# How much improvement in probability of power cells B wrt A?

1. Mar 4, 2015

### s3a

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!

2. Mar 4, 2015

### Ray Vickson

The improvement in success, from 0.90 to 0.99, looks modest, but the improvement in failure, from 0.099 down to 0.002 is much more impressive. Basically, failure is something to be avoided, so looking at project failure probabilities makes more sense. That being said, a valid comparison should try to assess the cost of failure vs the cost of batteries. In the absence of such failure cost information, all you can do is examine probabilities, and try to judge whether increasing the cost from $6 to$24 is worth it.

In principle, one might consider mixing up the two battery types (if technically allowed), so one could use some of type A and some of type B. That would give failure probabilities and costs that lie between the extremes above.

3. Mar 4, 2015

### s3a

Okay, I was expecting the ratio comparing successes to be equal to the ratio comparing failures, but I see that they aren't.

That comment you made about it being better to examine information about failure than success makes "partial sense", but could you please tell me in words (in a way that answers part c)) what the two ratios mean? Basically, what do 0.99777/0.90112 = 1.10725541548295454545 and 0.099/0.002 = 49.5 mean in words?

4. Mar 4, 2015

### Ray Vickson

You are just as capable as I am of interpreting them in words. Anyway, if I were analyzing the problem I would not bother with ratios---that is just something your book insists that you do.

5. Mar 4, 2015

### haruspex

There is no justification for using either ratio. The wording of the question clearly encourages you to use the ratio of probabilities of success, so you could defend that approach. There is no defence here for using the ratio of the failure probabilities. The question setter is being too tricky for his own good.
Let the value of succes (or, equivalently, cost of failure) be v. If the success probabilities are pA, pB, and the battery costs are cA, cB, what are the expected outcomes of the two strategies?