Probability of combination lock

In summary, we have a standard combination lock with a password consisting of 3 distinct numbers between 00 and 59. a) The number of valid passwords is calculated using the permutation formula, P(60,3). b) The probability that the password is a palindrome is 180 out of the total of 21600 possibilities. c) To calculate the probability of the product of the three numbers being a factor of 5, we can use the compliment principle and subtract the number of possibilities that are not a factor of 5 from the total number of possibilities, P(48,3). d) If we interpret the question as an arithmetic sequence of 3 two-digit numbers, the total number of possible sequences is 21600
  • #1
stunner5000pt
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2

Homework Statement


a standard combination lock consists of a password which is 3 distinct numbers between 00 and 59. a) compute the number of valid passwords
b) compute the probability that the password is a palindrome. eg. 03 - 22 - 30 is a palindrome (same backward as it is forward)
c) the product of the three numbers is a factor of 5
d) the password forms an arithmetic sequence with difference k>0



2. The attempt at a solution

a) Valid passwords = P(60,3) (since none can be repeated)
b) This is possible only if the first number is the first digit is 0 to 5 - 6 possilbities, however, we cannot use the same number twice - can't use 00 in the first and last place so only 5
Suppose the first digit is zero such as 00, the middle number must be a number whose units and tens place are identical such as 11, 22, 33 ... 55- 6 possibilities.
THe last number is decided by the first so 1 there
Each case has 30 possiblities and since we can only allow first digit 0, 1, 2, 3, 4, 5 - 6 so 180 possibilities

c) We can do it by compliment - calculate the number of possiblities whose numbers are NOT a factor of 5 (exclude 00, 05, ... 55 - 12 possiblities is total) leaving 48 possibilities
the total number of possibilities is P(48,3)

d) I thought this question was really hard
I did it by cases
When the first number is 0 to 9 , we can allow up 1 to 30 in the second digit. The 30 gets reduced by 1 for each number we add to the first so 10 allows 30 in the second, 11 allows 29 in the second and so on. 10 x 30!
first number 10 to 19, second number allows 15 numbers as it allows 11 through 25 in the second place 10 x 15!
first number 20 to 29, second number allows 15 as well?

According to this, the second number allows 15 as long as the number is 10 to 39 so
10 x 15! x 3
if the first number is 40 to 49, then 10 x 10! x 1
if the first number is 50 to 59, then 10 x 10! x 1

When we add all of these up we get the final answer and that is the total number of possibilites with sequences.

Is this all correct? Especially the last one?

Thanks
 
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  • #2
I confirm your answers for a, b and c, though I didn't follow your reasoning for b.
For d, you seem to be reading it as an arithmetic sequence of 6 digits. I read it as a sequence of 3 two-digit numbers, which is much easier.
 
  • #3
stunner5000pt said:
c) the product of the three numbers is a factor of 5
How many factors of 5 are there?

Or did you mean "multiple"?
 

FAQ: Probability of combination lock

1. What is the probability of guessing a combination lock correctly on the first try?

The probability of guessing a combination lock correctly on the first try depends on the number of digits in the combination and the number of possible combinations. For example, if a combination lock has 3 digits and each digit can range from 0-9, there are 1,000 possible combinations. Therefore, the probability of guessing the correct combination on the first try is 1 in 1,000 or 0.1%.

2. How many attempts does it usually take to crack a combination lock?

The number of attempts it takes to crack a combination lock depends on the number of digits in the combination and the method used to guess the combination. For example, if a combination lock has 4 digits and each digit can range from 0-9, there are 10,000 possible combinations. If a person tries one combination every second, it would take an average of 5,000 attempts to crack the lock. However, if a person uses a more efficient method such as narrowing down the possible combinations based on previous attempts, it may take fewer attempts.

3. Is it possible to calculate the exact probability of guessing a combination lock?

Yes, it is possible to calculate the exact probability of guessing a combination lock. The formula for calculating the probability is P = 1/n, where n is the number of possible combinations. For example, if a combination lock has 5 digits and each digit can range from 0-9, the probability of guessing the correct combination is 1 in 100,000 or 0.001%.

4. Does the order of the digits in a combination affect the probability of guessing correctly?

No, the order of the digits in a combination does not affect the probability of guessing correctly. For example, if a combination lock has the digits 1, 2, and 3, it does not matter if the correct combination is 123, 132, 213, etc. The probability of guessing correctly is still 1 in 1,000 or 0.1%.

5. Are some combination locks more difficult to crack than others?

Yes, some combination locks are more difficult to crack than others. This depends on the number of digits in the combination, the number of possible combinations, and the type of lock. For example, a lock with 5 digits and 10,000 possible combinations is more difficult to crack than a lock with 3 digits and 1,000 possible combinations. Additionally, some locks have features that make them more difficult to crack, such as anti-shim technology or a longer combination.

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