Combination Probability problem

[lakers2009]

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The phone has 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, and 3 in any order? Repititions are a okay.

Answer: 0.006

How did they get to this answer? Anybody knows?

Is the setup 10C1 * 10C1 * 10C1 / (10C3)?

 

[lakers2009]

I think I found the answer but not sure of the logic. Is this correct?

3! / (10C1 * 10C1 * 10C1) = .006

or


3C1 * 2C1 * 1C1/(10C1*10C1*10C1)

 

[kkmaths90]

I think I found the answer but not sure of the logic. Is this correct?

3! / (10C1 * 10C1 * 10C1) = .006

or


3C1 * 2C1 * 1C1/(10C1*10C1*10C1)

I think you are right. there are 6 different desired outcomes (123, 132,231,213,312,321) and 1000 possible outcomes (repititions are okay and order does not matter so you use nCr and here 10C1 and also it doesn't say without replacement means you can press that number again)

Right :-?