SOLUTION: #3 3. It is estimated that 4% of people who spend time in the woods will get Lyme disease. For people with Lyme disease, the test to determine if you have it is will give a positiv

Question 1170935: #3 3. It is estimated that 4% of people who spend time in the woods will get Lyme disease. For people with Lyme disease, the test to determine if you have it is will give a positive reading 97% of the time. Of people who do not have Lyme disease, the same test will give a negative rating of 92% of the time. Make a tree diagram for this problem and then answer the following questions. Find the probability that A) a person gets a positive reading B) a person gets a negative reading C) a person gets a positive reading and they have Lyme disease D) a person gets a negative reading and they have Lyme disease E ) a person gets a negative reading and they don’t have Lyme diseasePositive
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Absolutely! Let's break down this problem step by step.
**1. Tree Diagram**
* **First Branch:**
* Lyme Disease (L): Probability = 0.04 (4%)
* No Lyme Disease (¬L): Probability = 0.96 (96%)
* **Second Branch (for Lyme Disease):**
* Positive Test (+): Probability = 0.97 (97%)
* Negative Test (-): Probability = 0.03 (3%)
* **Second Branch (for No Lyme Disease):**
* Positive Test (+): Probability = 0.08 (8%) (Since negative is 92%)
* Negative Test (-): Probability = 0.92 (92%)
**2. Calculations**
Let's calculate the probabilities of each outcome:
* P(L+) = P(L) * P(+|L) = 0.04 * 0.97 = 0.0388
* P(L-) = P(L) * P(-|L) = 0.04 * 0.03 = 0.0012
* P(¬L+) = P(¬L) * P(+|¬L) = 0.96 * 0.08 = 0.0768
* P(¬L-) = P(¬L) * P(-|¬L) = 0.96 * 0.92 = 0.8832
**3. Answering the Questions**
A) **Probability of a positive reading:**
* P(+) = P(L+) + P(¬L+) = 0.0388 + 0.0768 = 0.1156
B) **Probability of a negative reading:**
* P(-) = P(L-) + P(¬L-) = 0.0012 + 0.8832 = 0.8844
C) **Probability of a positive reading and having Lyme disease:**
* This is P(L+), which we already calculated: 0.0388
D) **Probability of a negative reading and having Lyme disease:**
* This is P(L-), which we already calculated: 0.0012
E) **Probability of a negative reading and not having Lyme disease:**
* This is P(¬L-), which we already calculated: 0.8832
**Summary of Answers**
A) 0.1156
B) 0.8844
C) 0.0388
D) 0.0012
E) 0.8832

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