# 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.

• Shouldn't your second example be (3/6)*(3/5)*2? There are still 3 of the second fruit after successfully drawing the first. Nov 16, 2015 at 17:18

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 $30$.

• +1 for your coloured picture, which must have necessitated some time! Nov 17, 2015 at 0:17

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.

• Care to explain downvote? Nov 17, 2015 at 6:18
• I'm not the downvoter. I will say this, though... although your approach looks fine to me, please do not ever accuse someone of "overcomplicating" a probability problem. I have seen far too many cases where attempting to look at things in a simple way leads to an erroneous result. Probability problems must be handled with care, and doing everything the long, complex way is one way to ensure that you're doing it right.
– ajb
Nov 17, 2015 at 6:34
• @ajb so you can never over-complicate your approach to a probability problem? Nov 17, 2015 at 13:32

This answer is a sort of side-shot way of counting, based on Kemp's answer.

We know that we have $3$ oranges and $3$ apples. We assume that they are randomly chosen without replacement, with equal probability. Much like the binomial theorem, consider the following multiplication:

$$1 = 1\cdot 1 = \Big(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\Big)\Big(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\Big)$$

What does this represent? It is all the possible combinations we could have involving different ways of picking apples and oranges. This is a direct explanation regarding the probabilities. That is, the each of the $1/6$ in the first brackets tells us about the probability of choosing an apple or an orange, whereas the second group tell us the probability of picking an apple or an orange after picking the first.

Noting that there are $6$ terms on the left and $5$ on the right, we will have $30$ total terms before cancellation (as mentioned above). Likewise $3$ of the $6$ fractions, $1/6$ correspond to apples and $2$ of the fractions $1/5$ would correspond to an apple after first picking an orange. Thus, there are $3 \cdot 2$ of the fractions $\frac{1}{30} = \frac{1}{5 \cdot 6}$ which correspond to apples begin picked twice.

Although this amounts to the exact same things as Kemp mentioned, it is interesting to see that the sample space is preserved--this is indeed, always, the case. In addition, it then amounts to a simple counting argument; remembering that multiplication is commutative, we could have pick an apple out of a bag with three oranges and two apples then a bag with three of each. It doesn't make a difference, therefore, we don't need to multiply by a factor of two.