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Problem Involving Combinatorics and Probability

  1. Feb 3, 2013 #1
    1. The problem statement, all variables and given/known data
    The problem I am working on is:

    An ATM personal identification number (PIN) consists of four digits, each a 0, 1, 2, . . . 8, or 9, in succession.

    a.How many different possible PINs are there if there are no restrictions on the choice of digits?

    b.According to a representative at the author’s local branch of Chase Bank, there are in fact restrictions on the choice of digits. The following choices are prohibited: (i) all four digits identical (ii) sequences of consecutive ascending or descending digits, such as 6543 (iii) any sequence starting with 19 (birth years are too easy to guess). So if one of the PINs in (a) is randomly selected, what is the prob-ability that it will be a legitimate PIN (that is, not be one of the prohibited sequences)?

    c. Someone has stolen an ATM card and knows that the first and last digits of the PIN are 8 and 1, respectively. He has three tries before the card is retained by the ATM (but does not realize that). So he randomly selects the 2nd and 3rd digits for the first try, then randomly selects a different pair of digits for the second try, and yet another randomly selected pair of digits for the third try (the individual knows about the restrictions described in (b) so selects only from the legitimate possibilities). What is the probability that the individual gains access to the account?

    d.Recalculate the probability in (c) if the first and last digits are 1 and 1, respectively.


    2. Relevant equations



    3. The attempt at a solution

    For part a): The total number of pins without restrictions is 10,000

    For part b): The number of pins in either ascending or descending order is [itex]2 \cdot 7[/itex], because there are two choices (ascending or descending), and starting the pin with certain digits is prohibited--namely, 0,1, and 2 when you are descending, and 7,8, and 9 when you are ascending . The number of pins where each slot contains the same digit is [itex]10⋅1⋅1⋅1[/itex], because once the first digit is known there is only one option left to the rest of the slots. The number of pins that have their first and second slot occupied by 1 and 9, respectively, is 1⋅1⋅10⋅10⋅. So, if R is the set that contains these restricted pins, then [itex]|R|=124[/itex]; and if N is the set that contains the non-restricted ones, meaning R and N are complementary sets, then [itex]|N|=10,000−124=9876[/itex]. Hence, the probability is then P(N)=9786/10000=0.9876.

    I am currently stuck on part c). For part c): The sample space, containing all of the outcomes of the experiment that will take place, is |N|=9870. When it says that the thief won't use the same pair of digits in each try, does that not allow him trying the pin 8 5 2 1 in one try and the pin 8 2 5 1 in another try?
     
  2. jcsd
  3. Feb 3, 2013 #2

    haruspex

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    Yes, but you should note, as part of the reasoning, that there's no overlap between the prohibited sets.
    What are you counting as outcomes there? How do you get that number?
    I'm pretty sure they mean he won't try the same ordered pair, so he may well try the same pair in the other order. But I don't think it will make any difference to the answer.
     
  4. Feb 3, 2013 #3
    Yes, the set that contains the prohibited pin numbers is the complement of the set that contains the non-prohibited numbers, and vice-versa. For part c), completely disregard what I said; I see now where I went wrong.

    Here is my second attempt at part c):

    Since there are no restrictions on the digits we can use to fill slot 2 and 3 of the 4-digit pin, and we know the first and last, [itex]1⋅10⋅10⋅1=100[/itex] is the number of pins containing 8 as its first digit and 1 as its last digit. We can also think of this number as 100 different trys, but the thief only has 3; that is, he can only use 3 different pin numbers of the 100. Hence, [itex]3/100=.03[/itex].This is still wrong, though. The answer is .0333
     
  5. Feb 3, 2013 #4
    Could my textbook possibly be wrong? It has been wrong on other occasions.
     
  6. Feb 3, 2013 #5

    haruspex

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    I agree with your answer. I would get the book answer if the 1st and last digits were 1 and 8 respectively.
     
  7. Feb 3, 2013 #6
    Thank you so much for you help! I was able to solve the problem!
     
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