Question 1166442
The probability that the door will open, given that the light is green, is approximately **99.44%** (or **0.9944**).

This is a **Bayes' Theorem** problem, where we update our initial belief (prior probability) based on new evidence (the green light). 

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## 🔬 Probability Setup

Let $O$ be the event that the door **Should Open** (i.e., it is unlocked).
Let $L$ be the event that the door **Should Not Open** (i.e., it is locked).
Let $G$ be the event that the **Green Light** appears.

### 1. Prior Probabilities (What *Should* Happen)

* $P(O)$ (Probability the door is unlocked) $= 90\% = 0.90$
* $P(L)$ (Probability the door is locked) $= 1 - P(O) = 1 - 0.90 = 0.10$

### 2. Likelihoods (System Accuracy)

* **True Positive Rate ($P(G|O)$):** Probability of a green light when the door *should* open.
    $$P(G|O) = 98\% = 0.98$$
* **False Positive Rate ($P(G|L)$):** Probability of a green light when the door *should not* open (system error).
    $$P(G|L) = 5\% = 0.05$$

### 3. Goal

We want to find $P(O|G)$, the probability that the door is actually unlocked (and will open) given that we see a green light.

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## 🧮 Applying Bayes' Theorem

$$P(O|G) = \frac{P(G|O) \cdot P(O)}{P(G)}$$

### Step 1: Calculate the Total Probability of a Green Light ($P(G)$)

The green light can occur in two ways: correctly (True Positive) or incorrectly (False Positive).

$$P(G) = P(\text{Green and Open}) + P(\text{Green and Locked})$$
$$P(G) = [P(G|O) \cdot P(O)] + [P(G|L) \cdot P(L)]$$
$$P(G) = (0.98 \cdot 0.90) + (0.05 \cdot 0.10)$$
$$P(G) = 0.882 + 0.005$$
$$P(G) = 0.887$$

The overall probability of the system showing a green light is $88.7\%$.

### Step 2: Calculate the Posterior Probability $P(O|G)$

$$P(O|G) = \frac{P(\text{Green and Open})}{P(\text{Total Green})}$$
$$P(O|G) = \frac{0.882}{0.887}$$
$$P(O|G) \approx 0.99436$$

Rounding to four decimal places, the probability the door will open is **$0.9944$**.