Which expression can be used to determine the probability of the alarm code beginning with a number greater than 7?

Question

A security alarm requires a four-digit code. The code can use the digits 0–9 and the digits cannot be repeated.

Which expression can be used to determine the probability of the alarm code beginning with a number greater than 7?

Answer 1

There are 10,000 values from 0000 to 9999.

 

Of those, 5040 do not have multiples of the same digit.  That is calculated as:

      10*9*8*7 = 5040

 

The word "beginning with a number greater than 7" mean numbers in the range of 8012 to 9876  [remember, no repeats].  There are 1008 of those, calculated as 5040/5    [because 8-9 is 2 digits, which is 1/5th of the 10 digits from 0-9].

 

Now, "probability" is defined as (the number of successful outcomes) / (the number of possible outcomes).  There is no probability in this problem unless a 4-digit code is randomly selected from the allowable codes.  If that is the case, then:

   P(code beginning with 8 or 9) = 1008 = (1/5)(5040) = (1/5)(10!)/((10-4)!)

Answer 2

If the first digit must be greater than 7 it can only be 8 or 9, so you have two options.

The second digit can be any of the 10 numbers (0-9) EXCEPT whichever one you chose for the first digit, so you have 9 options

The third digit can be any of the 10 numbers (0-9) EXCEPT whichever two you chose for the first and second digit, so 8 options

The fourth digit can be any of the 10 numbers (0-9) EXCEPT whichever three you have already used for the first three digits, so 7 options.

 

Then multiply these numbers together

 

(2)(9)(8)(7)

 

That is the number of codes with the first digit above 7 (your numerator)

The total possible four digit non repeating codes would be (10)(9)(8)(7)...similar argument to above except now I have a full 10 choices for the first one.

 

I can also write that denominator in a factorial form (10!/(10-4)!) or 10!/6! or sometimes as a "permutation" 10p4