Probability

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