SOLUTION: 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

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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