Conditional Probability and the Multiplication Rule

Conditional Probability

• Definition: A conditional probability is a probability computed with the knowledge of additional information.
• Notation: The conditional probability of event B given event A is denoted as $P(B|A)$, which is calculated as: $P(B|A) = \frac{P(A \text{ and } B)}{P(A)}$.

Example: Education Levels of U.S. Adults

• Context: A table presents the number of U.S. men and women (25 years and older) with various levels of education.
• Data:
  • Total men: 94 million
  • Total women: 101 million
  • Total people: 195 million
• Problem 1: What is the probability that a randomly chosen person is a man?
  • Solution: $P(\text{Man}) = \frac{94}{195} = 0.4821$
• Problem 2: What is the probability that the person is a man with a bachelor's degree?
  • Given: 17.5 million men have bachelor's degrees.
  • Solution: $P(\text{Man and Bachelor's}) = \frac{17.5}{195} = 0.08974$
• Problem 3: What is the probability that the person has a bachelor's degree given that he is a man?
  • Solution: $P(\text{Bachelor's} | \text{Man}) = \frac{P(\text{Man and Bachelor's})}{P(\text{Man})} = \frac{17.5/195}{94/195} = 0.1862$

General Multiplication Rule

• Derivation: The general method for computing conditional probabilities leads to a way to compute probabilities for events of the form A and B.
• Rule: For any two events A and B, the probability of A and B is: $P(A \text{ and } B) = P(A) \cdot P(B|A)$.
• Alternative Form: $P(A \text{ and } B) = P(B) \cdot P(A|B)$.

Example: Job Application

• Given:
  • Probability of being granted an interview: $P(\text{Interview}) = 0.1$
  • Probability of being offered a job given an interview: $P(\text{Job} | \text{Interview}) = 0.25$
• Problem: Find the probability that an applicant is offered a job.
• Solution: $P(\text{Job}) = P(\text{Interview}) \cdot P(\text{Job} | \text{Interview}) = 0.1 \cdot 0.25 = 0.025$

Independent vs. Dependent Events

• Independent Events: Two events are independent if the occurrence of one does not affect the probability that the other event occurs.
• Dependent Events: If two events are not independent, they are dependent.
• Example 1 (Dependent):
  • Events: Being a freshman and being less than 20 years old.
  • Explanation: These are not independent because the probability of being less than 20 is greater for freshmen.
• Example 2 (Independent):
  • Events: Being born on a Sunday and taking a statistics class.
  • Explanation: These are independent because being born on a Sunday has no effect on the probability of taking a statistics class.

Multiplication Rule for Independent Events

• Condition: When two events A and B are independent, $P(B|A) = P(B)$.
• Rule: For any two independent events A and B: $P(A \text{ and } B) = P(A) \cdot P(B)$.

Example: People Under 18 in Different Cities

• Given:
  • Percentage of people under 18:
    • New York City: 23.5% (0.235)
    • Chicago: 25.8% (0.258)
    • Los Angeles: 26% (0.26)
• Problem: If one person is selected from each city, what is the probability that all of them are under 18? Is this an unusual event?
• Solution:
  • $P(\text{All under 18}) = P(\text{NYC}) \cdot P(\text{Chicago}) \cdot P(\text{LA}) = 0.235 \cdot 0.258 \cdot 0.26 = 0.0158$
  • Unusual Event: Yes, because 0.0158 < 0.05 (common cutoff point).

Example: Coin Tosses

• Setup: A fair coin is tossed twice.
• Part A: What is the probability that the second toss is a head?
  • Sample Space: HH, HT, TH, TT
  • Solution: $P(\text{Second toss is head}) = \frac{2}{4} = \frac{1}{2}$
• Part B: What is the probability that the second toss is a head given that the first toss is a head?
  • Solution: $P(\text{Second is H} | \text{First is H}) = \frac{P(\text{First is H and Second is H})}{P(\text{First is H})} = \frac{1/4}{2/4} = \frac{1}{2}$
• Part C: Are the answers to A and B different? Does the probability that the second toss is a head change if the first toss is a head?
  • Answer: The answers are the same, and the probability does not change. The events are independent.

At Least Once Probabilities

• Concept: Finding the probability that an event occurs at least once in several independent trials.
• Method: Use the rule of complements: $P(\text{At least one event occurs}) = 1 - P(\text{No events occur})$.

Example: Coin Tossed Five Times

• Problem: A fair coin is tossed five times. What is the probability that it comes up heads at least once?
• Solution:
  • $P(\text{At least one head}) = 1 - P(\text{No heads})$
  • $P(\text{No heads}) = P(\text{All tails}) = (0.5)^5 = 0.03125$
  • $P(\text{At least one head}) = 1 - 0.03125 = 0.96875$

Example: Inspectors Detecting Flaws

• Given:
  • Probability that an inspector detects a flaw: 0.8
  • Probability that an inspector fails to detect a flaw: 0.2
  • Three independent inspectors.
• Problem: If an item has a flaw, what is the probability that at least one inspector detects it?
• Solution:
  • $P(\text{At least one detects}) = 1 - P(\text{None detect})$
  • $P(\text{None detect}) = (0.2)^3 = 0.008$
  • $P(\text{At least one detects}) = 1 - 0.008 = 0.992$

Example: Semiconductor Wafers

• Context: A population of 600 semiconductor wafers categorized by lot (A, B, C) and conformance to thickness specification.
• Data:
  • Lot A: 88 conforming, 12 non-conforming (Total 100)
  • Lot B: 165 conforming, 35 non-conforming (Total 200)
  • Lot C: 260 conforming, 40 non-conforming (Total 300)
  • Total conforming: 513
  • Total non-conforming: 87
• Problem A: What is the probability that a wafer is from lot A?
  • Solution: $P(\text{Lot A}) = \frac{100}{600} = 0.167$
• Problem B: What is the probability that a wafer is conforming?
  • Solution: $P(\text{Conforming}) = \frac{513}{600} = 0.855$
• Problem C: What is the probability that a wafer is from lot A and is conforming?
  • Solution: $P(\text{Lot A and Conforming}) = \frac{88}{600} = 0.147$
• Problem D: Given that the wafer is from lot A, what is the probability that it is conforming?
  • Solution: $P(\text{Conforming} | \text{Lot A}) = \frac{P(\text{Lot A and Conforming})}{P(\text{Lot A})} = \frac{0.147}{0.167} = 0.88$
• Problem E: Given that the wafer is conforming, what is the probability that it is from lot A?
  • Solution: $P(\text{Lot A} | \text{Conforming}) = \frac{P(\text{Lot A and Conforming})}{P(\text{Conforming})} = \frac{0.147}{0.855} = 0.172$

Example: Car Repairs

• Given:
  • Probability that a car needs repairs in the first six months: 0.3
  • A dealer sells five such cars.
• Problem: What is the probability that at least one of them will require repairs in the first six months?
• Solution:
  • Probability that a car does not require repairs: 0.7
  • $P(\text{At least one car requires repairs}) = 1 - P(\text{None of the cars require repairs})$
  • $P(\text{None require repairs}) = (0.7)^5 = 0.16807$
  • $P(\text{At least one requires repairs}) = 1 - 0.16807 = 0.83193$