9/16: Probability Essentials: Negation, Addition, Independence, and Reversing Conditioning

Complement / Negation

• The complement rule: $P(A^c) = 1 - P(A)$

• Example: asteroid probability updates illustrate complements: if p(asteroid) = 0.013, then p(not asteroid) = 1 - 0.013 = 0.987 (98.7%). Updated p(asteroid) = 0.00004 (0.004%), so p(not asteroid) ≈ 0.99996 (99.996%).

Addition Rule (Union) and Mutual Exclusivity

• Addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

• If A and B are mutually exclusive (disjoint): $P(A \cap B) = 0 \Rightarrow P(A \cup B) = P(A) + P(B)$

• Mutual exclusivity concept: outcomes cannot occur together; e.g., drawing a diamond or a club are disjoint, so the joint probability is 0.

• Not mutually exclusive example (Ace and Spade): Ace and Spade overlap (Ace of Spades) so you must subtract the overlap to avoid double counting.

• Hogwarts example (year 4 or Gryffindor):

  • $P(year\ 4) = \frac{40}{280}$, $P(Gryffindor) = \frac{70}{280}$, $P(both) = \frac{10}{280}$

  • $P(year\ 4 \cup\ Gryffindor) = \frac{40}{280} + \frac{70}{280} - \frac{10}{280} \approx 0.357 \,(35.7\%)$

Independence and Dependence

• Independent events: knowing A occurs does not change the probability of B

  • $P(B|A) = P(B)$ and equivalently $P(A|B) = P(A)$

  • Joint probability if independent: $P(A \cap B) = P(A)P(B)$

• Dependence: one event conveys information about the other; e.g., drawing without replacement changes subsequent probabilities.

• Practical notes:

  • Use tables or data to test independence: compare $P(B|A)$ to $P(B)$ or $P(A|B)$ to $P(A)$, or compare $P(A \cap B)$ to $P(A)P(B)$.

  • Independence is a modeling choice; it may be approximately true or false depending on context.

Reversing Conditioning (Bayes-style / Conditional Probability)

• Reversing conditioning (Bayes-type thinking): given $P(A|B)$ or its components, find the reverse conditioning using the joint and marginals.

• Snow example (reversing):

  • Given: $P(OnTime|Snow) = 0.65$, $P(Snow) = 0.25$, $P(OnTime) = 0.80$

  • Joint: $P(Snow \cap OnTime) = P(OnTime|Snow)P(Snow) = 0.65 \times 0.25 = 0.1625$

  • Then $P(Snow|OnTime) = \dfrac{P(Snow \cap OnTime)}{P(OnTime)} = \dfrac{0.1625}{0.80} = 0.203125 \approx 20.3\%.$

• Key takeaway: often use the formula $P(A|B) = \dfrac{P(A\cap B)}{P(B)}$ and the law of total probability when needed.

Using Tables / Contingency Tables

• When information is partial, build a 2x2 table to organize:

  • Marginals: $P(A)$, $P(B)$, and the center cell $P(A \cap B)$

  • If given $P(A \cup B)$, use: $P(A \cap B) = P(A) + P(B) - P(A \cup B)$

• Example with dessert and menu items:

  • Given: $P(Dessert) = 0.60$ (Dessert = D ∪ E), $P(Dip) = 0.45$, $P(Entree) = 0.33$

  • Then $P(Dip \cap Entree) = P(Dip) + P(Entree) - P(Dessert) = 0.45 + 0.33 - 0.60 = 0.18$

Practice Problem Strategy (Quick Wins)

• Start from what is given; write in notation (A, B) as needed; choose the simplest rule first.

• If joint probability is unknown but marginals and an OR event are known, use the addition rule with algebra to solve for the missing joint.

• If data is sparse, contingency tables plus the negation rule can reveal the needed cell without heavy computation.

Practical Notes on Independence in Practice

• Independence is a useful simplifying assumption but must be stated and checked when possible.

• In real data, dependence often exists; quantify via table checks or model comparison.

• When decisions hinge on independence, acknowledge the approximation and potential impact of violations.

Quick Reference Formulas

• Complement: $P(A^c) = 1 - P(A)$

• Union: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

• Disjoint: if disjoint, $P(A \cup B) = P(A) + P(B)$

• Independent: $P(A \cap B) = P(A)P(B)$ and $P(B|A) = P(B)$

• Bayes (conditional): $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$

• Reversing conditioning (example): $P(A|B) = \dfrac{P(B|A)P(A)}{P(B)}$ (if needed via law of total probability)