You have 60 marbles in a jar, two of which are red. (so 2 red, 58 not red)
you pick 10 at random. what is the chance that you draw both red marbles?
explanation , not just the answer. brainliest answer if you calculate it for 2 out of 251 total marbles as well.

Here you have two proportions. 2/60 and 10/60. I would cross multiply. 
Giving you 20/120.
Simplify that to get 1/6. So you have a 1 in 6 chance of getting both the red marbles.
(I'm not 100% sure since it's been a while so you might want to get a second opinion.) 
Not sure what you mean by the second part, but hope I provided some help :)

Well, I agonized over this for a while, and I have something that we may want to consider.

The only probability formula I know how to use is

 Probability = (number of possible successes) / (total number of possibilities).

Can we work with that ?  Let's see . . .

The denominator of that fraction is the total number of ways to draw
10 marbles from a jar of 60 . . .

The first draw can be any one of 60 marbles. For each of those . . .
The second draw can be any one of the remaining 59.  For each of those . . .
The third draw can be any one of the remaining 58.
.
.
etc.

So the total number of ways to draw 10 from 60 is

       (60 x 59 x 58 x 57 x 56 x 55 x 54 x 53 x 52 x 51)

That's a very big number.  My calculator says something that rounds to
2.736 x 10¹⁷. But my calculator only shows 10 digits, so it can't show
all 18 digits in the number.  Fortunately, we don't need to see the whole
number written out.  We'll just write it in factorial notation, and go on
to do the numerator of the fraction, which is going to be much harder.

That number that we just found is equal to  (60!) / (50!) .  It's going to be
the denominator of the big fraction.

Now for the numerator.  That's going to be the number of ways that the
two red marbles can be included among the ten marbles drawn.

One red marble can be any one of the 10 marbles pulled out.
       For each of those . . .
The other red marble can be any one of the other 9 that are picked.

So there are (10 x 9) = 90 ways for the selected 10 to include both red ones.

SO !  Now, the probability that the ten that are drawn will include the two
red ones should be (it might be, it could be) . . .

             90 divided by  (60! / 50!)   .

Remember, that fraction on the right is just total number of ways
to pick 10 out of 60 .  The probability of including both red ones 
in the draw is  90 divided by that number.  It's very small.

Again from my calculator, it's    3.29 x 10⁻¹⁴ percent.

I have no confidence in my answer, but I invite you to look it over, along with
all the real gurus out there. 

If I'm wrong, then I've stolen only 5 points that I'm not entitled to, and at least
I did put some effort into it.

It has a slope of 1 nd a y intercept of -2