A symmetrical 8-sided dice has three red, one yellow and four green sides. Dice is thrown three times. What is the probability of at least one of the colours not appearing in the series of throws?
3 - red 1 - yellow 4 - green
I'm stuck with figuring out the complement properly - it is P("All the colours appear during three throws"), right? I tried to solve this drawing a 8x8 table with all the possible outcomes of all the colours appearing during the last throw - not getting it right...
Let X be the probability of all showing up out of 3 throws and Y the probability of at least one not showing up out of 3 throws. In this way you can write
$$P(Y) = 1 - P(X)$$ $$P(X) = \frac{3}{8}\frac{1}{8}\frac{4}{8}3!$$ Hence: $$P(Y) = 1 - \frac{12}{8^3}3!$$