zank wrote:Joshua and Jose work at an auto repair center with 4 other workers. For a survey on healthcare insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will boh be chosen.
a) 1/15
b) 1/12
c) 1/9
d) 1/6
e) 1/3

I have looked at other explanations which is very simple P(first person) x P (second person) = 2/6 x 1/5 = 1/15, which is the correct answer.

Combinatorics is the weakest part of my prep, and now probability, which im generally good at, is suffering because I've starting confusing combinatorics and simple probability questions.

How do we know that this question should be attempted with simple probability, vs combinatorics, which is what I tried to use and obviously got wrong.

The way i figured was total way that 6 different people can be chosen = 6!

Ways to chose 2 people from 6 = 6!/2!4! = 15

Total probability = 15/6!

Of course looking at the answers makes me realizing my method is not even making sense, but how do i attempt this question using combinatorics? Is it simply 1/# of ways to chose 2 people from 6 which is 15 thus 1/15?

according to me this problem can be solved in two ways..
first method : 2 persons have to be selected out of 6. it can be done in 6c2 ways.
now these 2 should be joshua and jose.
the can be select themselves in 2c2 ways..
so the answer is 2c2/6c2 which is 1/15

second method :
probability of joshua getting selected 1/6
probability of jose getting selected is 1/5
these two can be interchanged i.e first joshua and then jose or vuce versa in 2 ways.
so the final answer is 2X1/6X1/5
which is again 1/15

plz correct me if wrong..

zank wrote:Joshua and Jose work at an auto repair center with 4 other workers. For a survey on healthcare insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will boh be chosen.
a) 1/15
b) 1/12
c) 1/9
d) 1/6
e) 1/3

I have looked at other explanations which is very simple P(first person) x P (second person) = 2/6 x 1/5 = 1/15, which is the correct answer.

Combinatorics is the weakest part of my prep, and now probability, which im generally good at, is suffering because I've starting confusing combinatorics and simple probability questions.

How do we know that this question should be attempted with simple probability, vs combinatorics, which is what I tried to use and obviously got wrong.

The way i figured was total way that 6 different people can be chosen = 6!

Ways to chose 2 people from 6 = 6!/2!4! = 15

Total probability = 15/6!

Of course looking at the answers makes me realizing my method is not even making sense, but how do i attempt this question using combinatorics? Is it simply 1/# of ways to chose 2 people from 6 which is 15 thus 1/15?

Yes. There are 15 total pairs. Joshua and Jose are exactly one of those pairs. So if the process is random, they have a 1/15 chance of being chosen.

Using 6! doesn't make sense in this context, as you noted. You would use that if you were doing something like lining up all six of the people or something else involving all six people where order matters. But here we're only choosing 2 out of 6 people.

The method that i used is:
Probability of choosing Joshua = 1/6
Probability of choosing Jose after Joshua = 1/5
Since both of them are to be present,
Probability = 1/6 * 1/5
Now since Jose could have been chosen first,
Probability = 1/6 * 1/5 + 1/6 * 1/5 = 1/15.

Thanks pete, that does make sense. Now just a follow-up. Two points of confusion

-If i wanted to do this the simple probability way, would i need to work out P(Josh & Jose) OR (+) P (Jose & Josh) = 1/6*1/5 + 1/6*1/5 ?

- Secondly, say the question had said 3 are to be chosen from 6. Then the simple probability method might become error prone and tedious. Would the combinatorics method then be preferred and the answer would be 1/(6!/3!/3!) = 1/20?

GmatMathPro wrote:

zank wrote:Joshua and Jose work at an auto repair center with 4 other workers. For a survey on healthcare insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will boh be chosen.
a) 1/15
b) 1/12
c) 1/9
d) 1/6
e) 1/3

I have looked at other explanations which is very simple P(first person) x P (second person) = 2/6 x 1/5 = 1/15, which is the correct answer.

Combinatorics is the weakest part of my prep, and now probability, which im generally good at, is suffering because I've starting confusing combinatorics and simple probability questions.

How do we know that this question should be attempted with simple probability, vs combinatorics, which is what I tried to use and obviously got wrong.

The way i figured was total way that 6 different people can be chosen = 6!

Ways to chose 2 people from 6 = 6!/2!4! = 15

Total probability = 15/6!

Of course looking at the answers makes me realizing my method is not even making sense, but how do i attempt this question using combinatorics? Is it simply 1/# of ways to chose 2 people from 6 which is 15 thus 1/15?

Yes. There are 15 total pairs. Joshua and Jose are exactly one of those pairs. So if the process is random, they have a 1/15 chance of being chosen.

Using 6! doesn't make sense in this context, as you noted. You would use that if you were doing something like lining up all six of the people or something else involving all six people where order matters. But here we're only choosing 2 out of 6 people.

-If i wanted to do this the simple probability way, would i need to work out P(Josh & Jose) OR (+) P (Jose & Josh) = 1/6*1/5 + 1/6*1/5 ?

Yes. When you do it this way, you have to take order into account and include all of the probabilities of the different orders in which they can be selected. This is related to another way we could solve the problem using permutations: There are 2P2=2 ways to order the pair of Josh and Jose, and there are 6P2=30 ways to order 2 out of 6 people, giving us a probability of 2/30 or 1/15. Note that when you do probability problems like this one, it doesn't matter if you use combinations or permutations as long as you are consistent.

- Secondly, say the question had said 3 are to be chosen from 6. Then the simple probability method might become error prone and tedious. Would the combinatorics method then be preferred and the answer would be 1/(6!/3!/3!) = 1/20?

Are we still just trying to find the probability that Josh and Jose are included, or the probability that Josh, Jose, and some specific third person is included? If the latter, then your calculation is correct. But note that we could use simple probability by saying P(picking any of the three on the first draw)=3/6. P(picking any of the remaining two on the second draw)=(2/5). P(picking the last guy on the third draw)=1/4, so (3/6)(2/5)(1/4)=6/120=1/20, so if we do it that way, we don't have to worry about doing all the different orders separately. Personally, I'd still prefer the combinations way, but it's up to you.

A side note: Don't go overboard trying to master probability and counting if it's just not working out. Every reputable source I've seen says that it's a relatively minor topic on the GMAT, so do your best, but don't neglect other areas that are tested more frequently.

Well, I would say this question is indeed best solved using combinatorics (you ll need to be accurate though!), because, it yields a straight one step solution.
2 people have to be chosen from 6 people. This can be done in 6C2 unique ways.
Definitely, J & J get chosen in one set, and only one set, since in combinatorics each set will be unique.
So the probability will be 1/6C2, which resolves as 1/15.
Intuitively, you should get to saying inverse of 6C2, when you read the question.

Yes, this may involve more calculation, but saves you from any requirement to think or reconfirm your approach.