Basketball Teams - CAT problem

[memav7]

Glad it helped.

[RonPurewal]

Thanks..Now I get it. Just to complete the solution in 1-X method..

Case 1: When neither Peter nor John are selected = 7C5 = 21

Case 2: When either Peter or Jhon are selected(Not Both) =
2* 7C4 = 70
(This was the case I missed out earlier)

The total Sample Space is 9C5 =126.

Therefore, the probability that both Peter and John are selected:

1-(21+70)/126 = 35/126

Hope the logic is complete now !

nicely done.

[rustom.hakimiyan]

I'm not sure if I'm misunderstanding the question but the way I read it, the question states that "what is the probability that jon and peter" will be selected. Correct?

If that's the case, why isn't this the correct method:

(5c2)/(5c9) - I need to pick J and P from a team of 5 and the number of options for the teams are 5 out of 9 players. Why is that wrong?

Additionally, I tried the direct probability approach and that seemed to work out:

(2/9)(1/8)(5!/2!3!) - This implies that the prob of picking J or P is 2 out of 9, then only J or P is left so 1 out of 9 and then I could have picked them the 1st time, 2nd, 3rd etc...

This method seems to work but I actually have a question regarding the permutation part. It obviously doesn't matter if I pick John 1st, 2nd, 3rd etc...and the same goes for peter. If order doesn't matter, then why do I need to apply the permutation part of this?

[RonPurewal]

If that's the case, why isn't this the correct method:

(5c2)/(5c9) - I need to pick J and P from a team of 5 and the number of options for the teams are 5 out of 9 players. Why is that wrong?

The red part is not right.

You don't need to pick these two people FROM a team. You just need to pick a team that inculdes those two players.

I.e., you would want
(# of teams including these two players) / (total # of possible teams).

[RonPurewal]

Additionally, I tried the direct probability approach and that seemed to work out:

(2/9)(1/8)(5!/2!3!) - This implies that the prob of picking J or P is 2 out of 9, then only J or P is left so 1 out of 9 and then I could have picked them the 1st time, 2nd, 3rd etc...

This method seems to work but I actually have a question regarding the permutation part. It obviously doesn't matter if I pick John 1st, 2nd, 3rd etc...and the same goes for peter. If order doesn't matter, then why do I need to apply the permutation part of this?

When you multiply probabilities, order is ALWAYS taken into account.
It's not possible to perform an "order doesn't matter" calculation when you multiply probabilities. So, if you want to account for every possible order, you must explicitly account for every possible order.

[rustom.hakimiyan]

If that's the case, why isn't this the correct method:

(5c2)/(5c9) - I need to pick J and P from a team of 5 and the number of options for the teams are 5 out of 9 players. Why is that wrong?

The red part is not right.

You don't need to pick these two people FROM a team. You just need to pick a team that inculdes those two players.

I.e., you would want
(# of teams including these two players) / (total # of possible teams).

Hi Ron,

Understood. That being the case, if we want J and P both to be on the team of 5 players and the team is chosen from players, I tried the approach:

9c2/9c5 and that leads to 2/7, which is incorrect in this case. What's the problem with the equation above?

EDIT: P.S: Thanks for the clarity on the probability question.

[RonPurewal]

If that's the case, why isn't this the correct method:

(5c2)/(5c9) - I need to pick J and P from a team of 5 and the number of options for the teams are 5 out of 9 players. Why is that wrong?

The red part is not right.

You don't need to pick these two people FROM a team. You just need to pick a team that inculdes those two players.

I.e., you would want
(# of teams including these two players) / (total # of possible teams).

Hi Ron,

Understood. That being the case, if we want J and P both to be on the team of 5 players and the team is chosen from players, I tried the approach:

9c2/9c5 and that leads to 2/7, which is incorrect in this case. What's the problem with the equation above?

EDIT: P.S: Thanks for the clarity on the probability question.

It seems you might be just throwing numbers into formulas and seeing what falls out, rather than thinking about what the formulas MEAN.

"9c2" means "the number of different ways to pick 2 people out of 9".
You are not concerned with this number at all, since you already have one pre-specified pair of people whom you have to choose. Thus, if you think about the meaning of this quantity before tossing it into a formula, you'll know that it's inappropriate for the problem.

Rather than just giving you the correct work, here's an outline. See whether you can come up with the correct formula.

* Your denominator is correct"”The total pool of choices is the set of all ways of picking 5 of 9 people.

* You have a pre-specified set of 2 people. You need to pick both of those people.

* Of the REMAINING people (how many of them are there?"”think about it), you only have to pick the rest of the team (how many of them are there?).

See whether you can construct the correct formula from those hints.

[qqixiaofan]

Hi,

Maybe im missing something here but how could the probability of John selected is 5/9? shouldnt it be 1/9? the probability of Peter selected should be 1/8. Please let me know what i missed here.

Thank you,
Xiao

Hello Stacey,

Is the following method of solving this problem correct ?

The probability that John will be selected is: 5/9

The probability that Peter will be selected is: 4/8

So the probability that John and Peter both will be selected is: 5/9*4/8 = 5/18 ?

Please comment. Thanks

actually, yes, that will work for this problem. nicely done.

[RonPurewal]

You can get the 5/9 and 4/8 if you conceptualize this problem a bit creatively.
Normally, we think about selecting one person at a time from a group of people. In this problem, though, we can think about selecting placement (on the team or off the team) instead.

Here's what I mean:
Imagine a row of 9 chairs. Five of the chairs are black; four are red.
The five black chairs represent spots ON the team.
The four red chairs represent spots NOT on the team (= players cut from the team, or not chosen).

"- First, assign a chair to John.
The probability that John gets onto the team is the probability that he is assigned to a black chair: 5 out of 9, or 5/9.

"- Now, assign one of the remaining chairs to Peter.
The probability that Peter gets onto the team is the probability that he is also assigned to a black chair: 4 out of 8 remaining black chairs (since one is already occupied by John), or 4/8 = 1/2.

[rustom.hakimiyan]

Hi Ron,

Understood. That being the case, if we want J and P both to be on the team of 5 players and the team is chosen from players, I tried the approach:

9c2/9c5 and that leads to 2/7, which is incorrect in this case. What's the problem with the equation above?

EDIT: P.S: Thanks for the clarity on the probability question.

It seems you might be just throwing numbers into formulas and seeing what falls out, rather than thinking about what the formulas MEAN.

"9c2" means "the number of different ways to pick 2 people out of 9".
You are not concerned with this number at all, since you already have one pre-specified pair of people whom you have to choose. Thus, if you think about the meaning of this quantity before tossing it into a formula, you'll know that it's inappropriate for the problem.

Rather than just giving you the correct work, here's an outline. See whether you can come up with the correct formula.

* Your denominator is correct"”The total pool of choices is the set of all ways of picking 5 of 9 people.

* You have a pre-specified set of 2 people. You need to pick both of those people.

* Of the REMAINING people (how many of them are there?"”think about it), you only have to pick the rest of the team (how many of them are there?).

See whether you can construct the correct formula from those hints.

Here it goes:

Judging from the two bolded sentences above, it's pretty clear(at least I think it is) that it should be:

(7c3)(2c2 which is 1)/9c5 = (how many ways can i pick the remainder of the team(3 people)/total options). Am I right?

Assuming that the above is correct, I have two questions:

1) I was going to write -- "why wouldn't I account for the various ways I could pick J & P - 5c2" but as I typed it out, it made sense that we don't care HOW we pick J and P therefore we don't put it in there. Is that an accurate assessment? To be completely honest -- I would have never been able to distinguish that it's 7c3 without your two statements above.
2) You mentioned above that when I use the probability method, I am taking ORDER into account. Since order doesn't matter, I have to multiply it by the permutation of the various choice. Since now, I'm using a combination combination method, I don't multiply because with combinations, the order DOESN'T matter. Is that correct?

[qqixiaofan]

Make sense, thank you for the help.

You can get the 5/9 and 4/8 if you conceptualize this problem a bit creatively.
Normally, we think about selecting one person at a time from a group of people. In this problem, though, we can think about selecting placement (on the team or off the team) instead.

Here's what I mean:
Imagine a row of 9 chairs. Five of the chairs are black; four are red.
The five black chairs represent spots ON the team.
The four red chairs represent spots NOT on the team (= players cut from the team, or not chosen).

"- First, assign a chair to John.
The probability that John gets onto the team is the probability that he is assigned to a black chair: 5 out of 9, or 5/9.

"- Now, assign one of the remaining chairs to Peter.
The probability that Peter gets onto the team is the probability that he is also assigned to a black chair: 4 out of 8 remaining black chairs (since one is already occupied by John), or 4/8 = 1/2.

[RonPurewal]

You're welcome.