Permutations and Combinations Help! (1 Viewer)

[Doctor Jolly]

1. A basketball team of 2 men and 3 women is to be chosen at random from a group of 5 male and 7 female players. If George and his girlfriend both try out of the team, find the probability that neither will be chosen in the team.

Answer: 12/35

My working out:

4C1 x 6C2
= 60

5C2 x 7C3
=350

60/350
= 6/35

1- 6/36
= 29/35

Where did I go wrong?

 

[YO!]

AHHH NOOOO!! PERMS AND COMBSS!! NOOO!

Seriously, just the name gives me shivers.

I don't know the answer to your question, sorry.

Perms + Combs: Hardest Maths topic.

 

[Doctor Jolly]

AHHH NOOOO!! PERMS AND COMBSS!! NOOO!

Seriously, just the name gives me shivers.

I don't know the answer to your question, sorry.

Perms + Combs: Hardest Maths topic.

Agreed. You have to think so much

 

[roadrage75]

1. A basketball team of 2 men and 3 women is to be chosen at random from a group of 5 male and 7 female players. If George and his girlfriend both try out of the team, find the probability that neither will be chosen in the team.

Answer: 12/35

My working out:

4C1 x 6C2
= 60

5C2 x 7C3
=350

60/350
= 6/35

1- 6/36
= 29/35

Where did I go wrong?

probability that george is not chosen: 4C2/5C2 = 6/10 =3/5
probability that girl not chosen: 6C3/7C3 = 20/35= 4/7

3/5 x 4/7 = 12/35

 

[Doctor Jolly]

probability that george is not chosen: 4C2/5C2 = 6/10 =3/5
probability that girl not chosen: 6C3/7C3 = 20/35= 4/7

3/5 x 4/7 = 12/35

Ohh... so you have to work it out separately? You can't do it together like what I did?

 

[bored of sc]

1. A basketball team of 2 men and 3 women is to be chosen at random from a group of 5 male and 7 female players. If George and his girlfriend both try out of the team, find the probability that neither will be chosen in the team.

Answer: 12/35

My working out:

4C1 x 6C2
= 60

5C2 x 7C3
=350

60/350
= 6/35

1- 6/36
= 29/35

Where did I go wrong?

My working is:

(P) George and girlfriend both not in: 4C2 x 6C3 = 120

all (P): 5C2 x 7C3 = 350

(P) George and girlfriend not in = (P) George and girlfriend not in / all (P)
= 120/350
= 12/35

Your error was saying 6C2 (instead of 6C3) for the number of combinations in which the girls can be selected without George's girlfriend and also saying 4C1 (instead of 4C2) for the number of combinations in which boys can be selected without George.

 

[Aerath]

Is this 2U or 3U? =\

 

[bored of sc]

Is this 2U or 3U? =\

3U

 

[lyounamu]

My working is:

(P) George and girlfriend both not in: 4C2 x 6C3 = 120

all (P): 5C2 x 7C3 = 350

(P) George and girlfriend not in = (P) George and girlfriend not in / all (P)
= 120/350
= 12/35

Your error was saying 6C2 (instead of 6C3) for the number of combinations in which the girls can be selected without George's girlfriend and also saying 4C1 (instead of 4C2) for the number of combinations in which boys can be selected without George.

Good work, bored of sc.

 

[Doctor Jolly]

My working is:

(P) George and girlfriend both not in: 4C2 x 6C3 = 120

all (P): 5C2 x 7C3 = 350

(P) George and girlfriend not in = (P) George and girlfriend not in / all (P)
= 120/350
= 12/35

Your error was saying 6C2 (instead of 6C3) for the number of combinations in which the girls can be selected without George's girlfriend and also saying 4C1 (instead of 4C2) for the number of combinations in which boys can be selected without George.

But I thought that if I went 1 minus (P) of George and GF in / (P) all, then it would equal the (P) of George and GF out, which explains why I went '4C1 x 6C2' and then minus 1. Or does this only work for non-mutually exclusive events?

 

[lyounamu]

But I thought that if I went 1 minus (P) of George and GF in / (P) all, then it would equal the (P) of George and GF out, which explains why I went '4C1 x 6C2' and then minus 1. Or does this only work for non-mutually exclusive events?

No, she was not even included in the first draw. So we take her out completely out of the draw. (Same with her bf, what a poor couple)

 

[Doctor Jolly]

No, she was not even included in the first draw. So we take her out completely out of the draw. (Same with her bf, what a poor couple)

Ohh okay. So does that mean I can't go 1 minus (P) of something occurring if I'm doing Combinations?

 

[lyounamu]

Ohh okay. So does that mean I can't go 1 minus (P) of something occurring if I'm doing Combinations?

Not in that situation. According to different situations, you may need to alter your working out.

And that's basically why I hate permutations and combinations. :mad1:

 

[Doctor Jolly]

Not in that situation. According to different situations, you may need to alter your working out.

And that's basically why I hate permutations and combinations. :mad1:

So you can also do it roadrage's way too right?

gah, I hate it too

 

[tacogym27101990]

roadrages way is exactly the same as the other one
its just he did the boys and girls seperatly
its the same, just a little bit longerer

 

[Doctor Jolly]

roadrages way is exactly the same as the other one
its just he did the boys and girls seperatly
its the same, just a little bit longerer

Ah right, thanks for clearing that up for me!

And to everyone else, thanks for your input - much appreciated!