The question I have can be boiled down to this example problem:
There is a bag with 10 balls, 8 of which are white, 1 of which is red, and 1 of which is blue. In n trials, where you pull a ball, observe the color/result, and replace it back in the bag, what are the chances you see at least 1 blue ball and at least 1 red ball?
I'm struggling with my limited knowledge on how to approach or even describe this problem. If success were just seeing at least one colored ball, red or blue, it seems to be a simple cumulative binomial probability of n = 10, p = 0.20, chance equaling 89%.
However, if you break it down to blue and/or red, then the initial p is still 0.20, but once you observe either colored ball then the p drops to 0.10 to observe the remaining colored ball, thus breaking independence. I'm not sure if there's a way to use the binomial in such a way to work around this, or if there's another concept I need to learn that would accomodate this question's features. I'm more than happy to read up on the necessary materials myself (this is not an answer my homework question), I'm just not sure where to even begin on this.
There is a bag with 10 balls, 8 of which are white, 1 of which is red, and 1 of which is blue.
In n trials, where you pull a ball, observe the color/result, and replace it back in the bag, what are the chances you see at least 1 blue ball and at least 1 red ball?
enumerate the possibilities using rwb notation
possibilities for n = 2 are: ww with p=(8/10)(8/10), wr with p=(8/10)(1/10), wb with p=(8/10)(1/10), rw with p=(1/10)(8/10), rr with p=(1/10)(1/10), rb with p=(1/10)(1/10), bw with p=(1/10)(8/10), and br with p=(1/10)/(1/10)
the only two cases which count for success criteria are rb and br adding those probabilities you get (1/100)+(1/100)=0.02
now the case of n=3 ... here the cases which count are wbr, wrb, bbr, brb, rbr, and rrb
adding probabilities we get 2 times 8/1000 plus 4 times 1/1000
or (16+4)/1000 =20/1000 = 0.02
now for the case of n=4 etc