# Probability (Permutation Combination)

• Sai Alonzo

## Homework Statement

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

## 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?

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)

BvU said:
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)

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

Sai Alonzo said:

## Homework Statement

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

## Homework Equations

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

Is this right?

## 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?

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 ?

BvU said:
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 ?

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

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

## Homework Statement

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

## Homework 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?

## 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​

Sai Alonzo said:
would this be correct
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 ?

BvU said:
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 ?

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?​

If I follow you correctly, I get a probability of 2.16 10-7
Sai Alonzo said:
P(A) =
Number of possible ways to form group 1:

Number of ways Paula, Trina, Gia, Alex, Luis can be arranged in group 1:
5C5 = 1 ##\qquad## That would be 1. I don't agree: PTGAL is 1, PTGLA is 2 etc etc. How many of the groups 1 have these 5 persons in them ?

Neither of the 5 is in group 1:
17C5 = 6188 ##\qquad## Oh ? Why is that? I don't agree

Total possibilities of all groups:
22C5 x 17C5 x 12C6 x 6C6 ##\qquad## I agree. Do you see that this can also be written as ##\displaystyle {22!\over 5! \, 5!\, 6!\, 6! } ## ?

P(A) = 22C5 + 17C5 + 5C5
Sai Alonzo said:
is this it
Don't think so ...

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

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 ?

BvU said:
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 ?

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 ?

Sai Alonzo said:
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 ?

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 :(

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)

BvU said:
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)

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:
Hello Ray,

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

BvU said:
Hello Ray,

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

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.