Homework Help: Probability involving 4 digit password

[Pikachu1]

I am upgrading and this question i just can not process please help. I will be very greatful.

Abigail needs to create a four-digit password to access her voice mail. She can repeat some of the digits, but all four digits cannot be the same. Determine, to the nearest percent, the probability that her password will be greater than 6300. Show your work.

 

[MarkFL]

I would begin by finding the number of favorable outcomes, that is, the number of passwords that are greater than 6300. Suppose there are no restrictions...how many passwords would be greater than 6300?

edit: I have edited the thread title to make it descriptive of the question being asked, and I have moved the thread to our Basic Probability and Statistics forum as it is a better fit. :D

 

[Pikachu1]

Well the way i look at the number is that the only number that needs to be changed in order to get numbers greater than 6300 would be the last number.

So without restrictions the first number should still have 4 possibilites. 6 7 8 9

And for the second it should be 3 4 5 6 7 8 9

third can have 0 1 2 3 4 5 6 7 8 9

fourth the number would have to be greater than 0 soo 1 2 3 4 5 6 7 8 9

4*7*10*9 which would be 2520 numbers without restrictions.

In total the number of all possibilities would be 10*10*10*9 = 9000
because all 4 digits can not be the same.

but i feel i am doing this wrong. I have spent 4 hours on this.

 

[MarkFL]

This is how I would look at it:

The number of passwords $N$ (ignoring the restriction) greater than 6300 is:

\(\displaystyle N=9999-6300=3699\)

Now, we need to remove those passwords greater than 6300 which are not allowed (those in which all 4 digits are the same), which are:

\(\displaystyle 6666,\,7777,\,8888,\,9999\)

There are 4 of them, so the number of favorable outcomes $F$ is:

\(\displaystyle F=N-4=3695\)

Now, can you use similar reasoning to find the total number of passwords possible?

 

[Pikachu1]

OH wow. That makes so much sense. I was approaching this all wrong and over thinking it. I have used this method before. Geez i am so silly.

I am able to finish this problem now. You are amazing.

 

[Kaymajestic2]

This is how I would look at it:

The number of passwords $N$ (ignoring the restriction) greater than 6300 is:

\(\displaystyle N=9999-6300=3699\)

Now, we need to remove those passwords greater than 6300 which are not allowed (those in which all 4 digits are the same), which are:

\(\displaystyle 6666,\,7777,\,8888,\,9999\)

There are 4 of them, so the number of favorable outcomes $F$ is:

\(\displaystyle F=N-4=3695\)

Now, can you use similar reasoning to find the total number of passwords possible?

That should be the answer, isn't it?

 

[Prove It]

This is how I would look at it:

The number of passwords $N$ (ignoring the restriction) greater than 6300 is:

\(\displaystyle N=9999-6300=3699\)

Now, we need to remove those passwords greater than 6300 which are not allowed (those in which all 4 digits are the same), which are:

\(\displaystyle 6666,\,7777,\,8888,\,9999\)

There are 4 of them, so the number of favorable outcomes $F$ is:

\(\displaystyle F=N-4=3695\)

Now, can you use similar reasoning to find the total number of passwords possible?

Wouldn't it be 10 000 - 6300? Technically 0000 is a possibility as well.