Probability Demystified Problem

[ghettointlectual]

I am working my way through "Probability Demystified" published by McGraw Hill and written by Allen Bluman.
I am on Chapter 6 and doing the end of chapter quiz when I run into this problem that has me very confused because the answer I get is not found in any of the choices given.

Here is the problem:
A phone extension consists of 3 digits. If all the digits have a probability of being selected, what is the probability that the extension consists of the digits 1,2,3 in any order? Repetitions are allowed

This is how I approached it:
Since repetition is allowed, there are 10*10*10= 1000 combinations of 3 digits. For the digits 1,2,3 in any order with repetitions allowed there 3*3*3= 9 ways of getting those numbers. 9/1000= .009 however that is not one of the choices. Am I wrong? If am wrong, where did I go wrong?

Thank you

[jay_shark]

You must use all the digits 1,2,3 in some order .

There are 3 ways to select the first order .
Once this is fixed , there are two ways to select the second . Once the first two are fixed , there is 1 way to select the third .

{123,132,213,312,231,321} = 6 ways or 3*2*1 .

So we have 3*2*1/10^3 = 0.006

---------------------------------------------------------

It would be 3*3*3 =27 if we were allowed to use only 1 or 2 or 3 with repetition . This would include {111 , 112, 222, 123} etc.

[ghettointlectual]

Thanks for the reply... I can't believe I put 3*3*3= 9 [img]/images/graemlins/confused.gif[/img]

When the problem says that repetition is allowed its wasn't for 1,2,3 but for 0-9? If thats the case thats really confusing

[SheetWise]

[ QUOTE ]
You must use all the digits 1,2,3 in some order .

There are 3 ways to select the first order .
Once this is fixed , there are two ways to select the second . Once the first two are fixed , there is 1 way to select the third .

{123,132,213,312,231,321} = 6 ways or 3*2*1 .

So we have 3*2*1/10^3 = 0.006

---------------------------------------------------------

It would be 3*3*3 =27 if we were allowed to use only 1 or 2 or 3 with repetition . This would include {111 , 112, 222, 123} etc.

[/ QUOTE ]

You're saying that repetition is not allowed in the numerator (3*2*1) -- then why are you allowing it in the denominator (10^3)?

Wouldn't it make more sense to use permutations of 3 out of 10 as the universe to determine the probability?

[WhiteWolf]

[ QUOTE ]
[ QUOTE ]
You must use all the digits 1,2,3 in some order .

There are 3 ways to select the first order .
Once this is fixed , there are two ways to select the second . Once the first two are fixed , there is 1 way to select the third .

{123,132,213,312,231,321} = 6 ways or 3*2*1 .

So we have 3*2*1/10^3 = 0.006

---------------------------------------------------------

It would be 3*3*3 =27 if we were allowed to use only 1 or 2 or 3 with repetition . This would include {111 , 112, 222, 123} etc.

[/ QUOTE ]

You're saying that repetition is not allowed in the numerator (3*2*1) -- then why are you allowing it in the denominator (10^3)?

Wouldn't it make more sense to use permutations of 3 out of 10 as the universe to determine the probability?

[/ QUOTE ]
The original problem did not say repetition was not allowed, but it did specify that the digits 1, 2, AND 3 must appear in the 3-digit number.

[SheetWise]

I'm sorry -- this part I don't understand. The OP wrote -

[ QUOTE ]
A phone extension consists of 3 digits. If all the digits have a probability of being selected, what is the probability that the extension consists of the digits 1,2,3 in any order? Repetitions are allowed

[/ QUOTE ]

What does "repititions are allowed" mean if all three digits must appear in a three digit number?

I don't know what that means.

[WhiteWolf]

[ QUOTE ]
I'm sorry -- this part I don't understand. The OP wrote -

[ QUOTE ]
A phone extension consists of 3 digits. If all the digits have a probability of being selected, what is the probability that the extension consists of the digits 1,2,3 in any order? Repetitions are allowed

[/ QUOTE ]

What does "repititions are allowed" mean if all three digits must appear in a three digit number?

I don't know what that means.

[/ QUOTE ]
Yeah, the phrasing there is confusing. To me, it means the universe of possible numbers allows for repetitions (ie 556 is a legal number), and we are interested in calculating the probability that the number contains a 1, a 2, and a 3 in any order. The fact that our numbers of interest contains 3 unique digits, and the number is also 3 digits long, means that we can't see repetition in any of those numbers we are calculating the probability for.

I think the problem would have been better worded like this:

A phone extension consists of 3 digits (repetitions are allowed). If all the digits have a probability of being selected, what is the probability that the extension consists of the digits 1,2, and 3 in any order?