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I can't figure out how to find out following problem probability. I am in stuck but it is very important for me. Here is the problem description.

Box one contains $3$ white balls and $7$ black balls.

Box two contains $6$ white balls and $4$ black balls.

If the first ball is white, we then randomly pick the second ball from box one. If the first ball is black, we randomly pick second ball from box 2.

Question:

1) What is the probability that the second ball is white?

2) Given that the second ball is white. What is the probability that the second ball is picked from box two?

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    $\begingroup$ Is the first ball picked from box one? $\endgroup$ – meta Feb 8 '14 at 18:00
  • $\begingroup$ Which box will be selected that is not defined? That will under probability. $\endgroup$ – karim_fci Feb 8 '14 at 19:18
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You left out two things in the problem description: what is the chance the first ball comes from box 1 and is the first ball put back into the box?

There are four ways to get a white ball second. One of them is pick the first ball from box 1, get a black ball, pick a white ball from box 2. The probability of this P(first pick from box 1)$\cdot \frac 7{10} \cdot \frac 6{10}$ You can assess the other three and add them toget the total chance of a white second ball.

For the second, you have four routes that got you a white second ball. Two of them have the ball coming from box 2. The fraction you want is the sum of those two divided by the number you got in problem 1.

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  • $\begingroup$ Yes, the first ball put back into the box. First ball may be comes from both box. $\endgroup$ – karim_fci Feb 8 '14 at 19:20
  • $\begingroup$ Can you define the other three paths to a white second ball and compute their probabilities? Where are you stuck? $\endgroup$ – Ross Millikan Feb 8 '14 at 19:28

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