The question looks simple, but I couldn't find a way to solve it. It goes like this:
I have $10$ white and $2$ red balls in a box and I am picking them one by one. All of them have equal chance to be picked, and I'm not throwing them back in after I draw them. After how many picks will the probability of having drawn both red balls will be over $60\%$?
On
The probability you get two reds out of three draws is $3$ (ways to order the balls you draw) times the chance you get two reds and then a white, which is $\frac 2{12} \cdot \frac 1{11} \cdot 1$. This comes out to $\frac 6{132}=\frac 1{22}$, too small. Can you do four draws? Keep going until the probability gets high enough.
After the first two picks, the chance of having picked both red balls is: $\frac{2}{12}\times \frac{1}{11}=\frac{1}{66}$ or below $2\%$.
After the first three picks, the chance of having picked both red balls is the total of the chances of having picked both red balls on the 1st and 2nd, 1st and 3rd, and 2nd and 3rd picks:
$\frac{1}{66}+\frac{2}{12}\times \frac{10}{11}\times \frac{1}{10}+\frac{10}{12}\times \frac{2}{11}\times \frac{1}{10}=\frac{1}{22}$ or below $5\%$.
After the first four picks, the chance of having picked both red balls is the total of the chances of having picked both red balls on the first three picks, 1st and 4th picks, 2nd and 4th picks, 3rd and 4th picks:
$\frac{1}{22}+\frac{2}{12}\times \frac{10}{11}\times \frac{9}{10}\times \frac{1}{9}+\frac{10}{12}\times \frac{2}{11}\times \frac{9}{10}\times \frac{1}{9}+\frac{10}{12}\times \frac{9}{11}\times \frac{2}{10}\times \frac{1}{9}=\frac{1}{11}$ or below $10\%$.
After the first five picks, the chance of having picked both red balls is the total of the chances of having picked both red balls on the first four picks, 1st and 5th picks, 2nd and 5th picks, 3rd and 5th picks, 4th and 5th picks.
To make it shorter, the chances of having picked both red balls on 1st and 5th picks, 2nd and 5th picks, 3rd and 5th picks, 4th and 5th picks are actually the same (similar to first four picks) and it is $\frac{2}{12}\times \frac{10}{11}\times \frac{9}{10}\times \frac{8}{9}\times \frac{1}{8}=\frac{1}{66}$. As I am typing this, I found out that for any number of picks, the chance of having picked both red balls on two specific picks, for example 1st and 3rd picks,... is always 1 in 66.
So after the first five picks, the chance of having picked both red balls is $\frac{1}{11}+\frac{1}{66}\times 4=\frac{5}{33}$, still not enough.
After the first six picks, the chance of having picked both red balls is the total of the chances of having picked both red balls on the first five picks, 1st and 6th picks, 2nd and 6th picks, 3rd and 6th picks, 4th and 6th picks, 5th and 6th picks, calculated and applied similarly: $\frac{5}{33}+\frac{1}{66}\times 5=\frac{5}{22}$, still not enough.
After the first seven picks, the chance of having picked both red balls is the total of the chances of having picked both red balls on the first six picks, 1st and 7th picks, 2nd and 7th picks, 3rd and 7th picks, 4th and 7th picks, 5th and 7th picks, 6th and 7th picks, calculated and applied similarly: $\frac{5}{22}+\frac{1}{66}\times 6=\frac{7}{22}$, still not enough.
After the first eight picks, the chance of having picked both red balls is the total of the chances of having picked both red balls on the first seven picks, 1st and 8th picks, 2nd and 8th picks, 3rd and 8th picks, 4th and 8th picks, 5th and 8th picks, 6th and 8th picks, 7th and 8th picks, calculated and applied similarly: $\frac{7}{22}+\frac{1}{66}\times 7=\frac{14}{33}$, still not enough.
After the first nine picks, the chance of having picked both red balls is the total of the chances of having picked both red balls on the first eight picks, 1st and 9th picks, 2nd and 9th picks, 3rd and 9th picks, 4th and 9th picks, 5th and 9th picks, 6th and 9th picks, 7th and 9th picks, 8th and 9th picks, calculated and applied similarly: $\frac{14}{33}+\frac{1}{66}\times 8=\frac{6}{11}$, still not enough (below $55\%$)
After the first ten picks, the chance of having picked both red balls is the total of the chances of having picked both red balls on the first nine picks, 1st and 10th picks, 2nd and 10th picks, 3rd and 10th picks, 4th and 10th picks, 5th and 10th picks, 6th and 10th picks, 7th and 10th picks, 8th and 10th picks, 9th and 10th picks, calculated and applied similarly: $\frac{6}{11}+\frac{1}{66}\times 9=\frac{15}{22}$, over $68\%$.
The final answer is 10 picks required.