Good evening,
Imagine please I take 100 colored balls from a bag. Each ball can be either one of the 4 following colors: red, blue, violet or green (like magic; there are no specific amount of red balls, or blues, etc.)
So, the probability of a particular ball be red is of 1/4
After I retrieved the 100 balls, I found myself, for instance, with 40 red, 25 blue and the rest (35) green or violet (doesn't matter).
My question is: How do I do the math? What are the odds of having 40 red and 25 blue?
Certainly they are different if I had got 30 red, and 35 blue (closer of the probability average 25).
I think this is a multinomial problem. If that's the case, will p, the answer, be:
p = 100! / (40! * 25! * 35!) * (1/4)^40 * (1/4)^25 * (1 - (1/4)*2)^35
My doubt, if this is the right answer, is in the probability factors: are they all equal to 1/4,even the 3th one, or is this last one equal to 0.5 (1-(1/2)*2)?
Kind regards,
Kepler
Imagine please I take 100 colored balls from a bag. Each ball can be either one of the 4 following colors: red, blue, violet or green (like magic; there are no specific amount of red balls, or blues, etc.)
So, the probability of a particular ball be red is of 1/4
After I retrieved the 100 balls, I found myself, for instance, with 40 red, 25 blue and the rest (35) green or violet (doesn't matter).
My question is: How do I do the math? What are the odds of having 40 red and 25 blue?
Certainly they are different if I had got 30 red, and 35 blue (closer of the probability average 25).
I think this is a multinomial problem. If that's the case, will p, the answer, be:
p = 100! / (40! * 25! * 35!) * (1/4)^40 * (1/4)^25 * (1 - (1/4)*2)^35
My doubt, if this is the right answer, is in the probability factors: are they all equal to 1/4,even the 3th one, or is this last one equal to 0.5 (1-(1/2)*2)?
Kind regards,
Kepler
