Sum to infinity of a spinner with 3 colours Hey I'm stuck on a question which I think uses geometric series to solve, its for probability. The question is:
When a spinner is spun, the outcome is one of 3 colours, green, blue, or red. On anyone of the spins, the probability of each of those outcomes is pG, pB and pR respectively. Note that pG+pB+pR= 1. If the spinner is spun n times, what is the probability there is at least one blue outcome and one green outcome in the n spins.
My approach to this question is to find the probability of no blue and no green (2/3) then find the union (1/3 i think?, not sure on this) I did P(E) + P(F) - P(EnF) which is 2/3 + 2/3 - 3/3?? I don't know what to do from here though. I think I take the sum to infinity of (1/3)n ?

It's a variant of the Coupon Collector Problem
$G$=there is no green in n spins
$B$=there is no blue in n spins
$P(G\cup B)=P(G)+P(B) - P(G\cap B)$
Now:
$P(G)=(p_B+p_R)^n$ (only blue and red in n spins)
$P(B)=(p_G+p_R)^n$ (only green and red in n spins)
$P(G\cap B)=(p_R)^n$ (only red in n spins)
and what we want is $1-P(G\cup B)$
(the opposite of both green and blue means that either of them is missing)
