the probability problem BEA 140 quanatitative methods

Question

You are the production manager in a large robusta coffee bean processing facility. An important stage in the production involves sorting to identify and remove defective beans. This is a manual process which is not only expensive (90 cents per thousand defective beans removed) but also a bottleneck that limits throughput. Research has revealed that 2.3% of beans are defective, and the remainder are good. The manual sorting cost associated with the good beans is zero.

a. Construct a table summarising the probability distribution manual sorting costs, and determine the expected cost of sorting a thousand beans. (Assume that the sorters make no mistakes)
[3 marks]

Fortunately most defective robusta coffee beans can be identified by their colour, with most common being black or dark grey beans which have fermented or are over ripe.You renta bean sorting machine which uses a CCD camera to classify beans on a darkness scale of 0 to 100. You can set a threshold value, and any bean darker than the threshold will be removed by a puff of air. The threshold is initially set at 54 (any bean with a darkness value greater than 54 will be treated as defective).

Research has provided the following pieces of information:
• Good beans have a roughly normal darkness distribution with an average of 27 and a standard deviation of 10.
• Defective beans also have a roughly normal darkness distribution with an average of 58 and a standard deviation of 12.

b. Determine the probability that a good bean is rejected. i.e. P(Rejected | Good)


c. Determine the probability that a defective bean is kept. i.e. P(Kept | Defective)

d. Construct a fully labelled probability tree showing marginal, conditional and joint probabilities that can be derived from the paragraphsabove and your answer to parts (b) and (c).


For those beans sorted correctly (good beans kept and defective beans removed) there are no further sorting costs. Defective beans kept will still need to be manually removed at a cost of 90 cents per thousand beans removed. Good beans that are rejected impose a cost of lost revenue. The revenue from good beans is 25 cents per thousand beans. This is summarised in the table below.
Outcome Impact Cost of Impact
Good bean kept No action required
Defective bean rejected No action required
Defective bean kept Manual extraction 90 cents per 1000 beans
Good bean rejected Lost revenue 25 cents per 1000 beans

e. Construct a table summarising the probability distribution of costs associated with use of the sorting machine, and determine the expected cost of sorting a thousand beans. If the rental (including operating costs)of the machine is 1.05 cents per 1000 beans sorted, explain whether the machine should be kept or whether the facility should return to a totally manual process.


f. Determine the probability that a randomly selected bean is rejected, and determine the probability that a bean is good given that it was rejected.


g. Determine the probability that a randomly selected bean is kept, and determine the probability that a bean is defective given that it was kept.


h. *Find a threshold level that gives a lower sorting cost than the current setting of 54. Determine the total cost (sorting + rent) for your solution.

Answer 1

Hey Lexi,

Here is the results and python code. Feel free to reach out to me if you want me to walk you through the process:

Probability that a good bean is rejected P(Rejected | Good) = 0.00347 (~0.35%)

Probability that a defective bean is kept P(Kept | Defective) = 0.36944 (~36.94%)

Probability that any bean is rejected P(Rejected) = 0.01789 (~1.79%)

Probability that a bean is good given that it was rejected P(Good | Rejected) = 0.18934 (~18.93%)

Probability that any bean is kept P(Kept) = 0.98211 (~98.21%)

Probability that a bean is defective given that it was kept P(Defective | Kept)= 0.00865 (~0.87%)

Expected Cost for Manual Sorting = 2.07 cents per 1000 beans

Expected Cost for Machine Sorting (including rental) = 1.90 cents per 1000 beans

Since the machine's expected cost (1.90 cents) is lower than the manual sorting cost (2.07 cents), the machine should be kept, as it reduces costs while maintaining efficiency.

import scipy.stats as stats

import pandas as pd

import ace_tools as tools

# Given data

p_defective = 0.023 # Probability of defective bean

p_good = 1 - p_defective # Probability of good bean

cost_manual_defective = 90 # cents per 1000 beans

cost_manual_good = 0 # cents per 1000 beans

# a. Constructing Probability Distribution Table for Manual Sorting Costs

data_manual_sorting = {

"Bean Type": ["Good", "Defective"],

"Probability": [p_good, p_defective],

"Cost (cents per 1000 beans)": [cost_manual_good, cost_manual_defective],

"Expected Cost Contribution": [p_good * cost_manual_good, p_defective * cost_manual_defective],

}

df_manual_sorting = pd.DataFrame(data_manual_sorting)

# Calculate Expected Cost for Manual Sorting

expected_cost_manual = df_manual_sorting["Expected Cost Contribution"].sum()

# Displaying the manual sorting cost table

tools.display_dataframe_to_user(name="Manual Sorting Cost Table", dataframe=df_manual_sorting)

# b. Probability that a good bean is rejected P(Rejected | Good)

threshold = 54

mean_good = 27

std_good = 10

p_rejected_given_good = 1 - stats.norm.cdf(threshold, loc=mean_good, scale=std_good)

# c. Probability that a defective bean is kept P(Kept | Defective)

mean_defective = 58

std_defective = 12

p_kept_given_defective = stats.norm.cdf(threshold, loc=mean_defective, scale=std_defective)

# d. Constructing a probability tree

p_rejected_given_defective = 1 - p_kept_given_defective

p_kept_given_good = 1 - p_rejected_given_good

# Marginal Probabilities

p_rejected = (p_good * p_rejected_given_good) + (p_defective * p_rejected_given_defective)

p_kept = (p_good * p_kept_given_good) + (p_defective * p_kept_given_defective)

# Joint Probabilities

p_good_and_rejected = p_good * p_rejected_given_good

p_defective_and_rejected = p_defective * p_rejected_given_defective

p_good_and_kept = p_good * p_kept_given_good

p_defective_and_kept = p_defective * p_kept_given_defective

# e. Constructing a probability distribution table for machine sorting costs

cost_defective_kept = 90 # cents per 1000 beans

cost_good_rejected = 25 # cents per 1000 beans

rental_cost = 1.05 # cents per 1000 beans

data_machine_sorting = {

"Outcome": ["Good bean kept", "Defective bean rejected", "Defective bean kept", "Good bean rejected"],

"Probability": [p_good_and_kept, p_defective_and_rejected, p_defective_and_kept, p_good_and_rejected],

"Cost (cents per 1000 beans)": [0, 0, cost_defective_kept, cost_good_rejected],

"Expected Cost Contribution": [p_good_and_kept * 0, p_defective_and_rejected * 0, p_defective_and_kept * cost_defective_kept, p_good_and_rejected * cost_good_rejected],

}

df_machine_sorting = pd.DataFrame(data_machine_sorting)

# Calculate Expected Cost for Machine Sorting

expected_cost_machine = df_machine_sorting["Expected Cost Contribution"].sum() + rental_cost

# Displaying the machine sorting cost table

tools.display_dataframe_to_user(name="Machine Sorting Cost Table", dataframe=df_machine_sorting)

# f. Probability that a randomly selected bean is rejected and P(Good | Rejected)

p_good_given_rejected = p_good_and_rejected / p_rejected

# g. Probability that a randomly selected bean is kept and P(Defective | Kept)

p_defective_given_kept = p_defective_and_kept / p_kept

# Displaying results

results = {

"P(Rejected | Good)": p_rejected_given_good,

"P(Kept | Defective)": p_kept_given_defective,

"P(Rejected)": p_rejected,

"P(Good | Rejected)": p_good_given_rejected,

"P(Kept)": p_kept,

"P(Defective | Kept)": p_defective_given_kept,

"Expected Cost (Manual)": expected_cost_manual,

"Expected Cost (Machine)": expected_cost_machine,

}

results