# Weird probability question - Grapes & olives

Does anyone know how to solve this?

"A basket has 3 grapes and 2 olives. If two were taken out and random, what is the probability of picking both olives?"

-My first thought is that the total number of possible outcomes would be 3 grapes(G) + 2 olives(O) = 5, but once I start writing out the possible outcomes in a chart, it gets way more complicated.

Eg. -Grape 1: (G1,G2) (G1,G3) (G1,O1) (G1,O2) -Grape 2: (G2,G1) (G2,G3) (G2,O1) (G2,O2) ...etc. All the way through grape 3, olive 1 & olive 2.

The part that keeps confusing me, is where I get reoccurring picks, (G1,G2) or (G2,G1). Should I even be labelling each individual grape? Or just count them as 3 grapes in general?

You can label the grapes/olives $B=\{G_1,G_2,G_3,O_1,O_2\}$. You have a set of five distinct objects, and then can think of drawing two as choosing two of the five, so there are $C(5,2)=10$ possible "hands" you could draw. This is picking size two $\textit{subsets}$, so order doesn't matter. Only one of these hands has both olives, so that the probability is $\frac{1}{10}$.

You could take order into account, and get the same result. There are $5\cdot4=20$ ordered pairs of objects from the set $B$, and there are two hands with both olives $(O_1,O_2)$ and $(O_2,O_1)$, so the probability is $\frac{2}{20}=\frac{1}{10}$.

In either case labelling the grapes and olives is a good idea; they are all distinct objects, just falling into groups based on common properties (like being a grape).

It's not so weird a problem; it's quite common in fact.

You are counting equally-probable, distinct ways to obtain the favoured result and dividing by ways to obtain all results.   That is exactly what you should be doing.   Just make sure you count both in the same way -- with or without order mattering.

So in total there are $5\times 4$ ways to choose two pieces of fruit, when order matters.   There are $5\times 4 \mathop {/} 2$ ways to do so when order does not matter.

$$(G_1, G_2), (G_1, G_3), (G_1, O_1), (G_1, O_2), (G_2, G_3), ... (O_1, O_2) , \\ (G_2, G_1), (G_3, G_1), (O_1, G_1), (O_2, G_1), (G_3, G_2), ... (O_2, O_1)$$

For the favoured event: there are $2\times 1$ ways to either pick two olives when order matters, or if order does not matter, there are simply $1$ ways to do so.

$$(O_1, O_2) \\(O_2, O_1)$$

$$\dfrac{2\times 1}{5\times 4} = \frac{1}{5\times 4\mathop{/} 2} = \frac 1 {10}$$

Method 1 (with strict order): Label the round things g1, g2, g3, o1, o2.

You have 5 choices for the first item and 4 for the second. This is 5*4 =20 options. Two of them are both olives. o1:o2 and o2:o1. So the probability is 2/20 = 1/10.

Method 2 (without order): There are 5 choices for the first item and 4 for the second so that's 4*5 =20. But order doesn't matter. There are two ways to order to items so the total ways of choosing two items is 5*4/2 = 10. [ ${5} \choose {2}$$= \frac{5!}{2!(5-2)!}]$ Of these only 1 has both olives so the probability is 1/10.

I have a different take on your dilemma.

The question here is about probability, so it doesn't really matter whether you take or don't take order into account, i.e. whether you use permutations or combinations, the factors will cancel out (as many of the previous answers have demonstrated).

However, the question could instead have been "How many ways could you pick both olives if two were taken out at random ?"

Now you are faced with the dilemma whether the answer is 1 or 2, and the general guideline I would suggest is that if the data is classified, e.g. grapes and olives here, or men and women for a committee, you do not consider order, and use combinations, so here $\binom22 = 1$.

If each entity is to be considered as distinct, clear indications will normally be there.