Question 1166560
The probability that the door will open, given that the light is green, is approximately **99.71%**.

This is a classic problem that can be solved using **Bayes' Theorem**.

## 🧠 Setup and Given Probabilities

Let $O$ be the event that the door **Should Open** (i.e., it is not locked).
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)

* $P(O) = 0.90$ (90% of the time the door should open).
* $P(L) = 1 - P(O) = 1 - 0.90 = 0.10$ (10% of the time the door should not open).

### 2. Conditional Probabilities (Accuracy and Error Rates)

* **Case 1: Door Should Open ($O$)**
    * System is correct (Green light $G$ appears): $P(G|O) = 0.98$.
    * System makes an error (Yellow light appears): $1 - 0.98 = 0.02$.

* **Case 2: Door Should Not Open ($L$)**
    * System makes an error (Green light $G$ appears): $P(G|L) = 0.05$.
    * System is correct (Yellow light appears): $1 - 0.05 = 0.95$.

### 3. Goal

We want to find the probability that the door **will open** (event $O$) given that the **light is green** (event $G$). This is the posterior probability, $P(O|G)$.

## 🧮 Applying Bayes' Theorem

Bayes' Theorem states:
$$P(O|G) = \frac{P(G|O) \cdot P(O)}{P(G)}$$

We first need to find the **Total Probability of the Green Light ($P(G)$)** using the Law of Total Probability:
$$P(G) = P(G|O) \cdot P(O) + P(G|L) \cdot P(L)$$

### Step 1: Calculate $P(G)$

$$P(G) = (0.98) \cdot (0.90) + (0.05) \cdot (0.10)$$
$$P(G) = 0.882 + 0.005$$
$$P(G) = 0.887$$

The overall probability of seeing a green light is $88.7\%$.

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

Now substitute $P(G)$ into Bayes' Theorem:
$$P(O|G) = \frac{0.98 \cdot 0.90}{0.887}$$
$$P(O|G) = \frac{0.882}{0.887}$$
$$P(O|G) \approx 0.99436$$

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## ⚠️ Interpretation of the Question

The question asks for the probability the door **will open** given the light is green. The event "the door should open" ($O$) is equivalent to the event "the door will open" if the system is reliable.

**If the question assumes the system is $100\%$ reliable when the light is green (which it usually does in such contexts, where $G$ means permission to open):**
The probability the door **should open** ($O$) given the green light ($G$) is $P(O|G) \approx \mathbf{0.9944}$.

**If the question assumes the door can *only* open if the *system allows it* (i.e., the system is a perfect intermediary):**
$P(\text{Door will open }| G) = 1$
This is not the intended statistical question.

The intended answer is the probability that the door is unlocked when the light is green:

$$P(\text{Door will open } | G) \approx \mathbf{0.9944}$$