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This might be a little basic but I am not versed in statistics and I do not even know what to search for helping me answer this. I have a bag with some balls of different colors in it: 2 yellow, 4 red, 5 green, 1 blue, 3 half-red/half-green, 2 half-red/half-blue. I want to know the probability of each color if I randomly take one ball out of the bag. But if I take one of those with 2 colors they count as both. I mean if I choose one red/green that count as a success for both red and green. At first, I thought I could count those with 2 colors as if I had 1 of each of its colors per ball, but I don't know if that is correct.

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  • $\begingroup$ Why wouldn't you count that as the probability of being, say, half-red & half-blue? $\endgroup$
    – gung - Reinstate Monica
    Nov 16, 2016 at 20:50
  • $\begingroup$ Your approach is correct. The probabilities that you calculate won't add up to 1 since the two coloured balls count as both colours, that's not a problem with your calculation it's an expected result. $\endgroup$
    – Hugh
    Nov 16, 2016 at 20:55
  • $\begingroup$ well, I didn't count the probability of the double colored because I was asked specifically for the probability of each color, red, yellow, and blue. Thanks for your answers guys $\endgroup$
    – mikeF98
    Nov 16, 2016 at 22:07

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This is a way to do this problem. Set your information up as a table:

Yellow Red Green Blue
Yellow 2 2
Red 4 4
Green 5 5
Blue 1 1
red/green 3 3 3
red/blue 2 2 2
--------- ---- --------- ----- ------- ------
Totals 17 2 9 8 3
--------- ---- --------- ----- ------- ------

The row labels are the balls, while the column labels are the events.

So now calculate probabilities like $$ \DeclareMathOperator{\P}{\mathbb{P}} \P(\text{Green})=\frac{8}{17} \\ \P(\{Red\})=\frac{9}{17} $$ and so on. Note that these probabilities sums to more than one. Since one random selection of a ball can generate multiple outcomes, these experiment must be modeled as a multivariate random variable, say $$ X=\left(\text{ Yellow, Red, Green, Blue }\right) $$ where each coordinate random variable is binary.

The probabilities above is then describing the four marginal distributions, and there is no requirement they should sum to one.

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