I have a question regarding this subject. If we have replacement, technically the odds of pulling one colored ball remains he same form draw to draw, thus they're independent events, right? Here is a specific example.
There is an urn with 80 blue balls, 60 red balls, and 60 green balls.
If I draw from this urn twice, with replacement, what is the probability that at least one ball is blue given that at least one ball is either green or red?
I go back and forth on this... since it is replacement, part of me says that there will always be a 40% chance that any ball pulled is blue. However, if we have the scenario where there is two balls, and we know that for sure one of the balls chosen will be red or green (60%), would I multiply 80/200 * 120/200?
What is the probability that at least one ball is blue given that the SECOND ball is either green or red?
Does it matter that we know that the second ball is either green or red or is this question to be approached the same way we would answer the reverse - what is the probability that at least one ball is blue given that the FIRST ball is either green or red.... same question? My gut says that this is the same answer as above.
Any insight that could help me get this to click would be greatly appreciated. Thank You.