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In a game of bridge, choose 13 cards from a deck at random (use equally likely probability). Question: What is the probability you get NO spades? What is the probability you get no card higher than 9 (with aces considered high cards)? What is the probability neither of these occur?

I just wanted to know if I'm on the right track for this question, to calculate the probability that neither event occurs would it be equal to $P((A ∪ B)^c)$?

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  • $\begingroup$ Depends on what $A$ and $B$ are. $\endgroup$ – Thomas Andrews Oct 18 '15 at 18:45
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Let $A$ be the event "no spades"

Let $B$ be the event "no card higher than $9$"

We have $$P(A)=\frac{\binom{39}{13}}{\binom{52}{13}}$$

$$P(B)=\frac{\binom{32}{13}}{\binom{52}{13}}$$

$$P(A\cap B)=\frac{\binom{24}{13}}{\binom{52}{13}}$$

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

And finally $1-P(A\cup B)$ is the desired probability. The result is $0.9867$.

So your idea is absolutely correct.

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Yep, the probability that neither $A$ nor $B$ occurs is the probability that both $A$ and $B$ do not occur. The truth table for NOR is,

A B   A NOR B
0 0   1
0 1   0
1 0   0
1 1   0

Compare this to the truth table for not ($A$ and $B$) to confirm that you're right.

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