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One box contains one red and two white balls, the other box one white and two red balls. A box is selected at random and two balls are drawn from it with replacement. Assume that the boxes are equally likely to be selected, and that the balls in a box are equally likely to be drawn. Assume that the two drawings are independent. Is the event E = “red ball is obtained on the first draw” independent of the event F = “white ball is obtained on the second draw”?

1/2*1/3*2/3 = 1/9 p(ef) 1/2*1/3 = 1/6 Box A p(e) 1/3*2/3 = 2/9 Box A p(f) 3/54 = 1/18 =/ 1/9 so they are dependent?

Any help much be much appreciated.

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  • $\begingroup$ They are dependent. For intuition, imagine that the first box is all white, or almost all white, and the second is all red or almost all red. $\endgroup$ Feb 17 '16 at 18:44
  • $\begingroup$ And this is true even with replacement? Can you prove this using numbers? $\endgroup$ Feb 17 '16 at 18:45
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    $\begingroup$ Yes, one can do a computation. I will write it up with your numbers. $\endgroup$ Feb 17 '16 at 18:46
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They are dependent. For intuition, imagine that the first box is all white, and the second is all red. If we know the first ball was white, we can be sure that the second is white.

We do a formal calculation with your particular numbers. The probability that a red ball is obtained on the first draw, i.e. that $E$ happens, is $$(1/2)(1/3)+(1/2)(2/3),$$ which is $1/2$. (The $1/2$ is also obvious by symmetry.)

Similarly, the probability a white is obtained on the second draw is $1/2$.

Now we compute the probability of red on the first draw and white on the second. This can happen in two ways, (i) we are drawing from the first box and (ii) we are drawing from the second. The probability of $E\cap F$ is $$(1/2)(1/3)(2/3)+(1/2)(2/3)(1/3),$$ which is $2/9$. Note that this is not equal to $\Pr(E)\Pr(F)$, so the two events are not independent.

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