Discussion
[This problem requires a calculator.]
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 Math - Problem Set AA: Probability & Statistics, question #3
| (A) | 3/16 |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 08/04/2012 09:23
I am have trouble with the finally logic in the solution. I understand that there are 16 possible number combinations she could select on the first try. However after the first try wouldn't the probability go to 1/15, and then 1/14 for the 3rd try. Which means her probability would be 1/16 + 1/15 + 1/14? This is approx. 20%. Not the 18% from 3/16. Please help my logic.
Posted: 08/04/2012 13:12
Eddie, your logic would be correct; however the problem usually would explicitly say "without replacement, or something like: "she randomly
guesses, and remembers the digit combination that she already tries so that she doesn't try them the 2nd time"
guesses, and remembers the digit combination that she already tries so that she doesn't try them the 2nd time"
Posted: 09/24/2015 21:45
Reply: so you mean,she probably tries a same digit combination three times.
Posted: 08/05/2012 12:18
So the problem is slightly unrealistic because who would randomly select numbers without keeping track of which combinations were chosen each try. :) thanks for your timely response.
Posted: 07/25/2013 10:37
I'm not understanding the first part upto 1/16. Will you please explain sir? Thank you in advance!
Posted: 07/25/2013 16:48
Chandra, Sarah randomly guesses the last two digits. She can use 6, 7, 8, or 9. So we have (6,7), (6,8), (6,9) ... (9,9) -> it is a matrix of 16 choices, so each try has a probability of 1/16 of being correct. Since it is a random guess, Sarah does not remember the combination that she chose the first time (each guess is an independent event, statistically speaking). Then the probability of getting it right within 3 tries would be = P (get it right 1st time) + P (fail 1st time and right 2nd time) + P (fail 1st time, fail 2nd time, and right 3rd time)
= 1/16 + (15/16 * 1/16) + (15/16 * 15/16 * 1/16) = .176...
Another way to think of it: what is the probability of not failing in 3 tries? Since the max probability = 1, P (not failing in 3 tries) = 1 - (15/16 * 15/16 * 15/16) = .176...
A commenter (Eddie Kiep) above pointed out that realistically, Sarah should remember what she had punched in the first and second time if she did not get it right the first or second time. Hence there would be 15 combinations left in the second try, and 14 in the 3rd try. If the problem is phrased that way, then the probability of her getting it right within 3 tries = 1/16 + 1/15 + 1/14.
= 1/16 + (15/16 * 1/16) + (15/16 * 15/16 * 1/16) = .176...
Another way to think of it: what is the probability of not failing in 3 tries? Since the max probability = 1, P (not failing in 3 tries) = 1 - (15/16 * 15/16 * 15/16) = .176...
A commenter (Eddie Kiep) above pointed out that realistically, Sarah should remember what she had punched in the first and second time if she did not get it right the first or second time. Hence there would be 15 combinations left in the second try, and 14 in the 3rd try. If the problem is phrased that way, then the probability of her getting it right within 3 tries = 1/16 + 1/15 + 1/14.
Posted: 07/28/2013 17:31
i understand that we have to add the probabilities that she will get it right on the 1st try, 2nd try & 3rd try. but to get P(gets it on 2nd try), don't we need to multiply the probability that she will NOT get it on the 1st try by the probability that she will on the 2nd try? because her "access" to the 2nd try depends on her not getting it on the 1st try. likewise for getting it on the 3rd try. thanks.
Posted: 08/20/2013 01:50
I think there is a common mistake made by the contributor. If we have n chances to test, and the possibility of succeeding of each test is p, than it's false to say the overall possibility is np. Let's say Sarah has 20 chance. It would be ridiculous to say she have 125% of possibility to win the game, actually, even given she has 20 chance, it's still possible that she is so unlucky that she fails to access her account. The correct way of working out this problem is:
Sarah wins the game at the first attempt, the possibility is 1/16.
The possibility she wins at the second attempt is 15/16*1/16.
The ... at third attempt is (1-1/16-15/16*1/16)*1/16.
The overall result sums up all the three possibility above.
Sarah wins the game at the first attempt, the possibility is 1/16.
The possibility she wins at the second attempt is 15/16*1/16.
The ... at third attempt is (1-1/16-15/16*1/16)*1/16.
The overall result sums up all the three possibility above.
Posted: 08/30/2013 19:48
Elmar Chen, thank you for the correction. A minor note on English, it's probability and not possibility. Possibility is binary: 1 or 0, yes or no. So if you use possibility, then in this case it's 1. Thanks again.
Posted: 08/30/2013 14:55
I don't think this is correct. By the logic of the answer, if she had 16 tries, the probability that Sarah picks the right combination is 1. But there is replacement! There's a good chance that, since each guess is random, she'll pick a combination twice. Therefore there's a small chance that she'll miss the correct combination even after 16 tries. Correct answer is 1-(15/16)(15/16)(15/16). That is one minus the probability of getting it wrong three times. It's a little less than 3/16.
Posted: 08/30/2013 20:31
Patrick, you are right that the explanation is also incorrect. Your approach and Elmar's approach yield the same result.
P(make 1st) +
P(fail 1st and make 2nd) +
P(fail 1st, fail 2nd, and make 3rd)
=
1-P(fail1st, fail 2nd and fail3rd)
P(make 1st) +
P(fail 1st and make 2nd) +
P(fail 1st, fail 2nd, and make 3rd)
=
1-P(fail1st, fail 2nd and fail3rd)
Posted: 08/30/2014 19:54
I don't think this is correct. When the first try is incorrect, there are just 15 tries. And the next probability should be 1/15 if we succeed for two tries. So the final probability is 1/16+(15/16)(1/15)+(15/16)(14/15)(1/14).
Posted: 08/05/2015 01:10
I don't understand why the probability of choosing a combination for the second trial is remaining the same. You have 16 combinations in total. Let's say that she missed the first trial--> she still have 15 more combinations from which she can choose (I mean no one is that stupid to choose the same wrong combination again). This means that the probability of entering the wrong pin in the first trial but the right one in the second trial must be (15/16)(1/15)