Combinatorics!

[michele_valente]

Hi! Can anyone help me solve the following?

From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the possibility that Andrew will be among the 4 volunteers selected and Karen will not?

3/7
5/12
27/20
2/7
9/35

Thanks!

[RonPurewal]

Hello,

Is this problem from the free GMAT PREP software? If not, it doesn't belong in this folder.

- If the problem IS from the free software, please post a screen shot of it. (Sorry for the inconvenience, but this problem doesn't appear anywhere in any of the free GMAT PREP problem archives that I have.)

- If not, please re-post it in the appropriate folder. If it's a non-MGMAT problem, please post it in the General Math folder, with a citation of the original source.

Also"”When you re-post the problem, please note some specifics of what you've already tried.
How did you try to calculate the answer?
What methods have you tried?
What do you understand? What don't you understand?
Etc.


Thanks.

[michele_valente]

Hi!

The problem is from the GMAT Prep software, but it's not from the free one, I have bought the 2 additional tests and this is in one of them.

As to my thoughts on how to solve it, I actually found it quite difficult and I wouldn't have too many clues on how to answer it, that's why I posted it :)

[RonPurewal]

Well, those problems are technically off limits, since they're paid intellectual property.
However, we can make an exception this one time, since there is nothing particularly unique about this problem"”i.e., there are certainly others exactly like it in random math textbooks.

Here are two pathways to get you started. I don't want to just write out complete solutions, because that wouldn't stimulate your ability to solve the problem.

"- You can multiply probabilities together.
When you multiply probabilities, order is ALWAYS taken into account, so you need to account for each of the four possible orders: Andrew first, Andrew second, Andrew third, or Andrew fourth.

"- You can also find the total number of teams that include Andrew but not Karen. (You know you need to pick Andrew. Then, you need to pick 3 from a select group.)
Then you can divide this by the total number of possible groups of 4 people.

Give these a shot, and post back if there is still trouble.

[michele_valente]

Hi Ron,

Thanks for your reply.
I've tried to have a go with both methods.

With the first one I understand I need to calculate the 4 different scenario and try to understand what is the probability that Andrew will be selected first (1/8) + the probability that him and Karen won't be selected first but Andrew will be selected on a second shot (6/8 * 1/7) + probability that him & Karen won't be selected neither on the first nor on the second attempt but just Andrew on the third (?? 6/8 * 5/7 * 1/6??) + same concept with last shot (?? 6/8 * 5/7 * 4/6 * 1/5??). As you can see from my question marks and from the fact that what I get is not among the answer choices, there is something wrong in my logic...

I have tried to do something also with the second method.
I can easily calculate the total possible teams (8!/4!4!=70) and as you suggest this will be the denominator of my combination. But the problem is to pick up Andrew and the 3 remaining folks. I thought this could be something like: Andrew (8!/1!7!) + remaining folks excluding Karen (8!/6!2!)...

As you can see I'm kind of stuck.
Any help would be really appreciated!

Thanks again,


Michele

[RonPurewal]

With the first one I understand I need to calculate the 4 different scenario and try to understand what is the probability that Andrew will be selected first (1/8) + the probability that him and Karen won't be selected first but Andrew will be selected on a second shot (6/8 * 1/7) + probability that him & Karen won't be selected neither on the first nor on the second attempt but just Andrew on the third (?? 6/8 * 5/7 * 1/6??) + same concept with last shot (?? 6/8 * 5/7 * 4/6 * 1/5??). As you can see from my question marks and from the fact that what I get is not among the answer choices, there is something wrong in my logic...

The problem here is that, once Andrew has been selected, you’re neglecting your continued responsibility to the whole "not Karen" thing.

For instance, your first calculation should not just be 1/8. It should be 1/8 x 6/7 x 5/6 x 4/5.

Likewise, your second calculation should be 6/8 x 1/7 x 5/6 x 4/5.

The third, 6/8 x 5/7 x 1/6 x 4/5.

The fourth is exactly as you wrote it.

Conveniently, all four of these are the same product; they’re all 6x5x4 on the top and 8x7x6x5 on the bottom. (They MUST all be the same: Andrew is no more or less likely to be chosen in any particular place in the sequence.)
So, you only have to simplify the fraction once (= 1/14). So total probability is 4/14 = 2/7.

[RonPurewal]

I have tried to do something also with the second method.
I can easily calculate the total possible teams (8!/4!4!=70) and as you suggest this will be the denominator of my combination. But the problem is to pick up Andrew and the 3 remaining folks. I thought this could be something like: Andrew (8!/1!7!) + remaining folks excluding Karen (8!/6!2!)...

Andrew is just Andrew. As his grade-school teachers were fond of telling him, he is his own unique individual.
There’s only one way to pick him. Because he’s just Andrew.

Once you’ve selected Andrew, you just need to pick three of the remaining 6 people. (Not 7 people, since Karen is out.)
Since you know how to calculate the 70, you can do this in the same way; you’ll get 20.
1 x 20 = 20.
20 out of 70 = 2/7.

[RonPurewal]

PERHAPS MOST IMPORTANTLY

If you are stuck doing this problem in any way ...

... then why do anything other than MAKE A LIST?

Especially if you know how to calculate the denominator = 70. Then you have a very easy"”and not at all time-consuming"”task: just list out all the ways to pick your committee.
Let’s say the people are Andrew (A), Karen, "1", "2", "3", "4", "5", and "6". Then your possible committees are...
A123, A124, A125, A126
A134, A135, A136
A145, A146
A156
A234, A235, A236
A245, A246
A256
A345, A346
A356
A456
There are twenty committees in the list. 20 out of 70 = 2/7. If you can’t make this entire list in thirty or forty seconds, you should practice making lists.

[RonPurewal]

Apropos of the above...

Among the more difficult combinations problems on the GMAT, MOST"”almost all, actually"”can be solved by just listing out the possibilities.
That’s purposeful on the part of the test makers, by the way"”they want you to be able to make lists. (If they wanted to force you to use combinatorial methods, they could just write the problems with much, much bigger numbers of people/things. But they don’t... pretty much ever.)

[michele_valente]

Thanks Ron, this really helps.
I fully understand now the second approach.

I think here you have a very good point, doing lists is something I can quite easily exercise and I will definetely do.

I just need to get my head a little more around the first method.
I'll have a look at the books, would need to understand better when I should consider just the probability that an event occours, or when also the probability that a second or third event does not occur at the same time must be considered.

But thanks again for the explanations, they are really clear and helpful!

[tim]

Sounds good. Let us know if you have any further questions on this issue.