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What is the probability of drawing a heart and then a spade in 2 successive draws from a standard deck of cards? Do we consider these as independent events thus yielding:

$$\Pr(\text{Spade and Heart})=\Pr(\text{Spade})\times\Pr(\text{Heart})\rightarrow\frac{13}{52}\times\frac{13}{52}$$

or conditional so that:

$$\Pr(\text{Spade then Heart})=\Pr(\text{Spade})\times\Pr(\text{Heart})\rightarrow\frac{13}{52}\times\frac{13}{51}$$

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    $\begingroup$ As a friendly piece of advice, people here are more likely to help if you format your question correctly, and put the question in the body as well. See here: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – The Count Jun 24 at 0:12
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    $\begingroup$ I think the answer is within the phrase "successive draws". $\endgroup$ – ArsenBerk Jun 24 at 0:18
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A couple of clarifying points.

  1. "Drawing a heart and a spade" is a different statement than "Drawing a heart then a spade."
  2. One might assume that "successive draws" indicates that the draws are dong without replacement, but it's not explicitly stated.

If you're wanting to draw a heart and a spade, then you could get the heart first, or the spade first. The probability of doing this with replacement is $2(1/4)(1/4) = 1/8$. Doing it without replacement is $2(1/4)(13/51) = 13/102$. (The $2$ out front considers heart then spade, and spade then heart.)

If you're drawing a heart then a spade, the answers are half as much, because you must get the heart first.

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