7.3 Probability & Counting Principles

Foundations of Probability & Simple Events

Probability measures the mathematical likelihood that a specific event will occur. It is expressed as a number ranging strictly from 0 to 1 inclusive (or as a percentage from $0%$ to $100%$).

$\text{Probability Scale: } 0 \le P(E) \le 1$

• $P(E) = 0$ (or $0%$): The event is impossible (e.g., rolling an 8 on a standard six-sided die).
• $P(E) = 0.5$ (or $50%$): The event is equally likely to occur as not to occur (e.g., flipping a fair coin and landing on heads).
• $P(E) = 1$ (or $100%$): The event is certain to occur (e.g., drawing a red or black card from a standard deck).

Theoretical vs. Experimental Probability

Candidates must distinguish between theoretical calculations and real-world experimental results:

1. Theoretical Probability: Based on mathematical reasoning assuming all outcomes in the sample space are equally likely: $P(E) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes in Sample Space}}$

2. Experimental Probability (Empirical Probability): Based on actual observed data from conducting an experiment or trial: $P_{\text{exp}}(E) = \frac{\text{Number of Times Event Occurred (Successes)}}{\text{Total Number of Trials Performed}}$

The Law of Large Numbers: As the number of trials in an experiment increases significantly, the experimental probability approaches closer and closer to the theoretical probability. For example, flipping a coin 10 times might yield 7 heads ($70%$ experimental), but flipping it 10,000 times will result in an experimental percentage extremely close to $50%$.

Complementary Events

The complement of an event $E$ (denoted as $E'$ or $\text{not } E$) consists of all outcomes in the sample space that are not part of event $E$. Because an event must either occur or not occur, the sum of their probabilities equals 1:

$P(E) + P(\text{not } E) = 1 \implies P(\text{not } E) = 1 - P(E)$

Step-by-Step Worked Example (Complement Rule): A spinner is divided into 8 equal sectors numbered 1 through 8. What is the probability of spinning a number that is NOT a prime number?

1. Identify the total sample space: $S = {1, 2, 3, 4, 5, 6, 7, 8}$, so total outcomes $n = 8$.
2. Identify the prime numbers in $S$: Prime numbers are ${2, 3, 5, 7}$ (4 favorable outcomes).
3. Calculate $P(\text{Prime}) = \frac{4}{8} = \frac{1}{2} = 0.50$.
4. Use the complement rule: $P(\text{Not Prime}) = 1 - P(\text{Prime}) = 1 - \frac{1}{2} = \frac{1}{2}$.
5. Conclusion: The probability of spinning a non-prime number is $\frac{1}{2}$ (or $50%$).

Sample Spaces & Counting Principles

Sample Space & Tree Diagrams

A sample space ($S$) is the set of all possible outcomes of a random experiment. For multi-stage experiments, sample spaces can be visualized using tree diagrams or systematic lists.

For example, flipping a fair coin three consecutive times generates the following tree diagram branches:

• First Flip: H, T
• Second Flip: HH, HT, TH, TT
• Third Flip (Sample Space): ${HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}$
• Total Outcomes: $2 \times 2 \times 2 = 8$ outcomes.

The Fundamental Counting Principle

Listing sample spaces manually becomes unmanageable for large multi-stage events. The Fundamental Counting Principle states that if stage 1 can occur in $m$ ways, stage 2 in $n$ ways, stage 3 in $k$ ways, and so on, then the total number of possible outcomes for the combined event is:

$\text{Total Outcomes} = m \times n \times k \times \dots$

Step-by-Step Worked Example: An elementary school cafeteria offers a lunch combo consisting of 1 main dish, 1 side item, and 1 drink. The choices are:

• Main Dish (4 choices): Pizza, Chicken Tenders, Turkey Wrap, Veggie Burger
• Side Item (3 choices): Apple Slices, Carrots, Fries
• Drink (2 choices): Milk, Juice

How many distinct lunch combo combinations are possible?

1. Apply the Fundamental Counting Principle: $4 \times 3 \times 2$.
2. Calculate: $4 \times 3 = 12$, and $12 \times 2 = 24$.
3. Conclusion: There are 24 unique lunch combinations.

Compound Probability, Permutations & Combinations

Independent vs. Dependent Events

A compound event involves combining two or more simple events.

1. Independent Events: The outcome of the first event has no effect on the probability of the second event. This typically occurs in sampling with replacement or sequential coin/dice rolls. $P(A \text{ and } B) = P(A) \times P(B)$

2. Dependent Events: The outcome of the first event changes the probability of the second event. This occurs in sampling without replacement. $P(A \text{ and } B) = P(A) \times P(B \mid A)$ (where $P(B \mid A)$ represents the probability of event B occurring given that event A has already occurred).

Step-by-Step Worked Example (Without Replacement): A bag contains 5 red marbles and 3 blue marbles (total = 8 marbles). If two marbles are drawn randomly without replacement, what is the probability that both marbles drawn are red?

1. Calculate probability of first marble being red: $P(\text{Red}_1) = \frac{5}{8}$.
2. Since the marble is NOT replaced, there are now 4 red marbles left out of 7 total remaining marbles.
3. Calculate conditional probability of second marble being red: $P(\text{Red}_2 \mid \text{Red}_1) = \frac{4}{7}$.
4. Multiply the probabilities: $P(\text{Both Red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}$.
5. Simplify the fraction by dividing numerator and denominator by 4: $\frac{20 \div 4}{56 \div 4} = \frac{5}{14}$.
6. Conclusion: The probability is $\frac{5}{14}$.

Permutations vs. Combinations

When selecting items from a group, you must determine whether the order of selection matters.

Worked Example (Combination): A teacher selects 3 students from a group of 5 candidates to form a safety patrol committee. How many different committees can be formed?

1. Does order matter? No, all 3 committee members have equal status. Use Combination formula with $n = 5, r = 3$.
2. $C(5, 3) = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times (2 \times 1)} = \frac{120}{6 \times 2} = \frac{120}{12} = 10$
3. Conclusion: Exactly 10 distinct student committees can be formed.