Question 1089658: Someone has stolen a bank card and wants to use it at an ATM. PINS are designed to contain four numbers (0 through 9). ATM machines are designed in a way that after the first three attempts to log in using a wrong PIN, a bank card stays with a machine.
What is the probability that a thief will guess correctly the PIN at his third attempt?
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website! I am interpreting your question to ask what the probability is that the thief guesses the PIN precisely on the 3rd guess.
Since there are 10000 PINs (0000, 0001, …, 9999) and he/she presumably guessed two other unique PINs already, the probability of guessing correctly on the 3rd attempt is:
 or about 0.00010002
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As an aside/related note:
If you wanted to know the probability that he/she guesses the PIN using at most 3 guesses, then that is
1 - P(no guesses are correct) = 1-P(first guess is incorrect)*P(2nd is incorrect)*P(3rd is incorrect)
= 1 - (9999/10000)(9998/9999)(9997/9998)
= 1 - (9997/10000)
= 3/10000
= 0.0003
Although I calculated it as pick1 followed by pick2, then pick3, notice how the result is equivalent to 3 picks out of 10000 (if there were 10000 slips of paper, each with a unique PIN written on it, you could take 3 out all at once), as you'd expect intuitively.
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