How many possible ways can I choose a 5 card hand of 3 spades from a deck of 52 cards I have the answer to this question. But I'm not 100% sure how it works. My original way of approaching this was to get the sample space, 52 choose 5, and use bernoulli trials to get the probability of 3 spades out of 5 cards. Then multiply this probability times the sample space. However this didn't work since bernouli trials only work for independent events. I think Im one of those guys who just doesn't have an intuition for probability but if someone could lead me in the right direction that would be great. There is also one more question and I find it quite ambiguous. 
If I roll 3 fair dice, what size is the event where second largest value of all 3 dice is 5?
Does this allow for (6,6,5) or (6,5,5). Not sure.
Also if I have to post these as 2 separate questions let me know.
Thanks for any answers in advance. 
 A: 
I have the answer to this question. But I'm not 100% sure how it works. My original way of approaching this was to get the sample space, 52 choose 5, and use bernoulli trials to get the probability of 3 spades out of 5 cards. Then multiply this probability times the sample space. However this didn't work since bernouli trials only work for independent events. 

You're right.   This isn't a series of independent Bernouli events so it is not a Binomial Distribution.
The approach to use is called the Hypergeometric Distribution.
$$\frac { \binom{13}{3} \binom{52-13}{5-3} } { \binom{52}{5} }$$
The favoured space is selections of 3 from 13 spades, and 2 from the non-spades.   The total space is selections of any 5 cards from the deck of 52.   Divide and calculate.

There is also one more question and I find it quite ambiguous. If I roll 3 fair dice, what size is the event where second largest value of all 3 dice is 5? Does this allow for (6,6,5) or (6,5,5). Not sure. Also if I have to post these as 2 separate questions let me know. Thanks for any answers in advance. 

Yes, this is slightly ambiguous.   However, it is asking for the second highest value not the second highest distinct value.
So, in my opinion then, it would best be interpreted as asking for the second highest order statistic.   That is where you order the dice from lowest to highest and then find the value that is the second last.
So the second largest value of $(5,6,6)$ is $6$, while the second largest value of $(5,5,6)$ is $5$.
So you are counting permutations of $(x,5,6)$ and $(x,5,5)$ where $x\in\{1,2,3,4,5\}$.  (Be careful with the end case.)

But your measure may vary.   If you wish to read it as the second highest distinct value then you need to count permutations of $(x,5,6)$ where $x\in \{1,2,3,4,5\}$ (and, again, be careful with the end case).
