# What is the probability given another event is occurring?

I have Box A with $$3$$ red balls and $$1$$ blue ball. I have Box B with $$1$$ red ball and $$4$$ blue balls. I randomly take a ball from Box A and put it into Box B. I then randomly draw a ball from Box B and it happens to be a red ball. What is the probability that the ball taken from Box A was red?

What I have so far is defined two events

Event A: the ball taken from Box A was red Event B: the ball drawn from Box B was red

I was thinking we need to find P(A|B) = $$\frac{P(A \cap B)}{P(B)}$$. Would P(B) just be $$\frac {1}{6} + \frac {1}{3}$$ since there are two cases where the ball drawn from B can be red: If a blue ball was added from Box A to Box B -> giving us the $$\frac {1}{6}$$ or if a red ball was added from Box A to Box B -> giving us the $$\frac {1}{3}$$. I'm not sure if I'm proceeding the right direction and/or how to proceed.

To compute the probability of $$B$$: $$P(B)=P(A)P(B|A)+P(\overline{A})P(B|\overline{A})$$.
Then the probability of A knowing B: $$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B|A)P(A)}{P(B)}$$.
You should find $$P(A|B)=\frac{6}{7}$$.
• How did you expand the numerator for P(A$\cap$B) and then conclude with $\frac {6}{7}$ ? – krauser126 Oct 24 at 20:52
• You have that $P(A\cap B)=P(A)P(B|A)=P(A|B)P(B)$. – Quantic_Solver Oct 25 at 7:34