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)}.

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

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).

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 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.

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).

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).

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.

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}).

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

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

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

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