1. The problem statement, all variables and given/known data The full problem statement (I just need help for part c): You are determining which model of power cell to purchase for your project. Your project requires 6 power cells, at least 4 of which must work for the project to work. You would like your project to work for at least 100 hours. The two power cell characteristics are shown in the table below.: <I can't put a table here, so the data of the table will be given in a non-tabular format.> Probability of lasting over 100 hours for model A = 0.80 Probability of lasting over 100 hours for model B = 0.95 Cost per unit for model A = $1 Cost per unit for model B = $4 a) If you were to purchase 6 model A power cells, what is the probability that 4 or more cells will last over 100 hours? b) If you were to purchase 6 model B power cells, what is the probability that 4 or more cells will last over 100 hours? c) From the above information, does the higher cost produce a proportionally higher performance (i.e. the cost is 4 times larger, is the improvement in probability 4 times larger)? Justify your answer. d) Concisely explain in four or less sentences which power cell you would choose and justify your answer with probability concepts. Correct answers for parts a and b: a) 0.90112 b) 0.99777 2. Relevant equations Binomial distribution: nCr * p^r * (1 - p)^(n - r) 3. The attempt at a solution My trouble is with part c); the solution I'm looking at says to do (1 - 0.90112) / (1 - 0.99777) = 49.5 for the improvement in probability of model B cells (is "the improvement in probability" the correct terminology?), but why doesn't 0.99777 / 0.90112 work as well? Basically, why must one make a ratio involving the probabilities of the cells not lasting over 100 hours, and why can't one make a ratio (only) involving the probabilities of the cells lasting over 100 hours? Any input would be GREATLY appreciated!