One red, one green and one blue ball in a box I asked myself a question about probabilities and got pretty confused about it.
Let's say there is a box with 1 red ball, 1 green ball and 1 blue ball inside.
I pick randomly a ball from the box, then put it back, and repeat the process two more times (so three times at all).
So, from the first pick, my chance of picking the red ball is 1/3. And during the second picking, my chance of grabbing the ball is also 1/3. Same for the third time.
But if I ask myself this next question : what is my total chance of picking one time the red ball during the three random picks combined ?
And here I got confused. Shouldn't my chance of picking the red ball increase everytime I pick randomly one ball ?
Let's say the first time is 1/3, then the second time should be 1/3 (from the first time) +1/3 (from the second time) = 2/3 ?
And the third time 3/3 ? But if it's 3/3 it means that I have 100% chance to pick the ball after 3 times, which is not realistic in the real life : I could pick 3 times the green ball, or two times the green ball and one time the blue one, etc.
I hope you will be able to explain this to me.
Thank you!
 A: With only one draw (but let's call it a trial), there are only two possible outcomes with reference to picking exactly one red ball: picked $(R)$ or not picked $(N).$
With two trials, there are now four possible outcomes: $NN, NR, RN, RR.$
With three trials, eight possible outcomes: $NNN, NNR, NRN, RNN, NRR, RNR, RRN, RRR.$
Etc.
By letting $R=\frac13$ and $N=\frac23,$ in each case, we can calculate the required probability by adding up the probabilities of the relevant outcomes:

*

*$\frac13$

*$(\frac23\times\frac13)+(\frac13\times\frac23)=\frac49$

*$(\frac23\times\frac23\times\frac13)+(\frac23\times\frac13\times\frac23)+(\frac13\times\frac23\times\frac23)=\frac49$

*etc.

A: Probabilities do not add across draws, so adding $\frac13+\frac13+\frac13$ to get probability $1$ of drawing the red ball is simply absurd. However, independent events like "red on draw 1" and "not red on draw 2" do multiply, so the correct answer is $3×\frac13×\frac23×\frac23=\frac49$ (we multiply by $3$ because the red ball may be dran first, second or third).
