Probability of a Certain Number Occuring

If only five-digit phone numbers are possible, what is the probability of having a phone number that ends in 5? Am I just multiplying 10! and dividing by 10?

the total number of 5 digit phone numbers possible is 2 * 5!.

The total that end in 5 is (2 * 5!)/10

Why do I multiply by 2? Because the answer with 2*5! = 240 and dividing that by 10 is 24 and then I divide that into 5?

2partswater said:

If only five-digit phone numbers are possible, what is the probability of having a phone number that ends in 5?

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If that is the actual question, worded exactly that way then
there are \(\displaystyle 10^5\) possible phone numbers of which \(\displaystyle 10^4\) end in five.

why 10^5 total possiblilities and 10^4 for 5.

I understand that if you add 10^4 10 times you get 10^5 but each number has permutations of 10 * 9 * 8 * 7 * 6

Thus the total number of phone numbers possible is 10!/5! =/ 10^5

If you divide 10!/5! by 10 you will get 9!/5!

Thus the total phone numbers = 10!/5! and the total with 5 = 9!/5!

Sorry about my mistake the first time. I was only considering 1/5 the phone numbers and I am sorry for that.

Denis said:

Caters, pka gave the answer...

0000[5] to 9999[5] = 10000 = 10^4 ; OK?

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No it is 9!/5!. That is because of these reasons:

1) you aren't using all 10 digits in any particular number only half of it

2) you are dividing it by 10

Let me explain.

For 5 digits the 10 digits used have 5! permutations. You have 10 digits. That gives you 10!/5!.

You divide that by 10 and you get 9!/5! which is the correct answer and is not 10^4.

This whole thing was a permutations problem so everything had to be in terms of factorials.

Also when it comes to number of 1 to x digit numbers it is always a power of 10 but that does not mean that all 5 digit numbers add up to a power of 10 when you count them and it does not mean that the answer to this particular question is not 9!/5!.

caters said:

No it is 9!/5!. That is because of these reasons....

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Sorry, but no. The tutor really did give the correct answer.

caters said:

No it is 9!/5!. That is because of these reasons:

1) you aren't using all 10 digits in any particular number only half of it

2) you are dividing it by 10

Let me explain.

For 5 digits the 10 digits used have 5! permutations. You have 10 digits. That gives you 10!/5!.

You divide that by 10 and you get 9!/5! which is the correct answer and is not 10^4.

This whole thing was a permutations problem so everything had to be in terms of factorials.

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No, this is NOT a "permutations" problem. It would be if each of the 10 digits could be used only once in the number. But that is not true: "11111" would be valid number. If each of the 10 digits could be used in each of the 5 places there would be \(\displaystyle 10^5= 100000\): 00000 to 99999. However, if we add the (perfectly reasonable in my opinion) requirement that the number NOT begin with a 0, there are only 9 digits that can be used for the first so \(\displaystyle 9*10*10*10*10= 9*10^4=90000\) such numbers. Requiring that the last digit be fixed reduces that last "10"" to "1" leaving \(\displaystyle 9*10^3\). The probability is \(\displaystyle \frac{9*10^3}{9*10^4}= \frac{1}{10}\). (The "9"s cancel so that is not really an important point!)

Also when it comes to number of 1 to x digit numbers it is always a power of 10 but that does not mean that all 5 digit numbers add up to a power of 10 when you count them and it does not mean that the answer to this particular question is not 9!/5!.

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Permutations problems can either be permutations of an infinite set like this one is or permutations of a finite set.

What "infinite set" are you talking about?

infinite set as in infinitely many of what is in your set. In this case it is infinitely many 0's, 1's, 2's, 3's, 4's, 5's, 6's, 7's, 8's, and 9's

because you have 2! or 2 ways any group can be arranged with repetitions allowed. Since 11 and 11 are the same we are double counting all the ones with 2 of the same digit so that subtracts 5 from the total giving us 55-11=44+1=45 instead of the 50 that would result have we allowed those repetitions. However we can subtract 20 from that because we are not using any of the other digits

as to ending with 2 that occurs every 5th time and same for other digits.

So we have 5 ending with 1, 5 ending with 2 etc giving us 5^2 numbers.

So using 1-5 gives us powers of 5 as to the permutations of x digits if we don't use anything else and we have infinitely many 1's, 2's, 3's, 4's, and 5's in our set to choose from. This is what is behind base 5 except it is 0-4 for base 5.