Question 334673
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Replacement doesn't make any sense in the context of the probability of picking <i>a</i> red marble.  Now if you were interested in the probability of picking a red marble <i>on the second (or subsequent) trial</i>, THEN replacement makes a difference.


If you are only concerned with the first try, replacement doesn't matter.  The probability is simply the number of red marbles divided by the total number of marbles.


But if you want the probability of picking a red one on the second try, you have  to consider whether the one you picked in the first place 1. was red or not, and 2. went back in the bag.


Here's the deal.  The probability of picking a red marble on the first go is *[tex \Large \frac{16}{60}\ =\ \frac{4}{15}].  Now if you put the marble back in the bag, the probability of picking a red one on the second go is exactly the same, because you have the same total number of marbles and the same number of red marbles.  BUT, if you DON'T put the first one chosen back in the bag, then things change.


In this case, if the first marble chosen was white, then you would have 16 red marbles and 48 white marbles left in the bag for the second draw and the probability would be *[tex \Large \frac{16}{59}].  But if the first one was red, then the probability of picking a second red one is now *[tex \Large \frac{15}{59}]


See?


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