# Problem Involving Combinatorics and Probability

• Bashyboy
In summary, the total number of possible PINs without restrictions is 10,000. However, there are restrictions on the choice of digits, including all digits being identical, consecutive ascending or descending digits, and any sequence starting with 19. This leaves a total of 9876 legitimate PINs. The probability of selecting a legitimate PIN is 0.9876. In the case where the first and last digits are 8 and 1, the probability of gaining access to the account is 0.0333. However, if the first and last digits are 1 and 8, the probability is 0.03.
Bashyboy

## Homework Statement

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 $2 \cdot 7$, 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 $10⋅1⋅1⋅1$, 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 $|R|=124$; and if N is the set that contains the non-restricted ones, meaning R and N are complementary sets, then $|N|=10,000−124=9876$. 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?

Bashyboy said:
For part b): The number of pins in either ascending or descending order is $2 \cdot 7$, 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 $10⋅1⋅1⋅1$, 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 $|R|=124$; and if N is the set that contains the non-restricted ones, meaning R and N are complementary sets, then $|N|=10,000−124=9876$. Hence, the probability is then P(N)=9786/10000=0.9876.
Yes, but you should note, as part of the reasoning, that there's no overlap between the prohibited sets.
For part c): The sample space, containing all of the outcomes of the experiment that will take place, is |N|=9870.
What are you counting as outcomes there? How do you get that number?
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?
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.

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, $1⋅10⋅10⋅1=100$ 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, $3/100=.03$.This is still wrong, though. The answer is .0333

Could my textbook possibly be wrong? It has been wrong on other occasions.

Bashyboy said:
Could my textbook possibly be wrong? It has been wrong on other occasions.
I agree with your answer. I would get the book answer if the 1st and last digits were 1 and 8 respectively.

Thank you so much for you help! I was able to solve the problem!

## 1. What is combinatorics and how is it related to probability?

Combinatorics is a branch of mathematics that deals with counting and arranging objects. It is closely related to probability because it helps us calculate the number of possible outcomes in a given situation, which is important in determining the likelihood of an event occurring.

## 2. What are the basic principles of combinatorics?

The basic principles of combinatorics are multiplication, addition, and subtraction. Multiplication is used when the events are independent, addition is used when the events are mutually exclusive, and subtraction is used when we want to exclude certain outcomes from our calculation.

## 3. How do we calculate permutations and combinations?

Permutations are the number of ways in which a set of objects can be arranged in a specific order. This can be calculated using the formula n!/(n-r)! where n is the total number of objects and r is the number of objects being arranged. Combinations, on the other hand, are the number of ways in which a subset of objects can be chosen from a larger set. This can be calculated using the formula n!/r!(n-r)!

## 4. Can combinatorics and probability be applied in real-life situations?

Yes, combinatorics and probability are used in a variety of real-life situations, such as in gambling, insurance, and sports. They can also be used to solve problems in fields like genetics, computer science, and economics.

## 5. What are some common misconceptions about combinatorics and probability?

Some common misconceptions about combinatorics and probability include thinking that the outcomes are always equally likely, assuming that past events can predict future outcomes, and believing that the more times an event occurs, the more likely it is to happen again. These misconceptions can lead to errors in calculations and understanding of probability and combinatorics.

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