To my knowledge, "with replacement" means that rock selected for your first draw is put back in the lot. This makes the second draw independent from the first, as explained here:

https://math.youngzones.org/joint.html

The probability that the first draw is a slate rock is 15/45 (number of favorable cases/number of possible cases) or 1/3. The probability of the second rock being a slate rock will be 1/3 as well, since we have replaced the initial draw in the lot (i.e. we've put it back in the set).
The probability of both events taking place is 1/3 * 1/3 or 1/9.

HOWEVER

If you consider a "without replacement" probability, then you get something different. The first draw will have the same probability, i.e. 1/3. The second draw will not, since the number of favorable cases has dropped from 15 to 14 (since you've drawn a slate rock in your first pick) and the total number of cases has dropped from 45 to 44. So for your second draw, you get probability of = 14/44 = 7/22.
Multiply the two to get 7/66.

What is the source of this problem? Not all sources are 100% error-proof. Neither is my explanation, so I'm waiting for other opinions as well.

bburton11 wrote:There are 15 slate rocks, 20 pumie rocks, and 10 granite rocks randomly distributed in a certain field. If 2 rocks are to be chosen at random and with replacement, what is the probability that both rocks will be slate rocks?

1/3
7/22
1/9
7/66
2/45

Am I missing something here. I tired all the ways i know to solve this probability problem however, I am need help.

Hey bburton11,

are you sure the question says with replacement? or without replacement. I agree with DanaJ, b/c with replacement it would be (1/3)*(1/3) = 1/9

but w/o replacement... it would be 1/3*14/44 = 7/66.

also..If it does say with replacement, I am a bit suspicious b/c it doesn't say it is replaced with the same type of stone...it could have been replaced with a granite rock for all we know.

I agree, should be 1/9