Question 1166442: Due to the high incidence of crime, a company is giving its employees new and secured access to the company. Instead of leaving the doors open, they are installing a card system. To open the door, employees must insert a card into a slot. If a green light comes on, it is okay to turn the handle and open the door; if a yellow light comes on it indicates the door is locked from inside and you cannot enter. Suppose that 90% of the time when the card is inserted, the door should open because it is not locked from the inside. When the door should open, the system makes errors 2% of the time. That is, the green light will appear 98% of the time. When the door should not open, the system makes errors 5% of the time (green light appears). Suppose you inserted the card and the light is green, what is the probability the door will open?
Please help, thanks
Answer by CPhill(2030)   (Show Source):
You can put this solution on YOUR website! 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 **$0.9944$**.
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