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Combination / Probability

by fangtray » Sat Mar 03, 2012 7:20 am
Kate and her twin sister Amy want to be on the same relay-race team. There are 6 girls in the group, and only 4 of them will be placed at random on the team. What is the probability that Kate and Amy will both be on the team?

I can look at the OA and find the answer. and how the math works..the problem i'm having is why the math works that way. Without knowing, i can't apply it to other problems worded differently.

6!/4!2! = 15 possibile teams. this i get.

but then 4!/2!2! for the winning scenarios we want. Why is it like this? I don't understand. Please help! thanks!
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by LalaB » Sat Mar 03, 2012 8:24 am
4C2/6C4=2/5

Favorable outcome /total outcome =(select 2 girls out of 4)/ (select 4 girls out of six)
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by jzw » Sat Mar 03, 2012 11:59 am
fangtray wrote:but then 4!/2!2! for the winning scenarios we want. Why is it like this? I don't understand. Please help! thanks!
I had the same problem with these a couple of days ago, and then a nice beatthegmat person explained it to me. I will attempt to do the same. So as you said, to get 15 you understand. And, that is the first step of this problem. What you've found however, are all the possibilities withOUT the constraints of this problem, specifically, that the sisters want to be on the same team.

So, (using the bucket dash method, and I'll explain it if you're unfamiliar with it, just let me know), we only have two dashes because we're saving those two spots for Kate and Amy. Meaning, we had 4 dashes, and since we know that both sisters want to be on the team, we are saving two spots, one for each of them. Ok. Now, why do we start with 4 instead of 6? For the same reason - Kate and Amy are standing off to the side because in this scenario they are already on the team. So it's 4C2.

So, what it should be is: What are the chances that they will both be on the team? Well, that's the likelihood of the favorable outcome (4C2 which comes out to 6), over the total possibilities (6C4 which comes out to 15). 6/15 reduces to 2/5. Hope this helps!

What are the answer choices for this questions?
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by krusta80 » Sat Mar 03, 2012 5:11 pm
jzw wrote:
fangtray wrote:but then 4!/2!2! for the winning scenarios we want. Why is it like this? I don't understand. Please help! thanks!
I had the same problem with these a couple of days ago, and then a nice beatthegmat person explained it to me. I will attempt to do the same. So as you said, to get 15 you understand. And, that is the first step of this problem. What you've found however, are all the possibilities withOUT the constraints of this problem, specifically, that the sisters want to be on the same team.

So, (using the bucket dash method, and I'll explain it if you're unfamiliar with it, just let me know), we only have two dashes because we're saving those two spots for Kate and Amy. Meaning, we had 4 dashes, and since we know that both sisters want to be on the team, we are saving two spots, one for each of them. Ok. Now, why do we start with 4 instead of 6? For the same reason - Kate and Amy are standing off to the side because in this scenario they are already on the team. So it's 4C2.

So, what it should be is: What are the chances that they will both be on the team? Well, that's the likelihood of the favorable outcome (4C2 which comes out to 6), over the total possibilities (6C4 which comes out to 15). 6/15 reduces to 2/5. Hope this helps!

What are the answer choices for this questions?
Nice job! ;)
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by jzw » Sat Mar 03, 2012 5:27 pm
krusta80 wrote:Nice job! ;)
I could never have done it without your help from my problem :) You rock!
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by krusta80 » Sat Mar 03, 2012 10:14 pm
jzw wrote:
krusta80 wrote:Nice job! ;)
I could never have done it without your help from my problem :) You rock!
Appreciate the compliment...glad to see my help in action. :)
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by fangtray » Sun Mar 04, 2012 3:39 pm
jzw wrote:
fangtray wrote:but then 4!/2!2! for the winning scenarios we want. Why is it like this? I don't understand. Please help! thanks!
I had the same problem with these a couple of days ago, and then a nice beatthegmat person explained it to me. I will attempt to do the same. So as you said, to get 15 you understand. And, that is the first step of this problem. What you've found however, are all the possibilities withOUT the constraints of this problem, specifically, that the sisters want to be on the same team.

So, (using the bucket dash method, and I'll explain it if you're unfamiliar with it, just let me know), we only have two dashes because we're saving those two spots for Kate and Amy. Meaning, we had 4 dashes, and since we know that both sisters want to be on the team, we are saving two spots, one for each of them. Ok. Now, why do we start with 4 instead of 6? For the same reason - Kate and Amy are standing off to the side because in this scenario they are already on the team. So it's 4C2.

So, what it should be is: What are the chances that they will both be on the team? Well, that's the likelihood of the favorable outcome (4C2 which comes out to 6), over the total possibilities (6C4 which comes out to 15). 6/15 reduces to 2/5. Hope this helps!

What are the answer choices for this questions?
sorry mate, what is the bucket dash method?

The MGMAT booklet i'm using says there are 6 winning scenarios out of 15 possible teams. Ok this is what you were showing me up there.

But in the same section, they show the "winning scenario method" dicussed in a previous section... it says there are 12 winning scenarios, and the next step is to calculate the probability of each winning scenario and then add to all the probabilities. what does that mean? Why is there 6 winning scenarios, but then there are 12 winning scenarios that they list like

Kate is pick 1, Amy pick 2
Amy pick 1, Kate pick 2
Kate pick 2, Amy pick 3
Amy pick 2, Kate pick 3
so forth and so on..until you have 12 of these scenarios..then we are to...calculate the probability of each winning scenario?

They say this is a tedius method and to use the (4!/2!2!)/(6!/4!2!) , but... that then they say 12 winnign scenarios... whereas in the less tedius method, there are 6. What am i not getting here that everybody else is?
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by krusta80 » Sun Mar 04, 2012 7:13 pm
fangtray wrote:
jzw wrote:
fangtray wrote:but then 4!/2!2! for the winning scenarios we want. Why is it like this? I don't understand. Please help! thanks!
I had the same problem with these a couple of days ago, and then a nice beatthegmat person explained it to me. I will attempt to do the same. So as you said, to get 15 you understand. And, that is the first step of this problem. What you've found however, are all the possibilities withOUT the constraints of this problem, specifically, that the sisters want to be on the same team.

So, (using the bucket dash method, and I'll explain it if you're unfamiliar with it, just let me know), we only have two dashes because we're saving those two spots for Kate and Amy. Meaning, we had 4 dashes, and since we know that both sisters want to be on the team, we are saving two spots, one for each of them. Ok. Now, why do we start with 4 instead of 6? For the same reason - Kate and Amy are standing off to the side because in this scenario they are already on the team. So it's 4C2.

So, what it should be is: What are the chances that they will both be on the team? Well, that's the likelihood of the favorable outcome (4C2 which comes out to 6), over the total possibilities (6C4 which comes out to 15). 6/15 reduces to 2/5. Hope this helps!

What are the answer choices for this questions?
sorry mate, what is the bucket dash method?

The MGMAT booklet i'm using says there are 6 winning scenarios out of 15 possible teams. Ok this is what you were showing me up there.

But in the same section, they show the "winning scenario method" dicussed in a previous section... it says there are 12 winning scenarios, and the next step is to calculate the probability of each winning scenario and then add to all the probabilities. what does that mean? Why is there 6 winning scenarios, but then there are 12 winning scenarios that they list like

Kate is pick 1, Amy pick 2
Amy pick 1, Kate pick 2
Kate pick 2, Amy pick 3
Amy pick 2, Kate pick 3
so forth and so on..until you have 12 of these scenarios..then we are to...calculate the probability of each winning scenario?

They say this is a tedius method and to use the (4!/2!2!)/(6!/4!2!) , but... that then they say 12 winnign scenarios... whereas in the less tedius method, there are 6. What am i not getting here that everybody else is?
I'm not familiar with the specifics of the method, but it looks like they are simply listing all of the possible ORDERS in which Kate and Amy can be picked on the team, which is 12:

AK--
A-K-
A--K
-AK-
-A-K
--AK
KA--
K-A-
K--A
-KA-
-K-A
--KA

For each of these orders in which Amy and Kate can be chosen, there are 4*3 = 12 distinct ways to order two of the other four girls. So, 12*12 = 144 distinct winning orderings. BUT, since now we are considering the order in which the girls are chosen, the total number of possible orderings for four out of six girls is 6*5*4*3 = 360.

So, 144/360 = 6/15. Try to re-read the material carefully. From this and some of your other recent posts, it seems to me that you need to really nail down the basics of permutations vs. combinations, etc. Keep at it and don't get discouraged!

PM me if you really need to get more in-depth.
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