If she selects 66 for her first choice, then doesn't she really only have 1/15 choices for her next choice? 1/14 for her final? She wont pick the same number twice.

Hello Chris, it states random choices, there might be a button next to our new 2 digit ATM which picks any of the 4^2 possibilities. In this situation each try is equal to a 'last try' / 'random event' -> 1/16.
What you refer to is a different type of question -> She's allowed three tries but each must be unique.

Niels

The number is greater than 5, so the digits must be from 6,7,8,9. We have 16 choices: (6,6), (6,7), (6,8), (6,9), (7,6), (7,7), (7,8), (7,9), (8,6), (8,7), (8,8), (8,9), (9,6), (9,7), (9,8), (9,9).

Randomly, each try has 1/16 probability of being the right one. As Niels pointed out, the keyword is "random guess" for each try. BUT, I agree with Chris that the question is not well written for the real world. In the real world, if Sarah is not an airhead, the probability is 1/16 + 1/15 + 1/14. In fact, in the real world, Sarah would stop after the 2nd try, because she does not want to be locked out of her account, hence the probability is only 1/16 + 1/15. But that would be too advanced for the GMAT.

In fact, Niels came up with a perfect explanation: Sarah had a little too much Heineken, that's why she couldn't remember the last two digits, and that's why she forgot what she had used for the last try as soon as she tried the next one.