Basketball Teams - CAT problem

[shweezy]

Hi everyone,

I'm a bit confused on the best approach to the "Basketball Teams" problem on a recent MGMAT CAT. Here's the question:

"John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?"

I read through the approach, and while it makes sense, it isn't exactly the most intuitive solution (at least not to me). I know that the first part of the problem should be to do an anagram of YYYYYNNNN, but the second part of the solution (i.e. assume John and Peter are already on the team and come up with another anagram of 7 plays - YYYNNNN) is still unclear. Are there any other ways to tackle this problem?

Thanks,
Shuo

[tmmyc]

If you're not a fan of the anagram method, here is a more academic approach.

Probability is basically the number of possibilities we want divided by the number of total possibilities.


Let start with the number of total possibilities.

Number of total possibilities: Use Combinations.
Out of 9 players, we choose 5

9 choose 5
-> 9 C 5
--> 9! / (4! * 5!)

Note, you arrive at the same equation using your anagram YYYYYNNNN


Let's move on to the number of possibilities we want.

Number of possibilities we want: All possible teams that include John and Peter.
A team has 5 people, and 2 of them are John and Peter. Therefore, we need to find all possible combinations for the other 3 empty spots. Since John and Peter are already chosen, we have 9-2=7 possible players that can fill those 3 spots. Use Combinations again.

Out of 7 players, we choose 3

7 choose 3
-> 7 C 3
--> 7! / (3! * 4!)

Note, you arrive at the same equation using your anagram YYYNNNN


You can solve by using these two values. I hope this makes things more clear for you.

[StaceyKoprince]

Thanks tmmyc!

[shweezy]

i'm still a bit confused. i actually like the anagram method; it's just that i'm confused as to why you would have an anagram of YYYNNNN for the second portion of the problem. in other words, i'm unclear on why you would "assume" that the two spots have already been reserved for john and peter already and only consider the 7 other folks.

thanks guys.

[StaceyKoprince]

Ah, got it. So, we assume that John and Peter are given "Y"s because the question specifically asks us about the probability that they will both be on the team. Essentially, we're just setting up the anagram to match the question: the question indicates that we want John and Peter both to be on the team, so we assign them each a Y. If they had asked us, instead, for the probability that neither would be on the team, we would assign them each an N. Does that make sense?

[Spencer]

i'm still a bit confused. i actually like the anagram method; it's just that i'm confused as to why you would have an anagram of YYYNNNN for the second portion of the problem. in other words, i'm unclear on why you would "assume" that the two spots have already been reserved for john and peter already and only consider the 7 other folks.

thanks guys.

Here's a simple way to think of it:

In problems that order doesn't matter do the following

(Total # of people/things)! / (# of positions open)! x (Total # of people - # of positions open)!

First let's find the total # of teams of 5 people that can be selected from 9.

Using the equation from above we get 9! / 5! x 4! = 126

So now we know there are 126 teams to choose from and some of them have the two guys and some of them don't.

Since order doesn't matter a team with John, Peter, Nick, Mike, and Bob is no different from the team with Bob, Nick, Peter, Mike, and John. It's still the same team.

So right now we want to know how many different teams there are with the 2 guys on it.

We agreed above that the order of the players doesn't matter as long as John and Peter are on the team. So let's put them on the team and find out how many different way we can fill up the rest of the spots with the 7 players we have left.


John--Peter--(Open spot)---(Open spot)---(Open spot)


So remember our equation from above: (Total # of people/things)! / (# of positions open)! x (Total # of people - # of positions open)!

7! / (3! x 4!) = 35


35 out of the 126 different teams contain John and Peter.

Hope this helps!

[StaceyKoprince]

Thanks, Spencer. Yes, you can use either the anagram method or the classic comb/perm method we learned back in high school - whatever works best for you!

[pallaviamahajan]

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

[RonPurewal]

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.

[vijaykumar.kondepudi]

I understood the solution to this problem, but during the exam I tried the 1-(# of instances when both John and Peter are not selected ) approach.

1. Total nomber of events: 9C5=126.

2. Events in which both, Paul and John, are not selected:
In this case we need to pick the 5 players from a pool of only 7
players
7C5

3. Probability of both not selected is 7C5/9C5 =1/6

4. Probability that both, Peter and John, are selected: 1-1/6=5/6

It doesn't match :(
Plz. let me know what is wrong with this approach. More importantly, when should we use the "1-(Probability of non-favourable events)" approach?

[memav7]

The logic that you have used is not complete as you have eliminated the case in which one of the player is selected.

The question is to find the probability for which both the players are taken in a team.

I hope it helps!

Thanks

[vijaykumar.kondepudi]

Hi Memav7,
I am sorry, but I couldn't fully understand your reply. Could you please elaborate.
What is the exact mistake and how do I rectify it using the 1-(# of instances when both John and Peter are not selected ) approach ?

Thanks !!

[memav7]

Ok..

when you do
1 - (# of instances when both John and Peter are not selected)

will give you a ) Both John and Peter are selected together
b) Either of John or Peter are selected

But the question demands the value for the case a i.e when both John and Peter are selected together. So get the correct ans you will need to minus the b) from the ans you are getting.

Hope it helps.

Thanks

[tim]

Vijay,
In general your approach works, but you need to keep in mind that if you are using the 1-x method you MUST have x be the logical opposite of what you’re looking for. The opposite of "both being selected" is not "neither being selected" but rather "at least one is not selected". In other words, you have overlooked the vast majority of cases that fall outside the both/neither false dichotomy..

[vijaykumar.kondepudi]

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 !