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I cant get my head round the following exercise:

You are testing dice for a casino to make sure that sixes do not come up more frequently than expected. Because you do not want to manually roll dice all day, you design a machine to roll a die repeatedly and record the number of sixes that come face up. In order to do a Bayesian analysis to test the hypothesis that p = 1/6 versus p = .175 , you set the machine to roll the die 6000 times. When you come back at the end of the day, you discover to your horror that the machine was unable to count higher than 999. The machine says that 999 sixes occurred. Given a prior probability of 0.8 placed on the hypothesis p = 1/6 , what is the posterior probability that the die is fair, given the censored data? Hint - to find the probability that at least x sixes occurred in N trials with proportion p (which is the likelihood in this problem), use the R command :

1-pbinom(x-1,N,p)

The possible answers are 0.5, 0.684, 0.8 or 0.881.

I would really appreciate if someone could help me here! I need to understand this approach!

Cheers and thanks in advance! Markus ]]>

let t1 be the first arrival time which van be computed from min{X1,...,X5}

t2 is the time of the next and so on. and assuming independence of arrivals and each is ~exp(lambda)

suppose we have the data:

day 0, 1, 2, 3

# arrivals 0, 2, 3,5

means on day 0 there are 0 arrivals, day 1 there 2 arrivals,..., day 3 total arrivals 5.

if we let T=min{X1,...,X5} then the minimum is lower bounded by t1:

P(T>=t1)= P(min{X1,...,X5}>=t1)=P(X1>=t1).P(X2>=t2)...P(X3>= t5)=exp(-5*lambda*t1)

and our cdf is: F(t1)=1-exp(-5*lambda*t1)

so we can determine the arrival time of any,

my question is how to compute the probability of observing c people?

for example P(observing 2 persons)? ]]>