Question 1163964
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The probability of drawing a red card on the first draw is *[tex \Large \frac{26}{52}\ =\ \frac{1}{2}], because there are 52 cards to draw and 26 of them are red.  The probability of the second card being red, given that the first one was red and you didn't replace it, is *[tex \Large \frac{25}{51}] because there are only 51 cards to draw and only 25 of them are red.  The probability of the third card being red is, given that the first two were red and you didn't replace either one of them is *[tex \Large \frac{24}{50}\ =\ \frac{12}{25}] because now there are only 24 red cards left and only 50 cards to choose from.


Since the numbers of possibilities and the numbers of successes were changed to account for success of the previous draw, the events are independent.  Therefore the multiplication rule applies. You can do your own arithmetic.

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