Question 1170935
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