Probability at least one spade and card higher than a 7 There is a 52 card deck and you are dealt 13 cards.
Question: What is the probability neither of the following cases happen: no spades and no card higher than 7
Thus that means you get at least one spade and the card is higher than 7. I'm not sure if it's and or or with that wording though.
I understand that the $P(\text{at least one spade}) = 1 - P(\text{no spades})$.
$$P(\text{no spades}) = \frac{\binom{39}{13}}{\binom{52}{13}}$$
But how do I factor in the second requirement of the card being higher than a 7?
 A: No cards greater than 7 would include 2,3,4,5,6 and 7.  There are 4 suits Thus (6*4) = 24.  Subtract 6 spades because it is no spades too.  Thus you have 24-6 = 18 cards.These are the outcomes that should not happen and the complement is the required probabiity.  Thus the required probability =$1-\dfrac{{18\choose 13}}{{52\choose 13}}$
A: I think it'd be easier to not separate into two events.
There are $18$ "good" cards, the $2-7$ of diamonds, clubs, and hearts and $32$ bad cards, everything else.
So there are $18 \choose 13$ positive outcomes for this to happens and $52 \choose 13$ total outcomes.
So $P$ is $\frac {18 \choose 13}{52 \choose 13}=\frac {18!39!}{52!5!} $
It's probably straightforward to combine the probabilities of no spades, and lower than $7$ but.... why bother?
So $1 - \frac {18 \choose 13}{52 \choose 13}$
[This assumes an ace is higher than a king and a two is the lowest rank.]
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If you want to consider the events as two events (but I strongly urge you don't)
P(nothing higher than seven and no spaces)= P(nothing higher than seven| no spades)P(no spades) = P(no spades|nothing higher than seven)P(nothing higher than seven)
P(nothing higher than seven| no spades)=$\frac{18\choose 13}{39\choose 13}$
P(no spades) = $\frac {39\choose 13}{52 \choose 13}$
P(no spades|nothing higher than seven)=$\frac{18\choose 13}{24 \choose 13}$
P(nothing higher than seven) = $\frac {24\choose 13}{52\choose 13}$
Clearly these three different ways will all give the same result.
