Kate and her twin sister

[tim]

as i've said before, we have explained the method extensively, and you used it correctly!

the method of figuring out who was chosen first, second, third, and fourth is too complicated and unreasonable to use on the GMAT that i won't be discussing it here..

if you got a wrong answer by trying to figure out how many ways the girls can be selected, feel free to post it here and we can take a look at where you went wrong..

[ffearth]

(1/6x1/5)x 2 + (1/5x1/4)x2 +(1/4x1/3)x2 +(1/6x1/3)x2 +(1/6x1/4)x2 + (1/4x1/3)x2 = 119/180

What did I do wrong?

Thank you

[tim]

this is not an approach that reflects the WAYS the girls can be selected, and it does not arrive at the number 12 you described in the previous post. furthermore, if you want an analysis of what you did wrong, you have to explain your steps rather than just throwing a bunch of numbers onto the page..

[ffearth]

Alright.
If Kate is selected first and Amy second, the probability of them being in the same team will be (1/6)x(1/5).
Similarly, if Amy is selected first and Kate second, the probability will be the same. That's where my (1/6x1/5)x 2 comes from.

Then, if Kate is selected second and Amy third, the probability of them being in the same team is (1/5x1/4).Similarly, if Amy is selected first and first second, the probability will be the same. That's where my (1/5x1/4)x 2 comes from.

I don't think I need to explain the other cases, they follow exactly the same logic as the two first ones.

So indeed the ways the girls can be selected arrives to the number 12.

Please help.

Thank you

[tim]

no! you have still not logically connected how your probabilities relate to a number of ways the girls can be chosen..

[ffearth]

Could you please elaborate? What is my mistake exactly? What is the right way to do that?

[tim]

i don't know what you're trying to do exactly, so i can't point you in the right direction. you've done nothing to explain where your number of 12 comes from. i don't think it's relevant at any rate, so it might be best to abandon whatever alternate approach you're trying..

[ffearth]

As you can see on this picture there are 12 ways of selecting the two girls. I think that what I did is clear.

"so it might be best to abandon whatever alternate approach you're trying", very encouraging! MGMAT has great teachers really

[RonPurewal]

so i haven't read this entire thread, but the last post is the first in which this table appears, so i'll go ahead and comment on that.

* the list of possibilities is valid, and is constructed in a well-organized way.

* however, the calculation of probabilities in the right-hand column is incorrect for all but the first two rows of the table. in fact, every row of the table should give a probability of 1/30.
there are lots of different ways to see that this is true; since this thread seems to have become rather protracted, i'll give three.

1/
you can directly calculate, by multiplying consecutive probabilities.
i see that this is what you've tried to do here, but here's the problem: you aren't taking into account the choices in which neither amy nor kate is picked.
for instance, in the third row of your table, the first choice has to be someone other than amy or kate. you have ignored this component in your calculation, but it must be taken into account. therefore, the correct probability for the third row of the table is not just (1/5)(1/4), but, rather, (4/6)(1/5)(1/4).
the same thing happens in all of the other rows of the table, except for the first two (in which this is a non-issue because amy and kate are the first 2 picked).
if you do this, you'll find that it's 1/30 for every row.

2/
there's also no reason for, as one might say, "tyranny of chronological order".
i.e., you don't have to pick things in the order first, second, third, etc. you could just as well pick them in any order -- let's pick the third person at the beginning, then the fourth one, then the first one, then ... whatever.
so e.g., for the third row of the table, you can just say "this time i'll pick the second and third players first". in that case, the probabilities are 1/6 and 1/5 -- just as though you had chosen those players first and second in the original formulation -- and give a product of 1/30.

3/
you can also just use a symmetry argument.
it should be intuitively clear that it's no more (and no less) likely for amy and kate to be chosen, say, 2nd and 4th than for them to be chosen first and second. in fact, since there is nothing "special" (in the sense of "more or less likely") about any of the different places in the selection, it should be clear that amy and kate have exactly the same probability of being chosen in any two of the given places.
it thus follows that, if the probability in any one row of your table is 1/30, then the probability in every row of the table should be the same.

so, for any of those three reasons, each of your 12 rows gives a probability of 1/30. these are all different events with no overlap, so you can add them all to give 12(1/30) = 2/5. so there it is.

[tim]

ffearth, we're here to guide you toward appropriate solutions to questions; it's up to you whether you choose to accept that guidance. as you can see, your approach with the table is incredibly complicated and error-prone (as Ron pointed out). please, for the sake of your own success on the GMAT, abandon such unfruitful solution methods and concentrate instead on approaches that are more likely to yield quick, accurate answers on the GMAT..

[ffearth]

Many thanks Ron!! That's extremely helpful!
Could you PLEASE point out what is the logic behind the first method I used: 4/6 x 3/5 = 2/5 ?

To Tim: you are right, this mehtod would be inappropropriate for this problem. But how can one choose the most appropriate method for a problem if he doesn't even understand the methods he can use? It happens that this method is inappropriate for this problem, but it could be the best method for another problem. So it is important to understand it.
It seems that you want people to apply methods without understanding why these methods work, Ron's explanation definitely pointed out a lack of understanding I had regarding probability, a subject we do not study in high school in the country I come from.

[tim]

ffearth, this is exactly how we can help you; and Ron, Jamie, and I are all glad to have contributed to this process! You had a great method that worked, and we confirmed that. You tried another method that will probably never be beneficial for any GMAT problem, and we told you it wasn’t a good idea. If you are not fully convinced by now, please note that in general the GMAT will never expect you to set up such a complicated chart and perform a tedious series calculations, so just let it go. Your original method was great, and of course any of the methods Ron described will be great as well. Please focus on those and not on a solution method that involves extensive calculations and case-checking..

[hamsomwell]

I thinks the the best choice is 2/5,but I got a thing from all the discussion that it is not gonna happen that both the sisters will meet in on team.

[spam deleted]

[jnelson0612]

I thinks the the best choice is 2/5,but I got a thing from all the discussion that it is not gonna happen that both the sisters will meet in on team.

[spam deleted]

2/5 is indeed the answer, but I'm not sure if you are asking a question as I can't really understand what you are saying. Please let us know if you need further assistance.:-)

[tim]

keep your spam off our boards!!!