(There is already a similar question to this, but it has not an accepted answer…, so:)
Two boxes are given. There are 15 white and 12 black balls in the first box, and 14 white and 18 black balls in the second box. Anna provides the following experiment. Anna puts her hand in the first box, takes at once two balls and place them in the second box. Then, she takes one ball wihout looking from the second box.
What is the probability that she took a white ball from the second box?
My approach is to first calculate the probabilities of the 4 possible cases to move two balls: {(WW),(WB),(BW),(BB)}.
And afterwards to multiply the resulting probabilities with the new amounts of the second box. Finally I would add up the four resulting values.
For ex. for (WW) i would do:
- to draw WW from the first box: $\frac{15}{27} * \frac{14}{26} \approx 0.3$
- calculate the probability of drawing white from the second box: $0.3 * \frac{16}{34} \approx 0.14$
But I'm not even sure if I have to differentiate between (B,W) and (W,B)
Best Answer
Consider the following approach. After placing 2 additional balls in the second box, there are $14+18+2=34$ balls total there. So when you take the ball, you have $p_1=32/34$ probability that it will be the ball that was there before and $p_2=2/34$ that it was the ball from the first box. In that case, probability of it to be white is the same as picking white ball from the first box in the beginning. Thus: $$ p_w = \frac{14}{32}\times\frac{32}{34} + \frac{15}{27}\times\frac{2}{34}=\frac49. $$
Using your approach, we need to calculate the probability of WW, BW, and BB separately: $$ p_w = p_{WW}\frac{16}{34} + p_{BW}\frac{15}{34} + p_{BB}\frac{14}{34} = \frac{15}{27}\frac{14}{26}\frac{16}{34} + \left(\frac{15}{27}\frac{12}{26}+\frac{12}{27}\frac{15}{26}\right)\frac{15}{34} + \frac{12}{27}\frac{11}{26}\frac{14}{34} = \frac49 $$