Discussion
Sarah cannot completely remember her four-digit
ATM pin number. She does remember the first two digits, and she knows that
each of the last two digits is greater than 5. The ATM will allow her three
tries before it blocks further access. If she randomly guesses the last two
digits, what is the probability that she will get access to her account?
*This question is included in Nova Press: Set V - Probability & Statistics, question #3
| (A) | 1/2 |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 01/11/2012 09:49
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.
Posted: 01/11/2012 11:38
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
What you refer to is a different type of question -> She's allowed three tries but each must be unique.
Niels
Posted: 01/11/2012 13:33
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.
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.
Posted: 09/04/2013 11:46
Please take a look at the newly updated explanation. The answer is not as simple as adding 1/16 + 1/16 + 1/16 as it was originally explained.