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Calculating the expected value for a probability

  • #1

Homework Statement


An ice-cream store makes 150 ice-cream balls every day. The cost of making each ice-cream ball is $3. The price of an ice-cream ball is $8. The demand distribution is as follows: 100 ice-cream balls with probability 25%, 150 ice-cream balls with probability 50%, and 200 ice-cream balls with probability 25%
a. What is the store’s expected profit every day?
b. If the store decides to make 200 ice-cream balls every day, what is the store’s expected profit every day?

Homework Equations




The Attempt at a Solution


I'm confused whether i should use the probability for 200 ice-cream for part a since the store can only produced 150 ice-cream balls every day. But for part b, i should use the probability for 200 ice-cream balls, right?
 

Answers and Replies

  • #2
Stephen Tashi
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I'm confused whether i should use the probability for 200 ice-cream for part a since the store can only produced 150 ice-cream balls every day.
Use it. The demand for 200 is relevant. Demand is not necessarily satisfied by sales. If the demand is 200 and only 150 are produced, you assume 150 are sold.
 
  • #3
Thanks! If this is the case then, the two questions will have the same answer?
This is my solution:
profit=8-3=5
expected profit per day=5(100)(25%)+5(150)(50%)+200(5)(25%)
Is this right?
 
  • #4
Ray Vickson
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Thanks! If this is the case then, the two questions will have the same answer?
This is my solution:
profit=8-3=5
expected profit per day=5(100)(25%)+5(150)(50%)+200(5)(25%)
Is this right?
Why would you think both solutions have the same expected profit? In one case if customers order 200 you can only sell them 150; in the other case you sell them 200, so the expected revenue goes up. Of course, the "manufacturing" cost also goes up (because you make more), so some additional calculations are needed to assess the net effects.
 
  • #5
RPinPA
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Go back to first principles. What is the expectation value of a discrete random variable? It's the sum of the values of that variable times the probability it has that value. Don't blindly throw numbers around, THINK.

Expected profit per day = ##\sum \text{Profit}(n_i \text{ sales}) \text{Probability}(n_i \text{ sales})##

Now, here is where you are going wrong. You need to carefully think through what is the profit (revenue minus costs) in each of the possible cases.

First, it is assumed that they go ahead and make 150 at the beginning of each day, not knowing what they're going to sell. NO MATTER WHAT HAPPENS.

There could be a demand for 100, so they could sell 100. What is their revenue from sales? What did they spend at the start of the day? So what is their profit?
There could be a demand for 150. How many would they sell? What is their revenue from sales? What did they spend at the start of the day? So what is their profit?
There could be a demand for 200. How many would they sell? What is their revenue from sales? What did they spend at the start of the day? So what is their profit?

These are the questions which you need to think through. Also, these are the questions which do not necessarily have the same answers for parts (a) and (b).
 
  • #6
i get it now...
so for part a:
profit for demand 100 = 100(8-3)=500
for 150 = 150(8-3)=750
for 200 = 150(8-3)=750
so expected profit = 500(25%)+750(50%)+750(25%)=687.5

for part b:
expected profit=500(25%)+750(50%)+1000(25%)=750
thanks for the thorough explanation
 
  • #7
RPinPA
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Closer. But you've missed something about the costs. Reread the problem again, and reread what I said.
 
  • #8
part a:
(100x8-150x3)(25%)+150(5)(50%)+150(5)(25%)=650

part b:
(100x8-200x3)(25%)+(150x8-200x3)(50%)+200(5)(25%)=600
 

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