Probability Help: Calc. Non-Red, Marble or Glass

[efigen]

Heres the question.

A box contains twenty objects. Seven are red. Seven are marble. Seven are glass. there are 3 red marble objects. 3 red glass objects. 3 marble glass objects. There is exactly 1 red glass marble object. How many objects are neither red, marble or glass?

- So what i started out doing is making a chart for the 3 separate categories

7 red | 7 marble | 7 glass
3 red marble 3 red marble

4 red left 4 marble left
3 red glass 3 red glass
1 red left 4 glass left
3 glass marbles 3 glass marbles

1 marble left 1 glass left

1 object is all three so it leaves 0 left to chose from
- I added up all the combinations and subtracted it from 20.

20-(3redmarbles)-(3redglass)- (3marbleglass)-(1redglassmarble)= 10 objects

so in total there are 10 objects that are neither red, marble, or glass. I asked a friend and he said it was wrong, and i don't see why.

 

[Stephen Tashi]

It isn't clear to me how you are getting the results in your diagam. Draw a Venn diagram of 3 intersecting circles and put numbers in the 7 non-overlapping areas.

If you are going to approach this problem by making lists, I think you must pay attention to properties like not-red, not-marble and not-glass as well as the affirmative properties.

You need to fill out a chart with combinations like:

red and marble and glass
red and marble and not-glass
red and not_marble and glass
red and not_marble and not_glass
... etc.

Using relations like: number of (red and marble) = number of (red and marble and glass) + number of (red and marble and not-glass)

 

[dacruick]

The 10 makes sense to me.

EDIT: I lied, the 10 is wrong. There are 7A, 7B, and 7C. There are 3AB, 3AC, 3BC, and 1ABC. That 1ABC takes away 1 from each the AB, BC, and AC. So you probably have 13?

So in summary, it seems you have interpreted "red marble" as "red marble and not glass".

 

[eczeno]

i don't really understand your method, so i will not say if is a correct strategy or not, but your answer is certainly incorrect.

here is a way to look at it. start with all 20 objects in a bucket. let's take out all the ones that are red, a marble, or glass.

first, remove the one object that is all three. so, now 19 objects remain. however, only 6 will be r, 6 m and 6 g because the one object removed belongs to all three categories. likewise, there are now 2 rm, 2 rg, and 2 mg left.

so, let's take away the six remaining objects that belong to two categories. so, now there are 13 objects in the bucket. with some bookkeeping we can see that we removed 4 objects from each category.

that leaves us with 7-4-1 = 2 objects left from each category, and they can only belong to one category, so when we remove them we have to take six objects away again, leaving us with 7 objects left in the bucket. these will not be red, a marble or glass.

hope this helps

 

[efigen]

Thanks to everyone for the help!