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Suppose I have a deck of 26 black and 26 red cards. I randomly choose 26 cards from this deck and I look at the first two cards. What is the probability that both cards are black?

I think the probability is $$ \frac{1}{2}\frac{25}{51} = \frac{25}{102}$$ since we are choosing 26 cards randomly and this is equivalent to having a total deck. Is my reasoning correct?

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    $\begingroup$ Choosing the 26 cards randomly and checking the first two cards is same as checking the first two cards from the deck. The answer is correct. $\endgroup$ – tpb261 May 29 '14 at 11:04
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wrong there are two cases

first, withdrawing with replacement will be 26/52*26/52=0.25

Secondly, withdrawing without replacement will be 26/52*25/51=0.245

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YES!

The experiment is like picking one card at random from the deck, and then another.

How likely are both black? Use the rule that says P(A and B) = P(A)P(B|A) where A and B are the probabilities that the first is black, and that the second is black.

How likely is it that the first is black? 1/2

Given that the first is black, how likely is it that the second draw is too? only 51 cards left, of which 25 are black... 25/51.

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