Probability three digits chosen at random from 0 to 9....

Question:
A computer prints out three digits chosen at random from 0 to 9, inclusive. Find the probability that:

a)all three digits are different

my answer: Do I simply multiply 10 by 9 by 8 and divide by total possible to get 720/1000= 72 ? Or is the answer more likely solved by something that looks like this:

\(\displaystyle \L\\\frac{{C(10,1) \times C(9,1) \times C(8,1)}}{{C(10,3)}}\)

Part B)
The third digit differs from the first two digits

I wish I could say I was having as much luck at this one as I was for my first probablities post. But honestly I am lost here. Just a guess but would it be something like (10 * 10* 9)/ 1000 =.9 ?

Thank you for helping me out
sincerly
Louise

The question needs clarification!
Do we count 006 or 078 as three-digit numbers?
Some authors require the first digit to not equal zero.

From my meager probability experience I think that it is safe to assume that yes you can use the zero's in front of the other numbers as this question is just referring to a computer printing out random numbers. However I have had like a dozen questions now that refer to choosing 3 digits for things like phone area codes and house numbers which obviously can't have zero's in front and the question has stated that very clearly. Thats just what I am thinking anyways.
Thank you
Louise

\(\displaystyle \L a)\quad \frac{{\left( {10} \right)\left( 9 \right)\left( 8 \right)}}{{10^3 }}\quad \& \quad b)\quad \frac{{\left( {10} \right)\left( {10} \right)\left( 9 \right)}}{{10^3 }} + \frac{{\left( {10} \right)\left( 9 \right)\left( 8 \right)}}{{10^3 }}\)

Now you explain to us those answers.

quote="pka"]\(\displaystyle \L a)\quad \frac{{\left( {10} \right)\left( 9 \right)\left( 8 \right)}}{{10^3 }}\quad \& \quad b)\quad \frac{{\left( {10} \right)\left( {10} \right)\left( 9 \right)}}{{10^3 }} + \frac{{\left( {10} \right)\left( 9 \right)\left( 8 \right)}}{{10^3 }}\)

A) I had .72 for a which I could see plainly enough. Basically you have three spaces. The first space can be any of the ten digits. The second space can only be nine of the numbers because you have chosen one from the lot already. The final space you have a choice of 8 digits remaining. Multiply 10 9 and 8 to get 720. Now I had divide by 10 by 10 by 10 but writing 10^3 is musch more simple. To complete the answer you must divide your first part by the total number of possibilities which is 1000.

b) What I missed with part B of the question is that it is really two steps or parts. I am still confused at why you need to add the probability of all three different with the probability of the third digit differs from the first two digits. Are they not completely separate events? If anything wouldn't you take the first away from the second? To come up with an answer of .18 probability?
Thank you
Louise

Louise Johnson said:

I am still confused at why you need to add the probability of all three different with the probability of the third digit differs from the first two digits. Are they not completely separate events? If anything wouldn't you take the first away from the second? To come up with an answer of .18 probability?

Click to expand...

Because the first two digits could be the same and the third different OR all three could be different. These are disjoint events so add the probabilities.

seems simple now ... Huge weight now that you have clarified that!!
Thank you