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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.

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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 $P(A|B)=\frac{6}{7}$.

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  • $\begingroup$ How did you expand the numerator for P(A$\cap$B) and then conclude with $\frac {6}{7}$ ? $\endgroup$ – krauser126 Oct 24 at 20:52
  • $\begingroup$ You have that $P(A\cap B)=P(A)P(B|A)=P(A|B)P(B)$. $\endgroup$ – Quantic_Solver Oct 25 at 7:34

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