# Homework Help: Probability (Permutation Combination)

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1. Oct 2, 2017

### Sai Alonzo

1. The problem statement, all variables and given/known data
There are 22 students in a class. The professor will divide the class into 4 groups. Group 1 and 2 have 5 members each whilst Group 3 and 4 have 6. Given that the teacher forms the group at random, find the probabilities of :

A = event where Paula, Trina, Gia all belong in Group 1
B = event where Paula and Trina are in Group 1
C = event where the 2 students, Paula and Trina belong in the same group
2. Relevant equations

3. The attempt at a solution
I'm not sure how to work this problem since the number of people in each group is different? I'm confused so can someone please help me?

Group 1 = (22C5)
Group 2 = (17C5)
Group 3 = (12C6)
Group 4 = (6C6)

Do I multiply these to get all possible groups?

2. Oct 2, 2017

### BvU

Hello Sai,

With (22C5) I do not think you designate a probability ? Does it mean $22!\over 17!\,5!$ ?
Perhaps you should quote the relevant relationships under 2. -- with some explanation about the meaning (mainly for yourself)

3. Oct 2, 2017

### Sai Alonzo

Yes sorry. It's the combinations, but I'm not really sure about it.

4. Oct 2, 2017

5. Oct 2, 2017

### BvU

So there are (22C5) ways to pick a group of 5 from a sample of 22. What does that have to do with the probability that Paula is in that group ?

6. Oct 2, 2017

### Sai Alonzo

Sorry, I was actually trying to calculate the total number of possibilities for all groups, not yet of the part of Paula.

7. Oct 2, 2017

### BvU

In that case the answer to your original question in #1 is: Yes !

8. Oct 2, 2017

### Sai Alonzo

1. The problem statement, all variables and given/known data
There are 22 students in a class. The professor will divide the class into 4 groups. Group 1 and 2 have 5 members each whilst Group 3 and 4 have 6. Given that the teacher forms the group at random, find the probabilities of :

A = event where Paula, Trina, Gia, Luis, Alex all belong in Group 1
B = event where Paula and Trina are in Group 1
C = event where the 2 students, Paula and Trina belong in the same group
2. Relevant equations

Group 1 = (22C5) = 26334 possible ways for group 1
Group 2 = (17C5) = 6188
Group 3 = (12C6) = 924
Group 4 = (6C6) = 1

Is this right?

3. The attempt at a solution
I'm not sure how to work this problem since the number of people in each group is different? I'm confused so can someone please help me?

--------------------------------------------------------------------------------------------------
sorry! I didn't copy the given right. I've added Luis and Alex to event A.

For finding the probabilities, would this be correct?

P(A),
22C5 (which is the number of possible ways to form group 1) x 5C5 (which is the number of ways Paula, Trina, Gia, Alex, Luis can be arranged) / total number of ways

P(B),
22C5 x 5C2 (no. of ways to arrange Paula, Trina) x 5C3 (no. of ways to arrange the 3 others part of the group)
total number of ways​

9. Oct 2, 2017

### BvU

Hehe, statistics isn't a game of trial and error
You want to explain why you think this would be the correct answer. (and my guess is you'll then find that there's a mistake in your reasoning).

You position will be strong when you can come to an identical answer when you reason from the other end (very powerful!): what is the probability that neither of the five is in group 1 ?

10. Oct 2, 2017

### Sai Alonzo

P(A) =
Number of possible ways to form group 1:
22C5 = 26334

Number of ways Paula, Trina, Gia, Alex, Luis can be arranged in group 1:
5C5 = 1

Neither of the 5 is in group 1:
17C5 = 6188

Total possibilities of all groups:
22C5 x 17C5 x 12C6 x 6C6

P(A) = 22C5 + 17C5 + 5C5
total

is this it?​

11. Oct 2, 2017

### BvU

If I follow you correctly, I get a probability of 2.16 10-7
Don't think so ...

12. Oct 2, 2017

### Sai Alonzo

Number of ways Paula, Trina, Gia, Alex, Luis can be arranged in group 1:
5! = 5 x 4 x 3 x 2 x 1

Neither of PTGAL is in group 1:
17!

Sorry for being slow :( This topic gets me so confused

13. Oct 2, 2017

### BvU

Don't worry about confusion: you're definitely not the only one.
I may even have made it worse by commenting on the 5C5 =1 ; if so: sorry.

As you write: there are 22C5 different groups 1 (if you don't look at the order within the group -- that's why you divide $\displaystyle{22!\over 17!}$ by $5!$). How many of those satisfy the criterion that the 5 persons mentioned are the memebers ?

14. Oct 2, 2017

### Sai Alonzo

Without looking at the order within the group, there would only be 1 way that the 5 persons mentioned are the members of Group 1 ?

15. Oct 2, 2017

### Sai Alonzo

Since the submission of this homework is already done anyway, I would appreciate if you could explain to me step by step as I don't understand for the most part :(

16. Oct 2, 2017

### BvU

That's not how PF works, I'm afraid.
However, you have already provided the answer yourself: there are 26334 different groups 1 and only one of them contains our five members. All groups are equally likely, so the desired probability is 1/26334.

'Another' way to come to the same result:
One ordering of Paula, Trina, Gia, Alex, Luis is PTGAL

Line up the 22 classmates one by one. Probabilty P is in place 1 is 1/22
In such cases, probability T in place 2 is 1/21, etc. So PTGAL has a probability 1/(22*21*20*19*18) = 17! / 22!

There are 120 permutations of PTGAL, so the probability of all 5 in group 1 is 5! * 17! / 22! or 1/(22C5).

If you can follow this, then the next task is to find P(B)

17. Oct 2, 2017

### Ray Vickson

I get a different answer for P(A).

There are two groups of people. Group I, has the three members 1,2,3 and Group II has the remaining 19 members. We want to choose n = 5 members from the population of N = 22. The answer for P(A) is the probability that a random sample of 5 members has 3 members from Group I and 2 members from Group II. This is a classic problem in the so-called hypergeometric distribution.

We have a population of N = 22 objects consisting of sub-populations of sizes N1 = 3 and N2 = 19. If X is the number of Group-I members in the sample of size n = 5, its probability distribution is
$$P(X = k) = \frac{C(N_1,k) C (N_2, n-k)}{C(N,n)}.$$
For $N = 22, N_1 = 3, N_2 = 19, n=5$ and $k=3$ this gives $P(A) = P(X = 3) = C(3,3)\, C(19,2)/C(22,5) = 1/154.$

For more about the hypergeometric distribution, see, eg.,
https://en.wikipedia.org/wiki/Hypergeometric_distribution or
https://onlinecourses.science.psu.edu/stat414/node/58

Last edited: Oct 2, 2017
18. Oct 2, 2017

### BvU

Hello Ray,

Sai has changed the number of folks in group 1 to 5 instead of 3 ....

19. Oct 2, 2017

### Ray Vickson

Yes, I know that; that is why the chosen sample size is n = 5. However, he/she wants to know the probability that the three students Paula, Trina and Gia are chosen, and MYGroup I = {Paula, Trina, Gia}. Therefore, we need to choose 3 from MyGroup I (that is, all three) and 2 from MYGroup II = {everybody else} of size 19. (Note that my use of the word "group" is different from that of the OP, so maybe that is a cause of confusion.)

Another way to do the computation is to go back to basics, as follows. There are 22! permutations altogether, and we look at the contents in the first n = 5 positions. How many permutations have Paula, Trina and Gia among the first 5 locations? Well, choose other members x and y from the remaining 19, so we look at Paula, Trina, Gia, x, y in positions 1--5. There are 5! ways of permuting those 5, and 17! ways of permuting all the rest, so 5!*17! permutations with Paula, Trina, Gia, x, y in the first five. However, this applies to each distinct pair (x,y), and there are C(19,2) choices for that pair. Thus, the total number of relevant permutations is M = 5! * 17! * C(19,2). That means that P(A) = 5!*17!*C(19,2)/22! = 1/154, just as delivered by the hypergeometric distribution.

20. Oct 2, 2017