Probability Help?

Question

I think I'm doing this correctly, but I want to make sure before I proceed, since for some reason I'm feeling unsure.

A group of people are each given a bag of marbles. Marbles come in three colors: red, blue and green. Some bags are empty, some bags have all marbles of one color, some have multiple colors.

4% of bags have all three colors.
5% have only green.
15% have both red and green.
23% have both red and blue.
55% have red.
25% have green.
43% have blue.

Note that unless I said 'only', then I don't mean 'only'.

Given those odds, what is the chance that a randomly chosen person has only blue marbles? What is the chance that a randomly chosen person's bag isn't empty? What is the chance that a randomly chosen person has exactly two types of marbles?

I've got some answers, but I'm still confused - it seems like there's not enough information in the stats, since it didn't say anything about the odds of green and blue, only blue or only red.

Answer 1

You present three questions which I will enumerate:

1) What is the chance that someone has only blue?
2) What is the chance that a bag is non-empty?
3) What is the chance that a bag contains exactly two colors?

When confronted with a problem like this, it is often wise to begin by expanding the set of known probabilities. For example, 55% have red, 25% have green, 15% have red and green. Given that, we can ignore blue completely for a moment and, using Bayes rule, conclude that 40% have red but not gree, 10% have green but not red, 15% have green and red, and 100%-(40%+10%+15%) = 35% have nether green nor red. We conclude this by noting that if a bag contains red, it either does or does not contain green. If there are 100 bags, 55 have red and of those 15 also have green, therefore the remaining 40 do not contain green.

We then repeat the analysis for red and blue using the same math.

Red but not blue: 32%
Blue but not red: 20%
Red and blue: 23%
Niether red nor blue: 25%

Now we know that 23% have red and blue and 15% have red and green. 4% of the total is overlap (red, blue and green). Thus:

19% red, blue, not green
11% red, green, not blue
4% red, green, blue

Also:

5% green, not red, not blue

We now need one more key piece of info to unlock the whole thing. How many have green and blue?

A bag with green can be: green only; red and green only; green and blue only; or red, green, and blue (four conditions).

Of 25% containing green:
5% are just green (from problem)
11% are red and green but not blue (from above)
4% are red, green, and blue (from problem)
This leaves 5% green and blue only.

Let's start filling in a table with all 8 mutually exclusive conditions
-----------------------------
Empty: ?
Red only: ?
Green only: 5%
Blue only: ?
Red and green only: 11%
Red and blue only: 19%
Green and blue only: 5%
All three: 4%

Ok, three missing values now.
Red only = Red - Red, green, and blue - red and green ony - red and blue only
Red only = 55 - 4 - 11 - 19
Red only = 21%

Blue only = Blue - all three - blue and red only - blue and green only
Blue only = 43 - 4 - 19 - 5
Blue only = 15

The only remaining value (% of empty bags) can be derived by adding up all of the other values and subtracting from 100.
Empty = 100 - (21+5+15+11+19+5+4) = 100 - 80 = 20%

Now are completed table is as follows
----------------------------------------------
Empty: 20%
Red only: 21%
Green only: 5%
Blue only: 15%
Red and green only: 11%
Red and blue only: 19%
Green and blue only: 5%
All three: 4%

Your three questions should be easy to answer now.
1) only blue = 15%
2) isn't empty = 100% - is empty = 80%
3) exactly two = red, green, not blue + red, blue, not green + blue, green, not red = 11+19+5 = 35%

Answer 2

start by posting in "additional details" what you've found so far. I'll do the maths while you post that information.


1st question: how many blue alone?
For that, you need to start gathering information where you CAN:
In the following, my notations are (RB) means only red and blue and (RB?) means Red and Blue, and possibly something else (green). special notation for all 3 colors: (tri) for tri-color

you have
RG? = RG + tri => RG = 15-4 = 11%
RB? = RB + tri => RB = 23-4 = 19%
R? = R + RG + RB + tri => R = 55 - (4+11+19) = 21%
G? = G + RG + GB + tri => BG = 25 - (4+5+11) = 5%
B? = B + BG + BR + tri => B = 43 - (19+5+4) = 15%


bag not empty = G + B + R + RB + RG + BG + tri = 80%
exactly 2 types of marbles = RG + GB + BR = 35%

Answer 3

First one: 15%

Second one: 80%

3rd one: 35%

This is from figuring out the following (not sure if it's right, but I think it should be, through using logic) g=green r=red b=blue

g=5 gb=5 gr=11
grb=4 r=21 b=15 br=19

Answer 4

The best way to tackle this problem is to draw a Venn diagram. Once you do this, the problem is easy as cake:

Computations are as follows:
P(blue alone) = 0.15
P(not empty) = Sum of all probabilities = 0.76
P(two type of marbes) = P(G^B) + P(G^R) + P(B^R) = 0.01+0.15+0.23 = 0.39

Good Luck