September 1

Set Operations and Probability

The Addition Rule of Probability is a fundamental principle for finding the probability of the union of two events, $A$ and $B$. It states that the probability of either event $A$ or event $B$ occurring is the sum of their individual probabilities minus the probability of their intersection. This subtraction is crucial because it prevents double-counting the outcomes that are common to both events.

\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]

• Example: In a class of 30 students, 15 are in the band, 10 are on the debate team, and 5 are in both. What is the probability that a randomly selected student is in either the band or on the debate team?
  • $P(\text{Band}) = 15/30 = 0.5$
  • $P(\text{Debate}) = 10/30 \approx 0.333$
  • $P(\text{Band} \cap \text{Debate}) = 5/30 \approx 0.167$
  • $P(\text{Band} \cup \text{Debate}) = 0.5 + 0.333 - 0.167 = 0.666$ or $20/30$.

De Morgan’s Laws for Probability

De Morgan’s Laws provide a way to relate the union and intersection of events and their complements. The first law states that the complement of the union of two events is the intersection of their complements. The second law (which is the one you listed) states that the complement of the intersection of two events is the union of their complements.

\[P(A^{C}\cup B^{C}) = P((A\cap B)^{C}) = 1 - P(A \cap B)\]

• Example: Using the class example, what is the probability that a student is neither in the band nor on the debate team?
  • This is the complement of being in at least one of the groups, which we found to be $P(A \cup B) = 0.666$.
  • $P(\text{Band}^{C} \cap \text{Debate}^{C}) = 1 - P(\text{Band} \cup \text{Debate}) = 1 - 0.666 = 0.334$.
  • Alternatively, using the formula listed: $P(\text{Band}^{C} \cup \text{Debate}^{C}) = P((\text{Band} \cap \text{Debate})^{C}) = 1 - P(\text{Band} \cap \text{Debate}) = 1 - 0.167 = 0.833$.

Permutations and Combinations

Permutations and Combinations are methods for counting the number of possible arrangements of objects. The key difference lies in whether the order of selection matters.

• Combinations ($C(n,k)$): The number of ways to choose $k$ items from a set of $n$ items, where the order does not matter. This is used for committees, groups, or selections.

  \[C(n,k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}\]
  • Example: A committee of 3 people is to be chosen from a group of 10. How many different committees are possible?
    • $C(10,3) = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120$.

• Permutations ($P(n, k)$): The number of ways to arrange $k$ items from a set of $n$ items, where the order does matter. This is used for rankings, codes, or ordered arrangements.

  \[P(n, k) = \frac{n!}{(n-k)!}\]
  • Example: The first, second, and third-place winners in a race of 10 people. How many ways can the top three finishers be arranged?
    • $P(10,3) = \frac{10!}{(10-3)!} = \frac{10!}{7!} = 10 \cdot 9 \cdot 8 = 720$.

Conditional Probability and Independence

Conditional Probability is the probability of an event occurring given that another event has already occurred. It is denoted as $P(A

B)$, which is read as “the probability of A given B”.

\[P(A|B) = \frac{P(A \cap B)}{P(B)} , \quad P(B) > 0\]

• Example: A study of 100 people found that 40 are smokers, and 15 are smokers with a lung condition. What is the probability that a randomly selected person has a lung condition given they are a smoker?
  • $P(\text{Smoker}) = 40/100 = 0.4$
  • $P(\text{Smoker} \cap \text{Lung Condition}) = 15/100 = 0.15$
  • $P(\text{Lung} \mid \text{Smoker}) = \frac{P(L \cap S)}{P(S)} = \frac{0.15}{0.4} = 0.375$.

Independence is a special case where the occurrence of one event does not affect the probability of the other. Two events $A$ and $B$ are independent if and only if one of the following equivalent conditions is met:

• $P(A\mid B) = P(A)$
• $P(B\mid A) = P(B)$

• $P(A \cap B) = P(A) \cdot P(B)$

• Example: You flip a coin and roll a die. Let $A$ be the event of getting a head, and $B$ be the event of rolling a 6. Are these events independent?
  • $P(A) = 0.5$
  • $P(B) = 1/6$
  • $P(A \cap B) = P(\text{getting a head AND rolling a 6}) = (1/2) \cdot (1/6) = 1/12$.
  • Since $P(A \cap B) = P(A) \cdot P(B)$, the events are independent.

Corollary: If $A$ and $B$ are independent, so are:

• $A$ and $B^{C}$
• $B$ and $A^{C}$
• $A^{C}$ and $B^{C}$