Probability of Basketball Teams

A

anon2020

New member
Joined
Feb 2, 2020
Messages
12
24 people have come to play in a basketball tournament. The 24 individuals must be divided into four teams of six. Among the 24 people, 12 are male, and 12 are female. Label the four teams A,B,C,D.

a. Find the probability that at least one of the four teams is co-ed (e.g., has at least one male and at least one female).

b. Find the probability that team A has at least one male.

I've been stuck on how to do this for hours now, any advice would be greatly appreciated!
 
Hello, and welcome to FMH! :)

a. I would begin by computing how many teams are possible. We'll call this number \(T\). When we go to fill team A, we need to compute the number of ways to choose 6 from 24. When we got to fill team B, we need to compute the number of ways to choose 6 from 18, and when we go to fill team C we need to compute the number of ways to choose 6 from 12. Team D will consist of the 6 people left over. Using the fundamental counting principle, what do you determine is the total number of teams \(T\) that can be formed?
 
MarkFL said:
Hello, and welcome to FMH! :)

a. I would begin by computing how many teams are possible. We'll call this number \(T\). When we go to fill team A, we need to compute the number of ways to choose 6 from 24. When we got to fill team B, we need to compute the number of ways to choose 6 from 18, and when we go to fill team C we need to compute the number of ways to choose 6 from 12. Team D will consist of the 6 people left over. Using the fundamental counting principle, what do you determine is the total number of teams \(T\) that can be formed?
Click to expand...
Hi and thank you!

T would be 31104 since 24·18·12·6=31104
 
anon2020 said:
Hi and thank you!

T would be 31104 since 24·18·12·6=31104
Click to expand...

I get a somewhat larger number:

[MATH]T={24 \choose 6}\cdot{18 \choose 6}\cdot{12 \choose 6}=\frac{24!}{(6!)^4}=2308743493056[/MATH]
Read my first post again, and let me know if you have any questions about the above computation. Another way we can think of this is that there are 24! ways we can order the 24 people. Now, break off the first 6 and observe that there are 6! ways we could order them and have the same team. We find the same number of ways to order the individuals in each group of 6(team) and so we take 24! and divide by 6! to the 4th power (because there are 4 teams) to get the number of possible teams. Are you with me so far?
 
  • Like
Reactions: pka
MarkFL said:
I get a somewhat larger number:

[MATH]T={24 \choose 6}\cdot{18 \choose 6}\cdot{12 \choose 6}=\frac{24!}{(6!)^4}=2308743493056[/MATH]
Read my first post again, and let me know if you have any questions about the above computation. Another way we can think of this is that there are 24! ways we can order the 24 people. Now, break off the first 6 and observe that there are 6! ways we could order them and have the same team. We find the same number of ways to order the individuals in each group of 6(team) and so we take 24! and divide by 6! to the 4th power (because there are 4 teams) to get the number of possible teams. Are you with me so far?
Click to expand...
Oh, sorry! I misunderstood, but yes I understand where that number comes from now.
 
I have to leave shortly, so I will post an outline of how I would complete part a.

Next, I would compute the number of ways that all 4 teams will include members of the same sex. We'll call this \(T_S\) (for teams of the same sex). Let's line up the 24 people with the 12 women first, and the 12 men last. We can make \((12!)^2\) such lines, and then breaking them off in groups of 6, we obtain:

[MATH]T_S=\frac{(12!)^2}{(6!)^4}[/MATH]
Let \(X\) be the event that all 4 teams have members of the same sex...we may now state:

[MATH]P(X)=\frac{T_S}{T}[/MATH]
Let \(Y\) be the event that at least 1 team is co-ed. We know it is certain either all 4 teams have members of the same sex OR at least 1 team is co-ed, hence:

[MATH]P(X)+P(Y)=1[/MATH]
[MATH]P(Y)=1-P(X)[/MATH]
So, what do you find?
 
MarkFL said:
I have to leave shortly, so I will post an outline of how I would complete part a.

Next, I would compute the number of ways that all 4 teams will include members of the same sex. We'll call this \(T_S\) (for teams of the same sex). Let's line up the 24 people with the 12 women first, and the 12 men last. We can make \((12!)^2\) such lines, and then breaking them off in groups of 6, we obtain:

[MATH]T_S=\frac{(12!)^2}{(6!)^4}[/MATH]
Let \(X\) be the event that all 4 teams have members of the same sex...we may now state:

[MATH]P(X)=\frac{T_S}{T}[/MATH]
Let \(Y\) be the event that at least 1 team is co-ed. We know it is certain either all 4 teams have members of the same sex OR at least 1 team is co-ed, hence:

[MATH]P(X)+P(Y)=1[/MATH]
[MATH]P(Y)=1-P(X)[/MATH]
So, what do you find?
Click to expand...
I got P(Y)= 0.9999996302

Thank you so much for your help, I really appreciate it!
 
anon2020 said:
I got P(Y)= 0.9999996302

Thank you so much for your help, I really appreciate it!
Click to expand...
:
Yes, I also get:

[MATH]P(Y)=\frac{2704155}{2704156}\approx0.9999996302[/MATH]
 
anon2020 said:
24 people have come to play in a basketball tournament. The 24 individuals must be divided into four teams of six. Among the 24 people, 12 are male, and 12 are female. Label the four teams A,B,C,D.
b. Find the probability that team A has at least one male.
Click to expand...
I want to address part b. Let's select one of the twelve men to be on team \(A\). That can be done in twelve ways.
Now we assign the other twenty-three people to one of the four teams where team \(A\) gets five more and the other three get six each.
This can be done in \(T_{A_1}=12\left(\dfrac{23!}{5!(6!)^3}\right)\) ways.
Now \(T=\dfrac{24!}{(6!)^4}\) is the total number of possible teams.
Using these notations, what what is the answer to part b. ?
PLEASE explain your answer in detail!
 
pka said:
I want to address part b. Let's select one of the twelve men to be on team \(A\). That can be done in twelve ways.
Now we assign the other twenty-three people to one of the four teams where team \(A\) gets five more and the other three get six each.
This can be done in \(T_{A_1}=12\left(\dfrac{23!}{5!(6!)^3}\right)\) ways.
Now \(T=\dfrac{24!}{(6!)^4}\) is the total number of possible teams.
Using these notations, what what is the answer to part b. ?
PLEASE explain your answer in detail!
Click to expand...
I think I might be doing something wrong.
For TA1, I got 6926230479168. I divided that by the total number of possible teams (T) and got 3.
 
For part b., I would examine the number of ways that team A has no males. So, we need to choose 6 of the 12 females for team A. How many ways can we do this?
 
Yes, so take the product of those two numbers to get the total number of ways team A can have no males. What do you get?
 
MarkFL said:
Yes, so take the product of those two numbers to get the total number of ways team A can have no males. What do you get?
Click to expand...
924·17153136=15849497664
So, 15849497664 ways team A can have no males. Now would I divide that by the total number of possible teams and then subtract it from 1 to get my answer?
 
anon2020 said:
924·17153136=15849497664
So, 15849497664 ways team A can have no males. Now would I divide that by the total number of possible teams and then subtract it from 1 to get my answer?
Click to expand...
That seems reasonable.
 
anon2020 said:
So, 15849497664 ways team A can have no males. Now would I divide that by the total number of possible teams and then subtract it from 1 to get my answer?
Click to expand...
That number is correct: SEE HERE

Now subtract that number from the total. You should get (2292893995392). See Here
That is the number of ways team \(A\) has at least one male player.

Now divide by the total. You should get {0.993135} as the probability that team \(A\) has at least one male player.
 
pka said:
That number is correct: SEE HERE

Now subtract that number from the total. You should get (2292893995392). See Here
That is the number of ways team \(A\) has at least one male player.

Now divide by the total. You should get {0.993135} as the probability that team \(A\) has at least one male player.
Click to expand...
I also got that as my final answer. Thank you so much!
 
You must log in or register to reply here.
Top