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There are two boxes; the first one contains 10 white balls and 5 black balls; the other one contains 10 black balls and 5 white balls. King randomly selects a box (with equal probabilities) and then randomly takes a ball from this box (with equal probabilities). What is the probability that King selected the first box under the condition that the ball he selected is white?

So my approach is I thought that I should calculate the probability of the first box then the probability of the second and just add them but that's been wrong so can someone help me?

Thank you!

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    $\begingroup$ Do you know Bayes' Theorem? $\endgroup$ Commented Apr 22, 2020 at 16:12
  • $\begingroup$ Actually struggling with it, but I also thought about since they are with equal probabilities, I would just calculate the probability of the white balls in the first box but that didn't work either $\endgroup$ Commented Apr 22, 2020 at 16:18

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F = First Box, S = Second Box, W = White Balls, B = Black Balls

$$P(F|W)=\frac {P(F\cap W)}{P(W)}=\frac {P(F)\times P(W|F)}{P(F)\times P(W|F)+P(S)\times P(W|S)}=\frac {\frac 12 \times \frac 23}{\frac 12 \times \frac 23 + \frac 12 \times \frac 13}=\frac 23$$

Tree Diagram

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