Q:

4 apples, 2 worms - probability

Well I'm still going to school and I read that this forum is for ANY level so I hope you can answer my question.
I got probability in school and the task is the following (poorly translated) :
"You have 4 apples, 2 of them have worms. You take 2 apples at once. What is the probability that you just took the two apples with worms?"
I thought hmm.. 4 apples total? 2 Apples with worms so 
P(both apples have worms) = 2/4.
The correct answer is 1/6, but why?

A:

The probability of picking one apple with a worm is $\dfrac 24$ (since two of the four apples have a worm). Then, after having already picked an apple with a worm, there are then three apples left, only one of which has a worm. So the probability of picking a second apple with a worm is $\dfrac 13$. 
Multiplying the probabilities gives us the probability of picking both apples with worms 
$$\dfrac 12 \times \dfrac 13 = \dfrac 16$$

A:

Let's name the apples $A$, $B$, $C$, and $D$. How many different ways are there of choosing two apples?
$$A\text{ and }B\\\\ A\text{ and }C\\\\ A\text{ and }D\\\\ B\text{ and }C\\\\ B\text{ and }D\\\\ C\text{ and }D$$
There are $6$ ways of choosing two apples. Only $1$ of these ways will be the one in which both of the wormy apples are chosen. Therefore the probability is $1/6$.
I recommend reading the Wikipedia article on binomial coefficients. In general, the number of ways of choosing $k$ things out of $n$ things (where the order of the choice doesn't matter) is
$$\binom{n}{k}=\frac{n!}{(n-k)!k!}$$
where $m!$ denotes the factorial of $m$. We can see that this matches our result above: the number of ways of choosing $k=2$ things out of $n=4$ things is
$$\frac{4!}{(4-2)!2!}=\frac{4!}{2!\times 2!}=\frac{4\times 3\times 2\times 1}{(2\times 1)\times(2\times 1)}=6$$

A:

There are $\binom{4}{2}=6$ pairs of apples that may be chosen, and there is only one pair that both have worms.

