$21$ balls in a bag - Probability There is $21$ balls in a bag numbered $1$ to $21$. 
Two balls are picked at random, without replacement.


*

*What is the probability of the lowest number on a ball being $17$ or greater?
Is it simply $\frac{5}{21} \times \frac{4}{20}$? 
I'm confused because I'm not sure if it matters or not which one we choose first.


Also, if the ball with the number $8$ was chosen, what is the probability of this being the lowest number out of the two balls? 
 A: @the man The term "lowest" is the key, Think about it (Its a relative term)  
It shouldn't matter which order you pick them.
So there are 5 balls that match the criteria 17 or greater.So pick any two of them  $\binom{5}{2}$ over any two of 21 balls $\binom{21}{2}$ which will be 
$\frac {\binom{5}{2}}{\binom{21}{2}}$ which eauals to  $\frac {10}{21*20}$
Second question: If one ball is 8, for the second ball you have a choice from 9 to 21(not from 8, as 8 is already taken)  which is 13. so its gonna be 
$\frac {13}{20}$
A: User @Browning suggested drawing a tree.  Here's what it would look like with probabilities and outcomes:
First = 1-16 (P = 16/21) --> BAD

First = 17 (P = 1/21)
    Second = 1-16 (P = 16/20) --> BAD
    Second = 18,19,20,21 (P = 4/20) --> GOOD

First = 18 (P = 1/21)
    Second = 1-16 (P = 16/20) --> BAD
    Second = 17,19,20,21 (P = 4/20) --> GOOD

First = 19 (P = 1/21)
    Second = 1-16 (P = 16/20) --> BAD
    Second = 17,18,20,21 (P = 4/20) --> GOOD

First = 20 (P = 1/21)
    Second = 1-16 (P = 16/20) --> BAD
    Second = 17,18,19,21 (P = 4/20) --> GOOD

First = 21 (P = 1/21)
    Second = 1-16 (P = 16/20) --> BAD
    Second = 17,18,19,20 (P = 4/20) --> GOOD

The probability of the first ball being >= 17 is $(5*\frac{1}{21}) = \frac{5}{21}$.  For each of those five numbers, the probability of the second ball being >= 17 is $\frac{4}{20}$.  Multiplying gives us that the probability is $\frac{5}{21}*\frac{4}{20} = 4.76\%$.  You're correct.
You can use this same method to answer your second question.

Edit: Here's a crudely-drawn tree that illustrates this a little better.  I'm adding this because my answer is different than the other two already here and I think that the other two answers are half what they should be.

Green bubbles are good outcomes.  Red bubbles are bad outcomes.  The numbers next to the arrows indicate the probability of that outcome.  The numbers inside the bubbles represent the possible selection(s).
Following the arrows to all of the green bubbles gives this as the answer:
$(\frac{1}{21}*\frac{4}{20}) + (\frac{1}{21}*\frac{4}{20}) + (\frac{1}{21}*\frac{4}{20}) + (\frac{1}{21}*\frac{4}{20}) + (\frac{1}{21}*\frac{4}{20})$
Simplified:
$\frac{5}{21}*\frac{4}{20}$
A: First question:
For the first ball, you have 5 options: 17, 18, 19, 20, 21
However, for your second ball, the number of possibilities depends on the number of the first ball. From the way your question is worded, the order of the balls is not relevant.
Lowest number being 17: $\frac {1*4}{21*20}$
Lowest number being 18: $\frac {1*3}{21*20}$ etc.
Total would be $\frac {4+3+2+1}{21*20} = 2.38 \% $
Alternatively, if the lower number is 17 or more, then there are $\binom{5}{2}$ over a total of 21*20 outcomes, which yields $2.38\%$.
Second question:
Even easier. If the lower number is 8, it means that the higher numbered ball is 9 to 21. 13 outcomes out of 20 (not counting the 8)
$\frac {13}{20} = 65 \%$
