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Exercise 1.8 #5

@dsrasheed

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@dsrasheed

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

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