Market Researchers Inc. has been hired to perform a study to determine if the market for a new 
product will be good or poor. In similar studies conducted in the past, whenever the market 
actually was good, the market research study indicated that it would be good 80% of the time. 
On the other hand, whenever the market actually was poor, the market study incorrectly 
predicted it would be good 20% of the time. Before the study is performed, it is believed there is 
a 30% chance the market will be good. When Market Researchers, Inc. performs the study for 
this product, the results predict the market will be good. Given the results of this study, what is 
the probability that the market actually will be good?

We are given a scenario where a market research study predicts whether the market for a product will be good or poor. This is a classic example of **Bayes' Theorem**, which allows us to update probabilities based on new evidence.

Let’s define the events:
- \( G \): The market is good.
- \( P \): The market is poor.
- \( S_g \): The study predicts the market will be good.

We are asked to find the **probability that the market is good given that the study predicts it will be good**, which can be written as \( P(G \mid S_g) \).

### Given Information:
1. **Prior Probability:**
   - \( P(G) = 0.30 \) (The probability that the market is good before the study).
   - \( P(P) = 1 - P(G) = 0.70 \) (The probability that the market is poor before the study).

2. **Conditional Probabilities:**
   - \( P(S_g \mid G) = 0.80 \) (The probability that the study predicts a good market given that the market is actually good).
   - \( P(S_g \mid P) = 0.20 \) (The probability that the study predicts a good market given that the market is actually poor).

### Using Bayes' Theorem:
\[
P(G \mid S_g) = \frac{P(S_g \mid G) P(G)}{P(S_g \mid G) P(G) + P(S_g \mid P) P(P)}
\]

Now, we can plug in the known values:

\[
P(G \mid S_g) = \frac{(0.80)(0.30)}{(0.80)(0.30) + (0.20)(0.70)}
\]

### Step-by-step calculation:
\[
P(G \mid S_g) = \frac{0.24}{0.24 + 0.14} = \frac{0.24}{0.38}
\]

\[
P(G \mid S_g) = 0.6316
\]

So, the probability that the market is actually good given that the study predicts it will be good is approximately **63.16%**.

---

Do you want further details or have any questions?

### Related Questions:
1. How would the probability change if the prior probability of the market being good was 50%?
2. What happens to the probability if the study's accuracy improves to 90% for predicting a good market?
3. How would you calculate the probability if the study predicts a poor market instead?
4. Can this analysis be extended to a case with more than two possible market outcomes?
5. How does increasing the accuracy of predicting a poor market affect the overall result?

### Tip:
Bayes' Theorem is a powerful tool for updating probabilities based on new evidence, especially in fields like market research, medical testing, and even spam filtering!

Market Researchers Inc. has been hired to perform a study to determine if the market for a new product will be good or poor. In similar studies conducted in the past, whenever the market actually was good, the market research study indicated that it would be good 80% of the time. On the other hand, whenever the market actually was poor, the market study incorrectly predicted it would be good 20% of the time. Before the study is performed, it is believed there is a 30% chance the market will be good. When Market Researchers, Inc. performs the study for this product, the results predict the market will be good. Given the results of this study, what is the probability that the market actually will be good?

Learn how to apply Bayes' Theorem to determine the probability that the market will be good based on a market research study. With a prior probability of 30% for a good market and study accuracy rates of 80% for predicting a good market and 20% for incorrectly predicting a poor market, the final probability is approximately 63.16%. Suitable for advanced high school and college-level students.

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