Probability with combinations example: choosing groups | Probability & combinatorics

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Sal shows how we can use two combinations (when order does not matter) to find the probability of someone being included in the group that's chosen at random from a larger group.

[cariboux]

Why on God's green Earth would someone go through this? It's 3 out of 13, without any discussion. The question is the ANSWER. So confused.

[TasyaAdzkiya]

To make it more intuitive, I have a more elaborate (I guess) explanation that I've come up myself. I recommend you to visualize it with trees and branches.

First, we have to understand that 13 x 12 x 11 means that there are 13 initial scenarios that each have 12 branches because those are possible second outcomes for a given first outcome. So far, we have 13 x 12 = 156 different scenarios. Then, each of the the 156 scenarios branches into 11 branches because those are the possible third outcomes for a given second outcome. Now we've got 156 x 11 = 1716 total scenarios (and we care about the order because this is permutation). 1716 is our total number of permutations.

Let's take our total permutations and calculate the total number of groups of (in this case) 3, for which we call them the combination because combination is essentially a way to find out how many combinations/groups there are. We use combinations because there is no mention about the position (for example: leader, et cetera), so we assume that they don't care about which seat that a person sit on (order does not matter).

Number of combinations or groups = (total number of permutations [order matters])/(total number of ways to arrange the things in a single group [order matters]). This formula is intuitive because the total number of permutations (order matters) = total number of ways to arrange the things in a single group (order matters) x the number of groups or combinations. That is because the total number of permutations is just the TOTAL (not just of a single group) arrangements of things (order matters). So, because there will be 3 people in a group, the number of arrangements in a single group is just 3! (3 factorial). Basically we are counting the permutations of a single random group because all group will have the same number of permutations (that is because we have the same number of seat that they can take on in a single group).

So, let's think about why the number of teams with Kyra in it is 12C2. 1:36 "If we know that Kyra is on a team, then the possibilities are who is going to be the other two people on the team." So, we pick from 12 people (Kyra is excluded because she is definitely will be on the team and we are calculating the PROBABILITY of the combinations of the other 2 people that will go with Kyra) to choose the other 2 people that will be with Kyra on the team. If you calculate it mathematically (which we are), you can see that it's intuitive:

Let's first think about the total number of permutations with Kyra in it. It would be 12 x 11 because from those 12 people left (Kyra excluded because she is definitely is on the team), each of the 12 scenarios will branch into 11 branches because after choosing the person that will occupy the second seat (first seat is definitely Kyra's), we have to choose one person out of 11 people left to occupy the third seat. 12 X 11 = 132 different scenarios.

Or you can also imagine that the first scenario is just 1 scenario (which is Kyra's scenario) that will branch. Note there is a difference between total permutations without Kyra in it and total permutations with Kyra in it: the first scenario of the total permutations is 13 possible scenarios because we must choose one person out of 13 people to occupy the first seat. Just IMAGINE it as this: 1 x 12 x 11. This is the total number of permutations (order matters) with Kyra in it. I didn't write it because even if it doesn't change the result (it's still 132 scenarios), it shows us that there are 3 things/seat in a group. It differs from Sal's calculation, which is only 2 seats for the other 2 people that will be with Kyra on the team. It matters because the number of things/seat in a group will affect the number of ways to arrange things in a group or the number of arrangements considering the order (which is the denominator), so I don't want to mess with that.

The number of groups/combinations with Kyra in it = (the total permutations with Kyra in it [order matters])/(the total number of ways to arrange things in a group or the total number of arrangements in a group [order matters]). Because there will be 2 people in a group (people that will be with Kyra in a group), the number of ways to arrange the 2 people in a group is just 2! (2 factorial).

Lastly, we divide the number of combinations or groups with Kyra in it by the number of combinations or groups in total because it's just the formula for probability. That is just the total number of events that fit our constraints divided by the total number of possible outcomes.

Please correct me if I'm wrong.

Additional explanations from hints on Khan Academy website (different word problems, same drill, so I modify it to suit this exact problem):
We want to count all of the groups that include Kyra. This is equivalent to how many groups of 2 students are possible from her other 12 teammates. Kyra could then "join" any of those groups, and we have counted how many groups of 3 people including Kyra.

[martinoberto]

I solved it in one equation. The odds of three out of thirteen people chosen is 3/13

[ttgg]

this is a quite confusing question ...

[Hobbit]

wrong, it's actually 50%. She either gets chosen or not. ^^