# How to deal with a cat with 9 lives?

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This question is based on Tutorial 3 Question 1 (ii). It pertains to a cat that has 9 lives. My first guess was to apply a simple negative binomial model but on closer inspection such a model entails the deaths occurring discretely at the end of the weeks in question.

This would mean that all 9 lives being lost in 5 weeks is impossible. How do I go about tackling parts (a) and (b)?

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(a) For the cat to die during the next 10 weeks, it must have lost all 9 lives by then. Hence, the probability of the cat not losing a particular life is $$(0.8)^{10}$$, and correspondingly, the probability that the cat loses a particular life is $$1-(0.8)^{10} = 0.89263$$
Hence, the probability of losing all 9 lives over the next 10 weeks is $$(0.89263)^9 = 0.360$$
(b) By using similar logic to above, we find the probability that the cat is dead at the end of the 5th week is $$(1-0.8^{5})^9 = 0.02807$$
and continuing, we find the probability that the cat is dead at the end of the 4th week is $$(1-0.8^{4})^9 = 0.00872$$
Hence, we find the probability that the cat dies during the 5th week is equal to $$P(\text{Dead at the end of week 5})-P(\text{Dead at the end of week 4}) = 0.02807-0.00872 = 0.019$$