High school Math: confusion about the basic probability
I am confused about the following two scenarios:
Out of a bag of 3 apples and 3 oranges, you pick 2 items.
1) What is the probability that you will have 2 apples?
2) What is the probability that you will have 1 apple and 1 orange?
My attempt:
1) $$ \begin{aligned} P(\mbox{2 apples}) &= P(\mbox{1st apple}) \times P(\mbox{2nd apple}) \\ &= \frac{3}{6} \times \frac{2}{5}. \end{aligned} $$
2) $$ \begin{aligned} P(\mbox{1 apple and 1 orange}) &= P(\mbox{1st apple}) \times P(\mbox{2nd orange}) \\ & + P(\mbox{1st orange}) \times P(\mbox{2nd apple}) \\ &= \frac{3}{6} \times \frac{2}{5} \times 2. \end{aligned} $$
My confusion is with case number 1: why you don't need to multiply the result by 2? Since your first pick could be apple #1, #2, #3.
Consider all of the $6\times 5$ ways to pick two pieces of fruit. That's $30$: $$\boxed{\begin{array}{|l|ccc:ccc|}\hline ~ & A_1 & A_2 & A_3 & O_1 & O_2 & O_3 \\ \hline A_1 & \times & \color{green}{A_1,A_2} & \color{green}{A_1,A_3} & \color{blue}{A_1,O_1} & \color{blue}{A_1,O_2} & \color{blue}{A_1,O_3} \\ A_2 & \color{green}{A_2,A_1} & \times & \color{green}{A_2,A_3} & \color{blue}{A_2,O_1} & \color{blue}{A_2,O_2} & \color{blue}{A_2,O_3} \\ A_3 & \color{green}{A_3,A_1} & \color{green}{A_3,A_2} & \times & \color{blue}{A_3,O_1} & \color{blue}{A_3,O_2} & \color{blue}{A_3,O_3} \\ \hdashline O_1 & \color{indigo}{O_1,A_1} & \color{indigo}{O_1,A_2} & \color{indigo}{O_1,A_3} & \times & \color{red}{O_1,O_2} & \color{red}{O_1,O_3} \\ O_2 & \color{indigo}{O_2,A_1} & \color{indigo}{O_2,A_2} & \color{indigo}{O_2,A_3} & \color{red}{O_2,O_1} & \times & \color{red}{O_2,O_3} \\ O_3 & \color{indigo}{O_3,A_1} & \color{indigo}{O_3,A_2} & \color{indigo}{O_3,A_3} & \color{red}{O_3,O_1} & \color{red}{O_3,O_2} & \times \\ \hline \end{array}}$$
The ways to pick two apples are in the green quarter (upper left). There are $3\times 2$ of them; that is $6$ of $30$
The ways to pick an apple and an orange are in the blue quarter, but also in the indigo quarter (upper-right and lower-left). There are $3\times 3+3\times 3$ of them; that's $18$ of $30$
The final quarter are ways to pick two oranges. Again, just $3\times 2$ of these; that's $6$ of $30$.
The only reason you are multiplying by 2 in the second case is because you are using a shortcut due to the fact that the two scenarios that you are adding have a probability found with the same formula. You just need to add up the probabilities you are seeking.
Case 1) $\frac{3}{6}*\frac{2}{5}$
Case 2) $\frac{3}{6}*\frac{3}{5}+\frac{3}{6}*\frac{3}{5}$
Or, like I said, you could use a $*2$ in the second case, but only as an arithmetic simplification. Not because of any rules of probability.
When you state that the chance of getting an apple first is $3/6$ you are assuming that all the apples are interchangeable. If you want to label the apples you can. You can then say the chance of getting apple 1 followed by apple 2 is $(1/6)(1/5)=1/30$. Now you can find that there are six different permutations of two apples, so the total chance of getting two apples is $6(1/30)=1/5$, the same result as you give.
I think you are way over complicating the second one. The first pick is irrelevant, because no matter what, you will get one thing you need and leave five pieces of fruit, three of which are the second thing you need. Which leaves you with 3/5's.