Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a strange question, it might be easy for all you math wizards or it even may be impossible. If you don't understand what I mean let me know so I can change the way I post the question.

Product A can be built using Product B.

To get Product B I have to transform product C or Product D.

I have a 50% chance to get product B by transforming product C and 25% to get product B by transforming product D.

Is there any way to predict the cost of production for Product A knowing the price for Product C and D

Edit 1 There are no transformation costs, only the costs for the materials

share|improve this question
add comment

2 Answers

Suppose you go with the strategy "buy C and attempt a transformation to B until you get B". Then with probability 1/2, this will cost C, with probability $1/2(1-1/2)$, cost $2C$, with probability $1/2^n$ cost $nC$. Thus the expected price will be

$C\sum_{n=1}^\infty \frac{n}{2^n}=2C$

Similarly, the strategy "buy D and attempt a transformation to B until you get B" will have expected price

$B_D+D\sum_{n=1}^\infty \frac{n3^{n-1}}{4^n}=4D$

It's no coincidence that the expected price is just the cost divided by the probability of success; this follows from the properties of the binomial distribution.

You'll want to compare these two and see which is minimal and run with that strategy.

share|improve this answer
There is no cost of transformation, how would that change your calculation? Sorry for the newbiesh question. –  Nuno Furtado Nov 3 '10 at 17:19
As per their definitions, $B_C$ and $B_D$ will be equal to zero... –  j.c. Nov 4 '10 at 10:02
add comment

What i thinks is that if you go for Producing Product A from Product C it will cost you double the total price of Product C and if you Produce Product A from Product D it will cost you 4 times the total Price of Product D according to Probabilities respectively :)

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.