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.
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?