So we have $2$ boxes. In the first box, there are $4$ blue balls and $5$ red balls. In the second one, there are $5$ blue balls and $3$ red balls. From the first box, we move $2$ balls at the same time and place them to the second box. Then we draw one ball from the second box and it is red. What is the probability that we moved $2$ red balls from the first box to the second one?
So my understanding is that the chance is $$\frac{\dbinom{5}{2}} {\dbinom{9}{2}} = \frac{5}{18} = 0.277777…$$
Is it correct?
Best Answer
We wish to find the conditional probability that two red balls were transferred from the first box to the second box given that a red ball was selected from the second box after two balls were transferred from the first box to the second box.
Let $R$ be the event that a red ball was selected from the second box.
Let $BB$ be the event that two blue balls were transferred from the first box to the second box.
Let $BR$ be the event that one blue ball and one red ball were transferred from the first box to the second box.
Let $RR$ be the event that two red balls were transferred from the first box to the second box.
We wish to find $\Pr(RR \mid R)$. \begin{align*} \Pr(RR \mid R) & = \frac{\Pr(R \cap RR)}{\Pr(R)}\\ & = \frac{\Pr(R \mid RR)\Pr(RR)}{\Pr(R \mid BB)\Pr(BB) + \Pr(R \mid BR)\Pr(BR) + \Pr(R \mid RR)\Pr(RR)} \end{align*} The probability that two of the five red balls are transferred when two of the nine balls in the first box are transferred to the second box is $$\Pr(RR) = \frac{\dbinom{5}{2}}{\dbinom{9}{2}}$$ which is what you calculated.
If two red balls are transferred from the first box to the second box, then five of the ten balls in the box are red when a ball is drawn from the second box. Hence, $$\Pr(R \mid RR) = \frac{5}{10}$$
The probability that one of the four blue balls and one of the five red balls are transferred to the second box when two balls are selected from the nine balls in the first box is
If one blue ball and one red ball are transferred to the second box, then four of the ten balls in the second box when a ball is selected are red. Hence,
The probability that two of the four blue balls are transferred to the second box when two of the nine balls in the first box are transferred to the second box is
If two blue balls are transferred to the second box, then three of the ten balls in the second box are red when a ball is selected from the second box. Hence,