Question 610840
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Pretty obviously this represents dependent events.  If you get dealt a red card on the first card, you have lost already, but even presuming that you are dealt a black card on the first card (probability 1/2 if you don't have a joker -- neither red nor black -- in the deck), then the probability of getting a black card on the second draw is reduced to 25/51 since one of the 26 black cards in the deck is gone, namely your first card reducing both the denominator and numerator by 1.  Presuming that you got a black card on the first AND second draws, then the probability of a black card on the third draw is reduced again, this time 24/50.  And so on.


The overall probability of 4 black cards is the product of the four individual probabilities.


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