I am struggling with the following problem:
Consider two boxes with several blue and red colored balls in.
In the right box there are 4 red and 5 blue balls.
In the left one are 7 red and 3 blue balls.
Now you blindly draw 4 balls from the right box and put them into the left one.
What is the probability of getting one red ball when you grab into the left box?
This has for sure something to do with conditional probability but in my opinion the answer depends on how many red balls are transfered into the left box.
Case 1: all of the 4 tranfered balls are blue.
So the left box contains 7 red and 7 blue balls. The probability to get a red one is $\frac{1}{2}$. The probability that the transfered balls are all blue is $\frac{5}{9} \cdot \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6} = \frac{5}{126}$.
But how am I able to calculate P('red ball' | 'all transf. balls are blue')? By the Kolmogorov definition it is $\dfrac{P(\text{'red ball' } \cap \text{'all transf. balls are blue'})}{\frac{5}{126}}$ but surely P('red ball' $\cap$ 'all transf. balls are blue') is not simply $\frac{1}{2} \cdot \frac{5}{126}$.
Futhermore, do I need to differentiate between these different cases? The question sounds like the probability is always the same.
Best Answer
The probability that a red is drawn after transferring $k$ reds is $$\frac{7+k}{14}$$
The probability of drawing $k$ reds is $$\binom{4}{k}\frac{(4)_k(5)_{4-k}}{(9)_4}$$
where $(n)_k$ is the falling factorial, so for example $(9)_4=9\cdot8\cdot7\cdot6$.
So the final probability is $$\sum_k P(\text{draw k reds})P(\text{draw red after transfer})$$