$v$ is pulled from a binomial distribution, not geometric: $\mathbb{P}(red=0) = \mathbb{P}(v = 0.0) = {10 \choose 0}\mu^0(1-\mu)^{10}$ $\mathbb{P}(red=1) = \mathbb{P}(v = 0.1) = {10 \choose 1}\mu^1(1-\mu)^9$
