# 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)?

answered Apr 14, 2018 by (580 points)
selected Apr 15, 2018 by Daniel

(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 $$P(\text{Dead at the end of week 5})-P(\text{Dead at the end of week 4}) = 0.02807-0.00872 = 0.019$$