Multiplication of probabilities vs Combinations If I have a list of numbers from 1 to 100, what is the probability of choosing 5 random numbers which are equal or less than 10? Each time you choose a number, it is removed from the list, and you only have those 5 chances.
I tried doing the following:
(10/100) * (9/100) * (8/100) * (7/100) * (6/100)

The first time you choose a number, you have a 10/100 chance of getting a number which is equal or less than 10, the second time it is 9/100 and so on, 5 times.
That gives me $3.024 \times 10^{-6}.$
I also tried to use combinations with the following formula:
$$C(10, 5) / C(100, 5) = \frac{10 \choose 5}{100\choose 5}.$$
But the result is slightly different: $3.347 \times 10^{-6}.$
Could you tell me how to solve this problem correctly? In case that both solutions are correct, which is the reason for that slight difference?
I'm sorry if I'm not following some rule, this is my first time posting here. If it's the case, please let me know to edit my post.
 A: Your first calculation is wrong because when you go to pick the second you need to pick one of $9$ out of $99$, not out of $100$.  The difference is the denominator in the first is $100^5$ and in the second is $100\cdot 99 \cdot 98 \cdot 97 \cdot 96$.  The second is correct.  The small difference is because you aren't choosing that many of the numbers, so $96$ is not too different from $100$
A: Ordered. The correct version of your first method. as in the Answer of @RossMillikan, deals with ordered selections (without replacement)
in the numerator and denominator: 
Numerator: $10\cdot 9\cdot 8\cdot 7\cdot 6.\;$ This is sometimes 
denoted ${}_{10}P_5 = P(10,5) = \frac{10!}{5!}.$
Denominator: ${}_{100}P_5 = P(100,5) = \frac{100!}{95!}.$
Unordered. The second version version deals with unordered selections
in the numerator and denominator:
Numerator: ${10 \choose 5}.$
Denominator: ${100 \choose 5}.$
Both. Both answers are correct, and you can verify upon computation. that
they both give the same numerical answer. Many combinatorial probability problems can be solved either be considering a sample space or ordered outcomes
or a sample space of unordered outcomes. But it is important to pick one
point of view and use it in both numerator and denominator.

Notes: (a) This problem can also be solved using a 'hypergeometric distribution'; you may encounter them later in your course.
Imagine an urn with 100 balls of which 10 are red and 90 are green. If you 
draw 5 balls from the urn at random and without replacement, what is the
probability they are all red?
(b) In case you are interested in computing probabilities with R statistical software, I show three ways to use R to get the answer to your problem:
prod(10:6)/prod(100:96)     # ordered
## 3.347168e-06
choose(10,5)/choose(100,5)  # unordered
## 3.347168e-06
dhyper(5, 10, 90, 5)        # hypergeometric (which uses unordered)
## 3.347168e-06

