I own a bar by Raymond James Stadium when the Bucs play. I do a lot of business game day. If the Bucs win, 1000 people come in. If the Bucs lose, 300 people come in. This Sunday I give the Bucs a 80% chance of winning a 20% chance of losing. What is the expected value for the number of people that come into bar this Sunday?

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**Problem Statement:**

_I own a bar by Raymond James Stadium when the Bucs play. I do a lot of business on game day. If the Bucs win, 1000 people come in. If the Bucs lose, 300 people come in. This Sunday I give the Bucs an 80% chance of winning and a 20% chance of losing. What is the expected value for the number of people that come into the bar this Sunday?_

**Solution Explanation:**

To find the expected value, we use the formula for the expected value of a discrete random variable:

**Expected Value (E) = (Probability of Event 1 × Value of Event 1) + (Probability of Event 2 × Value of Event 2)**

Here, the two events are:
1. The Bucs win.
2. The Bucs lose.

We are given:
- Probability of the Bucs winning (P(Win)) = 80% = 0.8
- Probability of the Bucs losing (P(Lose)) = 20% = 0.2
- Number of people if the Bucs win = 1000
- Number of people if the Bucs lose = 300

Plugging these values into the formula:

**Expected Value (E) = (0.8 × 1000) + (0.2 × 300)**

Calculate each term:
- (0.8 × 1000) = 800
- (0.2 × 300) = 60

Therefore:

**Expected Value (E) = 800 + 60**
**Expected Value (E) = 860**

**Conclusion:**
The expected value for the number of people that come into the bar this Sunday is 860.
Transcribed Image Text:**Problem Statement:** _I own a bar by Raymond James Stadium when the Bucs play. I do a lot of business on game day. If the Bucs win, 1000 people come in. If the Bucs lose, 300 people come in. This Sunday I give the Bucs an 80% chance of winning and a 20% chance of losing. What is the expected value for the number of people that come into the bar this Sunday?_ **Solution Explanation:** To find the expected value, we use the formula for the expected value of a discrete random variable: **Expected Value (E) = (Probability of Event 1 × Value of Event 1) + (Probability of Event 2 × Value of Event 2)** Here, the two events are: 1. The Bucs win. 2. The Bucs lose. We are given: - Probability of the Bucs winning (P(Win)) = 80% = 0.8 - Probability of the Bucs losing (P(Lose)) = 20% = 0.2 - Number of people if the Bucs win = 1000 - Number of people if the Bucs lose = 300 Plugging these values into the formula: **Expected Value (E) = (0.8 × 1000) + (0.2 × 300)** Calculate each term: - (0.8 × 1000) = 800 - (0.2 × 300) = 60 Therefore: **Expected Value (E) = 800 + 60** **Expected Value (E) = 860** **Conclusion:** The expected value for the number of people that come into the bar this Sunday is 860.
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