Previously on Probability and Statistics Concepts…

• Population parameters vs Sample statistics

• Performing hypothesis testing using difference in proportions examples

• Computing confidence intervals (CIs) and how to interpret them

• Performing simple linear regression (SLR) and how to interpret the slope and intercept (sample statistics), and performing basic residual analysis for the SLR

• Learned about basic probability rules.

What’s ahead in this course?

• We will be focusing on the mathematical details of the concepts we learned.

• We will do more applications using “by-hand” computations and R computations.

• Putting probability and statistics together - the fusion of art and science!

• Statistical decisions are always subjective - frequentist vs bayesian approaches.

Roadmap for Today

• Why do we want to learn probability and Statistics?

• What is independence, again?

• What are Permutations and Combinations?

• Take-home message: Becoming an expert-beginner on basic probability.

Basic Probability Rules (1/2)

• Rule 1: For any event \(A\),

  \[0 \le P(A) \le 1.\]

• Rule 2: The sum of the probabilities of all possible outcomes in the sample space \(\Omega\) is \(1\).

  \[P(\Omega) = 1\]

• Rule 3: The Complement Rule.

  \[P(A^c) = 1 – P(A).\]

Basic Probability Rules (2/2)

• Rule 4: Involving events \(A\) and \(B\).

  If events A and B are disjoint (\(A \cap B = \emptyset\) meaning \(P(A \cap B) = 0\)), then \[P(A \cup B) = P(A) + P(B).\] If events A and B are joint (\(A \cap B \ne \emptyset\) meaning \(P(A \cap B) \ne 0\)), then in general \[P(A \cup B) = P(A) + P(B) - P(A \cap B).\]

• Rule 5: For independent events \(A\) and \(B\), \(P(A \cap B) = P(A)P(B).\)

Joint and Disjoint Events

Independence and Mutual Exclusivity (1/2)

Independence and Mutual Exclusivity (2/2)

Combinations and Permutations (1/2)

• A combination is a grouping of outcomes in which the order does not matter.

• A permutation is an arrangement of outcomes in which the order does matter.

• Question - Flip a coin 3 times, how many possible outcomes with exactly 2 heads? Is this a combination or permutation problem? Hint: \({n \choose x} = \frac{n!}{x!(n-x)!}\) or choose(n,k) in R.

Combinations and Permutations (2/2)

• Answer - The order of outcomes are not important, so this situation involves combinations. The number of possible outcomes with exactly 2 heads is \[{3 \choose 2} = \frac{3!}{2!(3-2)!} = 3.\]

• Why? You can use Pascal’s triangle as a bridge to the binomial theorem.

• Other Example: Taste Testing by Khan Academy

Pascal vs Pascal

Blaise Pascal

Image Source: Wiki

• French
• Born in 1623
• Mathematician and Physicist
• Brilliant and Talented

Pedro Pascal

Image Source: Wiki

• Chilean-American
• Born in 1975
• Actor and Artist
• Brilliant and Talented

Pascal’s Triangle

Blaise Pascal

Image Source: Wiki

• French
• Born in 1623
• Mathematician and Physicist
• Brilliant and Talented

Image Source: Pascal’s Triangle by Reaghan Willison & Caitlin Kinser

3.33-Minute Activity (1/3)

• Question - If we flip a coin 5 times, what is \(P(H_3)\), the probability that we get exactly 3 heads (in any order)?
  Hint: \({n \choose x} = \frac{n!}{x!(n-x)!}\) or choose(n,k) in R.

• Timer starts.

• Step 1- Count how many possible outcomes in the event space for 5 coin flips. That is \(2^5 = 32\).

3.33-Minute Activity (2/3)

• Step 2- Look at which outcomes involve 3 heads: (H,H,H,T,T), (H,H,T,H,T), (H,H,T,T,H), etc. There are 10 outcomes with 3 heads out of five flips. \[{5 \choose 3} = \frac{5!}{3!(5-3)!} = 10\]

• Step 3 - Each event has probability \(\frac{1}{32}\). Then the probability we get exactly 3 heads (in any order) is \[P(H_3) = \frac{10}{32} = \frac{5}{16}\]

Rolling a Six-Sided Die

Now, let’s consider another classic problem in probability: rolling a die with six sides, numbered from 1 to 6.

3.33-Minute Activity

• Question - What is the sample space and event space for rolling a six-sided die?

• Timer starts.

• Sample Space: \(\Omega = \{1,2,3,4,5,6\}\)

• Event Space: Any subset of outcomes in the sample space.
  An example:
  Roll an even number: \(\{2,4,6\}\)
  Roll an odd number: \(\{1,3,5\}\)
  With \(n\) outcomes, \(2^n\) possible events

Set Complements

• \(A\) and \(A^c\) are disjoint sets

• Example 1

  • \(A\): roll an odd number.
  • \(A^c\): roll an even number.

• Example 2

  • \(A\): roll a number greater than 3.
  • \(A^c\) : roll a number at most 3.

DeMorgan’s Laws

X-Minute Activity

Consider rolling a six-sided die and these defined events.

• \(A\) = Roll a 2 or 3, \(A^c\) = Roll a 1,4,5 or 6
• \(B\) = Roll a 2 or 4, \(B^c\) = Roll a 1,3,5 or 6
• Questions: (a) What is \((A \cup B)^c\)? (b) What is \((A \cap B)^c\)?

• Answers:
  \((A \cup B)^c = A^c \cap B^c\) = Roll a 1, 5 or 6
  \((A \cap B)^c = A^c \cup B^c\) = Roll a 1, 3, 4, 5, or 6

Summary

Today, we discussed the following:

• Revisited basic probability rules

• Combinations and DeMorgan’s laws

Next, we will discuss:

• More examples of probability and more on permutations and combinations

• Conditional probability