Why is this probability equation wrong? I have a problem stating

"There are 13 apples and 17 oranges. What is the probability of making a group of 4 with at least two apples?"

Using my logic (left side of the following equation), I first multiply a group of 2 that consists of 2 apples, and then another group of 2, which consists of the other 28 elements.
However, this is wrong; the right way to do it is to add all possible outcomes (AAOO, AAAO and AAAA) as shown on the right side of the equation
$$\frac{\binom{13}{2}\binom{28}{2}}{\binom{30}{4}} \neq \frac{\binom{13}{4}+17\binom{13}{3}+\binom{13}{2}\binom{17}{2}}{\binom{30}{4}}$$
Why is my way incorrect? Is there a smarter way to do this?
p.s.: the right answer is $\frac{1079}{1827}$ if it helps...
 A: You are counting the cases with more than three apples multiple times.  If three of the apples are $a,b,c$ you count picking $a,b$ for the first pair and picking $c$ plus an orange, then you count $a,c$ as the first pair and $b$ plus an orange, then you count $b,c$ as the first pair and $a$ plus an orange.  Similarly you count four apples six times.
A: You can use complements. That is, if $E$ is the event that there are at least two apples, then $E^c$ is the event that there are less than two apples - one or no apple, to be specific. The probability that there is only one or no apple is $$P(E^c) = \frac{\binom{17}{3}\binom{13}{1} + \binom{17}{4}\binom{13}{0}}{\binom{30}{4}} = \frac{11220}{27405}.$$
The probability of the event $P(E)$, given the probability of the complement $P(E^c)$, is given by \begin{align*}P(E) &= 1 - P(E^c) \\ P(E) &= 1 - \frac{11220}{27405} \\ P(E) &= \frac{16185}{27405} \\ P(E) &= \frac{1079}{1827}\end{align*} which is easier to compute than the event itself which has three cases.
