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How to deal with a cat with 9 lives?

+2 votes
55 views
asked Apr 13 in BUS 3018F - Models by Daniel (380 points)
This question is based on Tutorial 3 Question 1 (ii). It pertains to a cat that has 9 lives.

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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)?


1 Answer

+3 votes
answered Apr 14 by Michael (380 points)
selected Apr 15 by Daniel
 
Best answer
The important idea is that more than one death can occur in a week. Hence, in order to tackle question (a) and (b), the probability of losing a particular life over the time period must be found.

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

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