Chapter 3.3

Definition: Two events A and B are said to be mutually exclusive if they cannot occur at the same time. This means:
- A and B have no outcomes in common.
- For example:
- If Event A is rolling a 3 on a die, and Event B is rolling a 4 on a die, then these two events are mutually exclusive since one cannot roll a 3 and a 4 at the same time.
- If Event A is selecting a male student, and Event B is selecting a nursing major, then these two events are not mutually exclusive as a male student can also be a nursing major.
- In another instance, if Event A is selecting a blood donor with type O blood, and Event B is selecting a female donor, then these two events are also not mutually exclusive as a donor can be female with type O blood.

Formula: The probability that events A or B will occur is defined as:
P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)
For mutually exclusive events A and B, the rule simplifies to: P(A \text{ or } B) = P(A) + P(B)
- This rule can be extended to any number of mutually exclusive events.

Scenario: Selecting a card from a standard deck, finding the probability that the card is either a 4 or an ace.
Solution: The events are mutually exclusive because if the card is a 4, it cannot be an ace.
Scenario: Rolling a die, find the probability of rolling a number less than 3 or rolling an odd number.
Solution: The events are not mutually exclusive. Here, 1 is an outcome of both events.
- Apply the Addition Rule:
  P(\text{less than 3 or odd}) = P(\text{less than 3}) + P(\text{odd}) - P(\text{less than 3 and odd})
- Outcomes less than 3: {1, 2} → Total outcomes = 2/6
- Outcomes that are odd: {1, 3, 5} → Total outcomes = 3/6
- Outcomes that are both: 1 → Total outcomes = 1/6
- Thus,
  P(\text{less than 3 or odd}) = \frac{2}{6} + \frac{3}{6} - \frac{1}{6} = \frac{4}{6} = 0.667

Scenario: A frequency distribution of monthly sales in dollars over the past three years:
- Sales volume ranges and corresponding months:
- $0 - 24,999: 3 months
- $25,000 - 49,999: 5 months
- $50,000 - 74,999: 6 months
- $75,000 - 99,999: 7 months
- $100,000 - 124,999: 9 months
- $125,000 - 149,999: 2 months
- $150,000 - 174,999: 3 months
- $175,000 - 199,999: 1 month
Solution: Let A represent monthly sales between $75,000 and $99,999, and let B represent monthly sales between $100,000 and $124,999. A and B are mutually exclusive.

The data gathered by a blood bank on blood types over the last five days:
- Types of blood given:
- Type O: Rh-Positive = 156, Rh-Negative = 28, Total = 184
- Type A: Rh-Positive = 139, Rh-Negative = 25, Total = 164
- Type B: Rh-Positive = 37, Rh-Negative = 8, Total = 45
- Type AB: Rh-Positive = 12, Rh-Negative = 4, Total = 16
- Grand Total of donors = 409
Find the Probability: Probability that a donor has type O or type A blood. The events are mutually exclusive.
- Calculate:
  P(\text{Type O or Type A}) = P(\text{Type O}) + P(\text{Type A})
  = \frac{184}{409} + \frac{164}{409} = \frac{348}{409}

Types of Probability:
- Classical Probability: The number of outcomes in the sample space is known and each outcome is equally likely to occur.
- Empirical Probability: The frequency of each outcome in the sample space is estimated from experimentation.
Rule Ranges:
- The probability of an event is between 0 and 1, inclusive.
- Complementary Events: The complement of event E is denoted by E' and represents outcomes not included in E.
- Multiplication Rule:
- For dependent events: P(A \text{ and } B) = P(A) \cdot P(B|A)
- For independent events: P(A \text{ and } B) = P(A) \cdot P(B)
- Addition Rule:
- For mutually exclusive events: P(A \text{ or } B) = P(A) + P(B)
- For non-mutually exclusive events: P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)

Scenario: Find the probability that a randomly selected draft pick is not a running back or a wide receiver.
Define events:
- A: Draft pick is a running back.
- B: Draft pick is a wide receiver.
These events are mutually exclusive, therefore use the Addition Rule:
P(A \text{ or } B) = P(A) + P(B)
To find the complement: 1 - P(A \text{ or } B) to determine the probability of not selecting either position.
- Final Answer:
  P(\text{not running back or wide receiver}) = 1 - P(A \text{ or } B)
- Probability Number: Calculate using provided event probabilities.

[Note: Include necessary calculations based on the event probabilities mentioned earlier in the document.]