This assessment is focused on the material contained in week 4 related to probability and probability distributions.  Some information for this assessment can be found in the file Problem Solving Task 3.xlsx.  As usual, do the working in Excel, but copy all of your solutions to an answer document in Word.  Note that if your solution is a single value that was generated using an Excel formula, please include the formula itself on your solution document e.g. for a binomial distribution calculation you may include the formula binom.dist(x value, sample size, probability, cumulative) with the correct values entered for each of the four required inputs.  Where calculations are done by hand, please include all working  Note that if an answer is given with no working or excel output it will not be marked.  Once completed save your solutions as a PDF file using the title PST3.pdf e.g. Andrew_paltridge pst3.pdf.  Submit only the PDF and not the Excel worksheet. We will request that if required. Performance enhancing drugs in athletes is a big issue and has been around for some time. One estimate is that 5% of all athletes use drugs (D) which means 95% are clean (C). In professional sports athletes are tested regularly, and some athletes who are clean test positive (P), and some who use drugs test negative (N) – false positives and false negatives. In relation to the reliability of testing it is estimated that 3% of Clean athletes test positive and 8% of Drug Takiing athletes test negative. P(P/C) = 0.03 and P(N/D) = 0.08. Note this implies P(N/C) = 0.97 a. Calculate the probability that an athlete is a Drug user and they test Positive b. Calculate the probability that an athlete is Clean and they test Negative c. Calculate the probability that an athlete is Clean or they test Negative d. From the calculations so far, construct a joint probability table comparing athlete status and their testing status i.e. fill out the table below. Show all calculations Test Positive Test Negative Total Drug User Clean Total e. Calculate the probability that an athlete who tested positive was in fact a drug user. f. What is the probability that an athlete who tested negative was clean. Assessment Instructions This assessment is focused on the material contained in week 4 related to probability and probability distributions.  Some information for this assessment can be found in the file Problem Solving Task 3.xlsx.  As usual, do the working in Excel, but copy all of your solutions to an answer document in Word.  Note that if your solution is a single value that was generated using an Excel formula, please include the formula itself on your solution document e.g. for a binomial distribution calculation you may include the formula binom.dist(x value, sample size, probability, cumulative) with the correct values entered for each of the four required inputs.  Where calculations are done by hand, please include all working  Note that if an answer is given with no working or excel output it will not be marked.  Once completed save your solutions as a PDF file using the title PST3.pdf e.g. Andrew_paltridge pst3.pdf.  Submit only the PDF and not the Excel worksheet. We will request that if required. 2 | P a g e Applied Business Analytics g. Athletes that test positive usually have a second sample tested independently to the first. What is the probability that an athlete who was clean would test positive on both the first and second samples

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∻Solution by Steps∻ ‖ step 1 ⋮ Given P(D) = 0.05 and P(P|D) = 1 - P(N|D) = 1 - 0.08 = 0.92. To find P(D and P), we use the formula P(D and P) = P(D) * P(P|D). ‖ ‖ step 2 ⋮ Calculate P(D and P): P(D and P) = 0.05 * 0.92. ‖ ‖ step 3 ⋮ Simplifying the calculation: P(D and P) = 0.046. ‖ ∻Answer∻ ⚹ P(D and P) = 0.046 ⚹ ∻Key Concept∻ ⚹Joint Probability of Independent Events⚹ ∻Explanation∻ ⚹The joint probability of two independent events, A and B, is the product of their individual probabilities: P(A and B) = P(A) * P(B|A).⚹ ∻Solution by Steps∻ ‖ step 1 ⋮ Given P(C) = 0.95 and P(N|C) = 0.97. To find P(C and N), we use the formula P(C and N) = P(C) * P(N|C). ‖ ‖ step 2 ⋮ Calculate P(C and N): P(C and N) = 0.95 * 0.97. ‖ ‖ step 3 ⋮ Simplifying the calculation: P(C and N) = 0.9215. ‖ ∻Answer∻ ⚹ P(C and N) = 0.9215 ⚹ ∻Key Concept∻ ⚹Joint Probability of Independent Events⚹ ∻Explanation∻ ⚹The joint probability of two independent events, A and B, is the product of their individual probabilities: P(A and B) = P(A) * P(B|A).⚹ ∻Solution by Steps∻ ‖ step 1 ⋮ To find P(C or N), we use the formula P(C or N) = P(C) + P(N) - P(C and N), where P(N) = P(N|D) * P(D) + P(N|C) * P(C). ‖ ‖ step 2 ⋮ Calculate P(N): P(N) = 0.08 * 0.05 + 0.97 * 0.95. ‖ ‖ step 3 ⋮ Simplifying the calculation for P(N): P(N) = 0.004 + 0.9215 = 0.9255. ‖ ‖ step 4 ⋮ Calculate P(C or N): P(C or N) = 0.95 + 0.9255 - 0.9215. ‖ ‖ step 5 ⋮ Simplifying the calculation for P(C or N): P(C or N) = 0.954. ‖ ∻Answer∻ ⚹ P(C or N) = 0.954 ⚹ ∻Key Concept∻ ⚹Probability of Union of Two Events⚹ ∻Explanation∻ ⚹The probability of the union of two events, A or B, is given by P(A or B) = P(A) + P(B) - P(A and B).⚹ ∻Solution by Steps∻ ‖ step 1 ⋮ To construct a joint probability table, we use the probabilities calculated in the previous steps and the given probabilities. ‖ ‖ step 2 ⋮ Fill in the table with P(D and P), P(C and N), P(D and N), and P(C and P) using the formulas P(D and N) = P(D) - P(D and P) and P(C and P) = P(C) - P(C and N). ‖ ‖ step 3 ⋮ Calculate P(D and N): P(D and N) = 0.05 - 0.046 = 0.004. ‖ ‖ step 4 ⋮ Calculate P(C and P): P(C and P) = 0.95 - 0.9215 = 0.0285. ‖ ‖ step 5 ⋮ Complete the table with the calculated probabilities. ‖ ∻Answer∻ ⚹ Joint Probability Table: Test Positive | Test Negative | Total Drug User: 0.046 | 0.004 | 0.05 Clean: 0.0285 | 0.9215 | 0.95 Total: 0.0745 | 0.9255 | 1 ⚹ ∻Key Concept∻ ⚹Construction of a Joint Probability Table⚹ ∻Explanation∻ ⚹A joint probability table summarizes the probabilities of different combinations of two or more categorical variables.⚹ ∻Solution by Steps∻ ‖ step 1 ⋮ To calculate P(D|P), we use the formula P(D|P) = P(D and P) / P(P), where P(P) = P(D and P) + P(C and P). ‖ ‖ step 2 ⋮ Calculate P(P): P(P) = 0.046 + 0.0285. ‖ ‖ step 3 ⋮ Simplifying the calculation for P(P): P(P) = 0.0745. ‖ ‖ step 4 ⋮ Calculate P(D|P): P(D|P) = 0.046 / 0.0745. ‖ ‖ step 5 ⋮ Simplifying the calculation for P(D|P): P(D|P) = 0.6174. ‖ ∻Answer∻ ⚹ P(D|P) = 0.6174 ⚹ ∻Key Concept∻ ⚹Conditional Probability⚹ ∻Explanation∻ ⚹Conditional probability is the probability of an event occurring given that another event has already occurred, which is calculated using the formula P(A|B) = P(A and B) / P(B).⚹ ∻Solution by Steps∻ ‖ step 1 ⋮ To calculate P(C|N), we use the formula P(C|N) = P(C and N) / P(N), where P(N) was calculated previously. ‖ ‖ step 2 ⋮ Use the previously calculated P(N): P(N) = 0.9255. ‖ ‖ step 3 ⋮ Calculate P(C|N): P(C|N) = 0.9215 / 0.9255. ‖ ‖ step 4 ⋮ Simplifying the calculation for P(C|N): P(C|N) = 0.9957. ‖ ∻Answer∻ ⚹ P(C|N) = 0.9957 ⚹ ∻Key Concept∻ ⚹Conditional Probability⚹ ∻Explanation∻ ⚹Conditional probability is the probability of an event occurring given that another event has already occurred, which is calculated using the formula P(A|B) = P(A and B) / P(B).⚹ ∻Solution by Steps∻ ‖ step 1 ⋮ To calculate the probability that an athlete who was clean would test positive on both the first and second samples, we use the formula P(P and P|C) = P(P|C) * P(P|C). ‖ ‖ step 2 ⋮ Given P(P|C) = 0.03, calculate P(P and P|C): P(P and P|C) = 0.03 * 0.03. ‖ ‖ step 3 ⋮ Simplifying the calculation: P(P and P|C) = 0.0009. ‖ ∻Answer∻ ⚹ P(P and P|C) = 0.0009 ⚹ ∻Key Concept∻ ⚹Independent Events Probability⚹ ∻Explanation∻ ⚹The probability of two independent events both occurring is the product of their individual probabilities, assuming the events are independent.⚹◊What is the formula to calculate the probability that an athlete who tested positive was in fact a drug user?⍭ Generate me a similar question◊

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