Picking Random Elements from Set Let $S$ be a set consisting of $6$ positiver integers and $8$ negative integers. Choose a 4-element subset of $S$ uniformly at random, and multiply the elements in this subset. Denote the product by $x$. Determine the probability that $x>0$.
Thinking about this I have 3 possible events;


*

*Event A: All of the 4 integers chosen are positive. (ex. [1, 1, 1,
1])

*Event B: The 4 integers chosen contain a odd amount of negative
numbers (ex [-1, 1, 1, 1] or [1, -1, 1, 1] or [1, 1, -1, 1] or [1, 1, 1, -1] OR  or [-1, -1, -1, 1] or [-1, -1, 1, -1] or [-1, 1, -1, -1] or [1, -1, -1, -1])

*Event $(A \cap B)$: Even amount of negative numbers (ex. [-1, -1, -1, -1] OR [-1, -1, 1, 1] or [-1, -1, 1, 1] or [-1, 1, -1, 1] or [-1, 1, 1, -1] or [1, -1, 1, -1] or [1, 1, -1, -1] or [1, -1, -1, 1])


For event A I get 1/16, event B 8/16, and event $(A \cap B)$ also 9/16. 
Not sure if I am on the right track with this question.
 A: The product is positive if and only if there are an even number of negative integers in the subset. So add the number of 4-element subsets that have no negatives, 2 negatives, and 4 negatives. Each of those counts is pretty easy.
Let's calculate the probability by dividing the number of successes by the number of trials. The trials are the 4-element subsets of $S$, which has $14$ elements, so the number of trials is
$$\text{#Trials}={14 \choose 4}$$
The successes have a positive product, which can be done by the subset of $S$ having no negative numbers, having exactly two negative numbers, and having exactly four negative numbers. Those subsets of the trials are mutually exclusive, so we can just add their counts.
The successes with no negative numbers have $0$ negative numbers (out of $8$ possibilities) and $4$ positive numbers (out of $6$ possibilities). The choices of positive and of negative numbers are independent, so we can multiply their choices to get the total number in this kind of success. This gives us
$$\text{#Successes}_0={8 \choose 0}{6 \choose 4}$$
Similarly, the count of successes with two negative numbers and thus two positive numbers is
$$\text{#Successes}_2={8 \choose 2}{6 \choose 2}$$
And the count of successes with four negative numbers and thus no positive numbers is
$$\text{#Successes}_4={8 \choose 4}{6 \choose 0}$$
Thus your desired probability is
$$P_{\text{success}}=\frac{\text{#Successes}}{\text{#Trials}}$$
$$=\frac{\text{#Successes}_0+\text{#Successes}_2+\text{#Successes}_4}{\text{#Trials}}$$
$$=\frac{{8 \choose 0}{6 \choose 4}+{8 \choose 2}{6 \choose 2}+{8 \choose 4}{6 \choose 0}}{{14 \choose 4}}$$
$$=\frac{1\cdot 15+28\cdot 15+70\cdot 1}{1001}$$
$$=\frac{505}{1001}\approx 0.504495504496$$
