Question 991118
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Since you are replacing the item drawn on the first draw, then the two events are independent.  Therefore, the probability of both events is the product of their individual probabilities.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

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*[tex \LARGE \ \ \ \ \ \ \ \ \ \