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.

  • $\begingroup$ Why wouldn't you count that as the probability of being, say, half-red & half-blue? $\endgroup$ Commented 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
    Commented 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
    Commented Nov 16, 2016 at 22:07

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.