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Two cards are drawn without replacement, the find the probabilility that the first card is a non face and the second is a black:

Non face would be 40/52 but how would I get the second part?

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The probability the first pick is a red non-face is $\frac{20}{52}$. That leaves $26$ black out of $51$. So the probability of a red non-face followed by a black card is $\frac{20}{52}\cdot \frac{26}{51}$.

The probability the first pick is a black non-face is $\frac{20}{52}$. In that case the probability the next pick is black is $\frac{25}{51}$. So the probability of a black non-face followed by a black is $\frac{20}{52}\cdot \frac{25}{51}$.

Add. There is much simplification.

Another way: By symmetry, the probability of non-face followed by black is the same as the probability of non-face followed by red.

The probability of non-face is $\frac{40}{52}$. The event "first is non-face" is the disjoint union of the events "first is non-face, second is black" and "first is non-face, second is red." Each has equal probability, so each has probability $\frac{20}{52}$.

We can also express this in the language of conditional probability: the probability the second is black given first is non-face is $\frac{1}{2}$. The events "first is a non-face card" and "second is black" are independent. So to find the proability of first non-face and second black, we multiply $\frac{40}{52}$ by $\frac{1}{2}$.

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