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### Course: High school statistics > Unit 6

Lesson 7: Probability using combinatorics- Probability using combinations
- Example: Lottery probability
- Example: Different ways to pick officers
- Probability with permutations & combinations example: taste testing
- Probability with combinations example: choosing groups
- Probability with combinations example: choosing cards
- Probability with permutations and combinations
- Mega millions jackpot probability

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# Probability with combinations example: choosing groups

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. Created by Sal Khan.

## Want to join the conversation?

- Is it just a coincidence that there were 13 members on the team and 3 people were selected per team and the odds of any individual being selected was 3/13? Or is there some tantalizing shortcut that we haven't been introduced to yet?(15 votes)
- It's not a coincidence. If one person out of 13 was being picked at random, then Kyra would have a 1/13 chance of being picked. Since the manager is picking three people, Kyra's chance of getting picked is tripled, so P(Kyra getting picked) = 3*(1/13) = 3/13(21 votes)

- 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.(11 votes)- Your explanation is correct! I appreciate the effort you put into making it more intuitive by visualizing it with trees and branches. Your explanation of permutations and combinations is also accurate, as well as your calculation of the total number of permutations and the number of groups with Kyra in it. Your explanation of the formula for probability is also correct. Overall, great job!(4 votes)

- I'm not sure why this doesn't make sense to me, but I don't understand what Sal means when he says that "if we know that Kyra's on the team, the possibilities are the other two people on the team." Why do the other two people matter?(8 votes)
- One slight correction, Sal says, "if we know that Kyra's on the team, then the possibilities are who's gonna be the other two people on the team, and who are the possible candidates for the other two people?" Adding the who's (even though transcript uses whose) turns it into a question rather than a statement as you have it. So if Kyra is one, what possibilities do we have for the other two? is how I would paraphrase what Sal is saying.(5 votes)

- I'm with bobbysundstrom in not getting the reason "why the other two people matter?"(5 votes)
- well, it's weird that the answer is so obvious while taking more time(3 votes)
- I'm confused at1:36. He says something similar to "if we know that Kyra is on a team, then what are the possibilities for the other two people?" And for that, he uses 12 choose 2.

But wouldn't that be the same thing as saying "if we know that <someone else> is on a team, then what are the possibilities for the other two people?" Why does this only work for Kyra's possibilities?(1 vote)- It doesn't only work for Kyra.

If we for example knew that Charlie was on the team, then there are 12C2 possibilities for the other two members.

However, this doesn't mean that there are 12C2 + 12C2 combinations where either Kyra or Charlie are on the team, because this double-counts the combinations where Kyra and Charlie are both on the team and we would also have to subtract 1C11.(2 votes)

- Why doesn't (1/13)+(1/12)+(1/11) work?(1 vote)
- Because this expression not enough strict to describe probability of Kyra being chosen (algorithmically speaking).

You want something like this:`Event 1 Kyra chosen on the first attempt (`

probability of Kyra being chosen

) OR

Event 2 Kyra chosen on the second attempt (

probability of Kyra not being chosen first time AND

probability of Kyra being chosen on the second attempt

) OR

Event 3 Kyra chosen on the third attempt (

probability of Kyra not being chosen first time AND

probability of Kyra not being chosen on the second attempt AND

probability of Kyra being chosen on the third attempt

)

As math expression: (1/13) + (12/13 * 1/12) + (12/13 * 11/12 * 1/11) = 1/13 + 1/13 + 1/13 = 3/13

Sorry for wording and grammar, English isn't my first language.(2 votes)

- i get 13 choose 3 and 12 choose 2 but why did we not do 11 choose 1?(2 votes)
- Where would this go? The chance that Kyra was on the committee (which is the question) is number of committees Kyra is on/total possible number of committees. 12 choose two tells number of committees that Kyra is on, thus the numerator, and total committees is 13 choose 3 which is the denominator.(0 votes)

- Why is twelve divided by two but not eleven? It goes from (12*11)/2 to 6*11(1 vote)
- What if Kyra wasn't at the 13 members team first and we choose 3 people from 12? The outcomes will not include Kyra(Whatever the name) definitely and we can calculate the possibility of the chosen teams that is not included Kyra, which if you use 1 to minus, I guess it will still be the possibility of Kyra was in.But the answer just seems different to Khan's.So does the name matter or I just can't write expression of that Kyra wasn't in the team in that way?(1 vote)
- Sorry is my fault. I check the answer they are the same. Hopefully not misleading someone(1 vote)

## Video transcript

- [Instructor] We're told that Kyra works on a team of 13 total people. Her manager is randomly
selecting three members from her team to represent
the company at a conference. What is the probability
that Kyra is chosen for the conference? Pause this video and
see if you can have a go at this before we work
through this together. All right, now let's work
through this together. So we wanna figure out this probability. And so one way to think about it is, what are the number of
ways that Kyra can be on a team or the number of possible teams, teams with Kyra, and then over the total
number of possible teams, total number of possible teams. And if this little hint
gets you even more inspired. If you weren't able to
do it the first time, I encourage you to try to pause it again and then work through it. All right, now I will
continue to continue. So first let me do the denominator here. What are the total
possible number of teams? Some of y'all might've found that a little bit easier to figure out. Well, we know that we're
choosing from 13 people and we're picking three of them and we don't care about order. It's not like we're saying
someone's gonna be president of the team, someone's gonna be vice-president and someone's gonna be treasurer. We just say there are
three people in the team. And so this is a
situation where out of 13, we are choosing three people. Now, what are the total number of teams, possible teams that could have Kyra in it? Well, one way to think
about it is if we know that Kyra is on a team,
then the possibilities are who's gonna be the other
two people on the team, and who are the possible candidates for the other two people? Well, if Kyra is already on the team then there's a possible
12 people to pick from. So there's 12 people to choose from for those other two slots. And so we're gonna choose two. And once again, we don't care about the order with which
we are choosing them. So once again, it is
gonna be a combination. And then we can just go ahead and calculate each of
these combinations here. What is 12 choose two? Well, there's 12 possible people for that first nine Kyra's seat. And then there would be 11 people there for that other non Kyra's spot. And of course it's a combination. We don't care what order we picked it in. And so there are two ways
to get these two people. We could say two factorial but that's just the same
thing as two or two times one. And then the denominator here. For that first spot, there's
13 people to pick from , then in that second spot, there are 12. Then in that third spot, there are 11. And then once again, we
don't care about order, three factorial ways to
arrange three people. So I could write three times two, and for kicks I could
write one right over here, and then we can, let's go down here. This is gonna be equal to my numerator over here is gonna be six times 11. And then my denominator is
going to be 12 divided by six right over here is two. So it's gonna be 13 times 11 times two. Just to be clear, I divided
both the denominator and this numerator over here by six to get two right over there. Now this cancels with that. And then if we divide the
numerator and denominator by two, this is gonna be three here. This is gonna be one. And so we are left with a probability of 3/13 that Kyra is
chosen for the conference.