jawad wrote:Kurt, a painter, has 9 jars of paint: 4 are Yellow, 2 are Red, and the remaining jars are Brown.Kurt will combine 3 jars of paint intoa new container to make a new color, which he will name according to the following conditions :

Brun Y if the paint contains 2 jars of brown paint and no yellow

Brun X if the paint contains 3 jars of brown paint

Jaune X if the paint contains at least 2 jars of yellow

Jaune Y if the paint contains exactly one jar of yellow

What is the probability that the new color will be Jaune?


5/42
37/42
1/21
4/9
5/9

I assume that he's randomly selecting the 3 jars of paint, because if it isn't random then there's no way to answer the question without knowing his selection process. Real GMAT questions would never be ambiguous in this manner.

We want the probability that the new color will be Jaune, which happens whenever:

1) there's exactly 1 jar of yellow; or
2) there are at least 2 jars of yellow.

In other words, if there's any yellow at all, the color is "Jaune".

A great equation to know for probability is:

(Prob of what you want) + (Prob of what you don't want) = 1

and rearranging that:

Prob of what you want = 1 - (Prob of what you don't want)

Here, we want the probability of some yellow. So what we don't want is no yellow at all, which will be much easier to calculate.

Prob (no yellow jars) = 5/9 * 4/8 * 3/7 = 5/42

(For the first pick, the chance of non yellow is 5/9; for the second, we're down to 4/8; for the third, we're down to 3/7.)

(Note that we could also solve with combinations: 5C3/9C3 gives us the same result.)

So, Prob (some yellow) = 1 - 5/42 = 37/42... choose (b)

gr8 explanation for the question... but it made me realise that how much i still have to go thru just to understand some of the questions still (not even considering the answers).....

Kovinsky wrote:

jawad wrote:Kurt, a painter, has 9 jars of paint: 4 are Yellow, 2 are Red, and thStuart e remaining jars are Brown.Kurt will combine 3 jars of paint intoa new container to make a new color, which he will name according to the following conditions :

Brun Y if the paint contains 2 jars of brown paint and no yellow

Brun X if the paint contains 3 jars of brown paint

Jaune X if the paint contains at least 2 jars of yellow

Jaune Y if the paint contains exactly one jar of yellow

What is the probability that the new color will be Jaune?


5/42
37/42
1/21
4/9
5/9

I assume that he's randomly selecting the 3 jars of paint, because if it isn't random then there's no way to answer the question without knowing his selection process. Real GMAT questions would never be ambiguous in this manner.

We want the probability that the new color will be Jaune, which happens whenever:

1) there's exactly 1 jar of yellow; or
2) there are at least 2 jars of yellow.

In other words, if there's any yellow at all, the color is "Jaune".

A great equation to know for probability is:

(Prob of what you want) + (Prob of what you don't want) = 1

and rearranging that:

Prob of what you want = 1 - (Prob of what you don't want)

Here, we want the probability of some yellow. So what we don't want is no yellow at all, which will be much easier to calculate.

Prob (no yellow jars) = 5/9 * 4/8 * 3/7 = 5/42

(For the first pick, the chance of non yellow is 5/9; for the second, we're down to 4/8; for the third, we're down to 3/7.)

(Note that we could also solve with combinations: 5C3/9C3 gives us the same result.)

So, Prob (some yellow) = 1 - 5/42 = 37/42... choose (b)

Dear Stuart,

Excellent solution.

But could you please explain why there could be 9C3 possible combinations ? The 9 items are not different. They have 4 yellow jars of the same type and similarly the red and brown ones.

regards,
Ankit

saxenankit wrote:Dear Stuart,

Excellent solution.

But could you please explain why there could be 9C3 possible combinations ? The 9 items are not different. They have 4 yellow jars of the same type and similarly the red and brown ones.

regards,
Ankit

Hi Ankit,

the 9 items are different, it's just that some of them are the same colour; there are 9 physical jars from amongst which to choose, so 9 is the total number of objects.

The only time (at least on the GMAT) that we worry about objects being distinct is for permutations - i.e. we're arranging them. However, even on permutations questions the fact that some items are identical doesn't change the total number of items, just how many arrangements we can make.

For example:

How many different arrangements can be made from the letters in the word BASE?

Here, all 4 letters are distinct, so the answer would be simply 4!.

Now let's add a duplicate:

How many different arrangements can be made from the letters in the word BASES?

Here, we have 5 letters in total, 2 of which are identical. So, we need to use the formula for duplicate objects:

Total # of arrangements of n objects, r of which are identical, is n!/r!

So: 5!/2!

Let's add another duplicate:

How many different arrangements can be made from the letters in the word ABASES?

When we have multiple duplicates, we need to divide by all of them.

Accordingly, there are: 6!/2!2! possible arrangements.

Here's one for you to try:

How many different arrangements can be made from the letters in the word DESSERTS?

Stuart Kovinsky wrote:

saxenankit wrote:Dear Stuart,

Excellent solution.

But could you please explain why there could be 9C3 possible combinations ? The 9 items are not different. They have 4 yellow jars of the same type and similarly the red and brown ones.

regards,
Ankit

Hi Ankit,

the 9 items are different, it's just that some of them are the same colour; there are 9 physical jars from amongst which to choose, so 9 is the total number of objects.

The only time (at least on the GMAT) that we worry about objects being distinct is for permutations - i.e. we're arranging them. However, even on permutations questions the fact that some items are identical doesn't change the total number of items, just how many arrangements we can make.

For example:

How many different arrangements can be made from the letters in the word BASE?

Here, all 4 letters are distinct, so the answer would be simply 4!.

Now let's add a duplicate:

How many different arrangements can be made from the letters in the word BASES?

Here, we have 5 letters in total, 2 of which are identical. So, we need to use the formula for duplicate objects:

Total # of arrangements of n objects, r of which are identical, is n!/r!

So: 5!/2!

Let's add another duplicate:

How many different arrangements can be made from the letters in the word ABASES?

When we have multiple duplicates, we need to divide by all of them.

Accordingly, there are: 6!/2!2! possible arrangements.

Here's one for you to try:

How many different arrangements can be made from the letters in the word DESSERTS?

Thanks Stuart for the explanation. The question I asked is because I did various problems, on permutation and combination, asking for selections and arrangements for duplicate and/ or distinct items. So every time I hit such a problem I get confused. I am unable to make out whether to consider items as duplicate or distinct.
Do you have any suggestion for me so that I can solve problems with confidence ?

Moreover, I at times get confused between solving for combination or permutation, especially on a probability problem. Also on a probability problem I waste some time to decide on the approach -
whether solve it by calculating individual probability and then add or multiply them
OR
go by calculating the total outcomes and then calculate the favorable ones and divide them.
Do you have any particular inputs which could help me out ?

And the answer for the problem should be 8! / 2!.3!

Thanks again for your help and time

-Ankit

saxenankit wrote:
Thanks Stuart for the explanation. The question I asked is because I did various problems, on permutation and combination, asking for selections and arrangements for duplicate and/ or distinct items. So every time I hit such a problem I get confused. I am unable to make out whether to consider items as duplicate or distinct.
Do you have any suggestion for me so that I can solve problems with confidence ?

Moreover, I at times get confused between solving for combination or permutation, especially on a probability problem. Also on a probability problem I waste some time to decide on the approach -
whether solve it by calculating individual probability and then add or multiply them
OR
go by calculating the total outcomes and then calculate the favorable ones and divide them.
Do you have any particular inputs which could help me out ?

For combinations vs permutations, the important question to ask is "does order matter".

If order doesn't matter, use combinations; if order does matter, use permutations.

Here's another way you can think about it: After I'm done selecting the subgroup, am I going to do anything else with the entities?

If not, then it's a combinations question; if you're doing something with the entities after choosing them, then it's a permutations question.

Choosing combinations and permutations vs probability is really a matter of personal preference. Here's a good general rule for the GMAT: when doing timed practice, use whatever method jumps out at you first; when doing untimed practice, use every method that occurs to you.

The reason that you should try every method you can think of when doing untimed practice is because doing so forces you to use the methods with which your less comfortable, but which ultimately may be quicker than your current method of choice. The most important methods to master are the strategic ones: picking numbers, backsolving and strategic guessing. Especially if you have a math background it's likely that you default to algebra, but for many high-level GMAT questions an algebraic approach is just too time consuming.

saxenankit wrote:And the answer for the problem should be 8! / 2!.3!

-Ankit

Correct - great job!

Thanks Stuart for this prompt help. I will try to follow your tips.

Regards,
-Ankit

Is this a GMAT question