you would do
1 1 1
----------- + ----------- + ----------
10^5 10^5 - 1 10^5 - 2

But I'm not completely sure

Answer 2

since there are 10^5 total combinations and you use up one each time

Answer 3

\[\frac{ 10! }{5!*5!} = 36\]

Answer 4

There are 36 Combination

Answer 5

don't you have 10 different numbers to choose for each digit
so 10*10*10*10*10 combinations?

Answer 6

\[\binom{10}{5} = \frac{10!}{5!*5!}\]

Answer 7

what?!! @Yahoo!
u have 3 trial, in first case, if u guess it correctly, its 1/(10^5)
in 2nd trial, U HAVE NOT GUESSED IT CORRECTLY, so (1-1/10^5)*(1/(10^5-1))

Answer 8

U have 10 Digit and u r taking 5 at a time

Answer 9

in third trial , u have not guessed it correctly both times , so (1-1/10^5)*(1-1/(10^5-1)*(1/(10^5-2)))

Answer 10

add all these 3

Answer 11

where does this come from (1-1/10^5)?

Answer 12

u will try for 2nd time only if your first trial was unsuccessful
so
unsuccessful = 1-1/10^5

Answer 13

to guess it correct 2nd time u need to multiply that by 1/(10^5-1)

Answer 14

what is that 1 from in the 1-(1/10^5)?

Answer 15

(1st guess correct) OR (1st guess incorrect AND 2nd guess correct) OR (1st guess incorrect AND 2nd guess incorrect AND 3rd guess correct)

Answer 16

correct = p, then not correct = 1-p

Answer 17

what is the p?

Answer 18

but r u sure about 10^5

Answer 19

probability of guessing correct

Answer 20

I am pretty sure. There was a similar example in my text

Answer 21

p=1/10^5

Answer 22

what if you just assume the second one is your first guess with just 1 less answer
and just find the probability of 1/(10^5 - 1)

Answer 23

but why would u take a 2nd guess if your 1st one is correct?

Answer 24

the question doesn't say that you stop when you get the right pin, only that you make 3 guesses

Answer 25

they just want to know, of the 3 guesses you made, how likely is it that one of those choices would be the correct pin

Answer 26

3 cases
(correct guess in 1st try itself) OR (correct guess in 2nd try only) OR (correct guess in 3rd try only)
which comes out to be
(1st guess correct) OR (1st guess incorrect AND 2nd guess correct) OR (1st guess incorrect AND 2nd guess incorrect AND 3rd guess correct)

Answer 27

notice 'one of the..'

Answer 28

only one must be correct

Answer 29

so for 2nd guess, 1st must be incorrect

Answer 30

well only 1 must be able to be correct since you need to choose different digits each time

Answer 31

assume that you make randomly make 3 different pins. What is the probability of 1 of those 3 different pins being the correct one?

Answer 32

1/3?

Answer 33

well there is no guarantee any of them are correct

Answer 34

3 choices out of 10^5 possible combinations

Answer 35

and each time you have one less option because they all must be different

Answer 36

i would still stick to my (1-...) thingy
different, thats why 10^5 in first denom, 10^5-1 in 2nd denom and 10^5-2 in 3rd denom...

Answer 37

I just don't really understand that 1- thingy

Answer 38

if it's there because you want to stop and check after each guess I don't think the question is asking to do that

Answer 39

(1st guess correct) OR (1st guess incorrect AND 2nd guess correct) OR (1st guess incorrect AND 2nd guess incorrect AND 3rd guess correct)
u didn't understand this, right ?

Answer 40

yeah ok I understand that part. Just not what you are putting it into cases at all. As far as I can see it want you to see what the probability is that of 3 randomly chosen pins, one of them matches up to the correct pin

Answer 41

ok,ok....when i solved such problems......i always did this....if the question is really just about guessing , then u are correct.....sorry for wasting your time...

Answer 42

no no, I appreciate your input, it really helps me find the correct solution.

Answer 43

I was never taught the 1- thingie yet so I don't think they would put it on an assignment as well