# Combinatorics select 3 cards out of 52, probability all 3 spades and not. You get 3 cards out of 52

Combinatorics select 3 cards out of 52, probability all 3 spades and not.
You get 3 cards out of 52, the order doesn't matter.
a) What is the probability all 3 cards are spades?
b) What is the probability none of the 3 cards are spades?
a) I think it is
$\frac{\left(\genfrac{}{}{0}{}{13}{3}\right)}{\left(\genfrac{}{}{0}{}{52}{3}\right)}\approx 0.012941$
b) I know it is
$=\frac{\left(\genfrac{}{}{0}{}{39}{3}\right)}{\left(\genfrac{}{}{0}{}{52}{3}\right)}\approx 0.41353$
I know $P\left({A}^{c}\right)=1-P\left(A\right)$
And if a is $A$ and b is ${A}^{c}$ the equation above is not correct..
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Karla Hull
Let P(n) be the probability that n cards are drawn. P(3) is the answer for a), and you got that right. The answer for b) is P(0).
$P\left(0\right)+P\left(1\right)+P\left(2\right)+P\left(3\right)=1$, so $P\left(0\right)+P\left(3\right)<1$
$P\left(n\right)=\frac{\left(\genfrac{}{}{0}{}{13}{n}\right)\left(\genfrac{}{}{0}{}{39}{3-n}\right)}{\left(\genfrac{}{}{0}{}{52}{3}\right)}$
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Ximena Skinner
The issue here is that the complement of the event "all of the 3 cards are spades" is "at least one of the 3 cards is not a spade" not "none of the 3 cards are spades".
So, $P\left(‘‘\text{all of the 3 cards are spades}"\right)+P\left(‘‘\text{none of the 3 cards are spades}"\right)<1$