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The first box contains $50$ white balls and $20$ black balls and second box contains $60$ white and $40$ black. If we take one ball randomly from the first box and put in the second box. What is the probability that when we withdraw a ball from second box it is white?

My teacher did it like this:

First they calculated probability when withdrawing from the first box $$P(\mathrm{WhiteBox1}) = \frac{50}{70} = \frac{5}{7}, \quad P(\mathrm{BlackBox1}) = \frac{20}{70}= \frac{2}{7}.$$

Then they calculated the probability given the first box was white $$P(\mathrm{Box2|Whiteball1}) = \frac{61}{101},$$ and given the first box was black $$P(\mathrm{Box2|Blackball1}) = \frac{41}{101} .$$

And in the end $$P(\mathrm{Total}) = \frac{5}{7} \times \frac{61}{101} + \frac{2}{7} \times \frac{41}{101} = 0.54.$$

I am confused why are we calculating even for black ball probability if we care only if the ball is white?

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    $\begingroup$ The second term is not correct. You want to draw a white ball from the second box, so you should consider $\frac {60}{101}$ not $\frac {41}{101}$. $\endgroup$ – lulu Feb 17 at 16:16
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    $\begingroup$ Note: the answer to the stated question should be very close to $.6$ since it is $.6$ before you transfer the ball and the one extra ball (of either color) doesn't have all that big an effect. $\endgroup$ – lulu Feb 17 at 16:18
  • $\begingroup$ instead of 41/101 I have to do 60/101 , thanks @lulu $\endgroup$ – Hiken No Ace Feb 17 at 16:21
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    $\begingroup$ Possibly worth noting: The given formula would be correct if the problem were asking "what's the probability that the ball you select from the second box has the same color as the one you drew from the first box." $\endgroup$ – lulu Feb 17 at 16:24
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    $\begingroup$ Yes. Same problem, I just want the second draw to match the first. $\endgroup$ – lulu Feb 17 at 16:34
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$P({W \ ball \ from \ box \ 2})$

$=P({W \ ball \ from \ box \ 1}) \times P({W \ ball \ from \ box \ 2 } \mid {W \ ball \ from \ box \ 1})$

$ + \ P({B \ ball \ from \ box \ 1}) \times P({W \ ball \ from \ box \ 2} \mid {B \ ball \ from \ box \ 1})$

$=\bigg(\frac{50}{70}\times \frac{61}{101}\bigg)+\bigg(\frac{20}{70}\times \frac{60}{101}\bigg)=0.601$

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