MATH 2421: Lecture 5

Date: 2024-09-16 12:03:51

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Notes

Conditional Probability (cont.)

Theorems
- multiplication rule
  - if $P (F) > 0$ , then
  - $P (EF) = P (E ∣ F) P (F)$
- general multiplication rule
  - for events $A_{1}, A_{2}, \dots, A_{n}$
  - $P (A_{1} A_{2} \dots A_{n}) = P (A_{1}) P (A_{2} ∣ A_{1}) P (A_{3} ∣ A_{1} A_{2}) \dots P (A_{n} ∣ A_{1} A_{2} \dots A_{n - 1})$
  - 👨‍🎓 more like, nested use of multiplication rule
  - explanation $R H S = P (A_{1}) \frac{P ( A _{1} A _{2} )}{P ( A _{1} )} \frac{P ( A _{1} A _{2} A _{3} )}{P ( A _{1} A _{2} )} \dots \frac{P ( A _{1} \dots A _{n} )}{P ( A _{1} \dots A _{n - 1} )}$
    - and cancel, cancel
    - or, as mentioned above, nested use $P (A_{1} A_{2} \dots A_{n}) = P (A_{n} ∣ A_{1} A_{2} \dots A_{n - 1}) P (A_{1} A_{2} \dots A_{n - 1}) = P (A_{n} ∣ A_{1} A_{2} \dots A_{n - 1}) P (A_{n - 1} ∣ A_{1} A_{2} \dots A_{n - 2}) P (A_{1} A_{2} \dots A_{n - 2}) = \dots$
Examples
- bin: with 25 light bulbs
  - A: 5: good & functional for more than 30 days
  - B: 10: partially defective; fail in the second day of use
  - C: rest: totally defective
  - if: a random bulb initially lights up ( $A \cup B$ )
    - what is the probability of it working after 1 week? ( $A$ ) $P (A ∣ (A \cup B)) = \frac{P ( A ) \cap P ( A \cup B )}{P ( A \cup B )} = \frac{P ( A )}{P ( A \cup B )} = \frac{5/25}{15/25} = \frac{1}{3}$
- Celine: $1/2$ chance of getting A in French, and $2/3$ chance for chemistry
  - is her decision: based on fair coin
  - what is the chance that she gets an A in chemistry?
  - $A :=$ Celine getting A
  - $C :=$ Celine taking chemistry
  - $P (A C) = P (A ∣ C) P (C) = \frac{2}{3} \frac{1}{2} = \frac{1}{3}$
- suppose: an urn w/ 8 red balls & 4 white balls
  - we draw 2 balls, one at a time, without replacement
  - assuming fair chance, what is the probability that both balls drawn are red?
  - w/ conditional probability
    - $R_{1} :=$ first ball drawn: red
    - $R_{2} :=$ second ball drawn: red
    - $P (R_{1} R_{2}) = P (R_{2} ∣ R_{1}) P (R_{1}) = \frac{7}{11} \frac{8}{12} = \frac{14}{33}$
  - w/ combinatorial analysis
    - i.e. counting no. out interested outcomes / all possible outcomes
    - no. of choosing 2 red balls / no. of choosing arbitrary balls
    - $= \frac{( 2 8 )}{( 2 12 )} = \frac{8 * 7/2}{12 * 11/2} = \frac{14}{33}$
- three cards: selected successively at random, without replacement
  - 52 playing cards
  - calculate probability of receiving in order: K,Q, and J
  - conditional probability
    - $P (J Q K) = P (K) P (J Q ∣ K) = P (K) P (Q ∣ K) P (J ∣ Q K) = \frac{4}{52} \frac{4}{51} \frac{4}{50}$
  - combinatorial analysis
    - no. of outcome w/ K-Q-J / no. of total outcomes
    - $= \frac{4 * 4 * 4}{52 * 51 * 50}$
  - box of fuses w/ 20 fuses
    - 5 are defective
    - 3 fuses: selected randomly and removed from all
    - what's the probability that all 3 are defective?
    - conditional probability
      - $P (R_{1} R_{2} R_{3})$ $P (R_{1}) P (R_{2} ∣ R_{1}) P (R_{3} ∣ R_{1} R_{2}) = \frac{5}{20} \frac{4}{19} \frac{3}{18} = \frac{1}{19 * 6} = \frac{1}{114}$
    - combinatorial analysis
      - $(3 5) / (3 20) = \frac{5 * 4 * 3/3 !}{20 * 19 * 18/3 !} = \frac{1}{114}$
  - If six cards are selected at random (without replacement) from a standard deck of 52 cards, what is the probability there will be no pairs?
    - conditional probability
      - $= \frac{52}{52} \frac{48}{51} \frac{44}{50} \frac{40}{49} \frac{36}{48} \frac{32}{47}$
      - each turn:
    - combinatorial analysis
      - $52 \cdot 48 \cdot 44 \cdot 40 \cdot 36 \cdot 32/ \frac{52 !}{( 52 - 6 )!}$

Total Probability

Introduction
- how can we compute probability under different conditions / cases?
- e.g. probability of getting 3 on a fair die
  - case 1: 4-faced fie is chosen: $1/4$
  - case 2: 6-faced fie is chosen: $1/6$
  - assuming both cases happen with equal chance
  - $P (X = 3) = \frac{1}{4} /2 + \frac{1}{6} /2 = \frac{5}{24}$
Theorem
- let $A, B$ be any two event, then
  - $P (B) = P (B ∣ A) P (A) + P (B ∣ A^{c}) P (A^{c})$
  - 👨‍🏫 or: prob. of $B$ occurs = weighted average of the conditional probabilities of $B$
    - conditioning on either $A$ occurs or $A$ does not occur
  - proof: w/ Venn diagram
    - $B = B A \cup B A^{c}$ , $B A \cap B A^{c} = \emptyset$ $P (B) = P (B A) + P (B A^{c}) = P (B ∣ A) P (A) + P (B ∣ A^{c}) P (A^{c})$
- $A_{1}, A_{2}, \dots A_{n}$ partitions the sample space $S$ if:
  - they are mutually exclusive: $A_{i} \cap A_{j} = \emptyset, \forall i \neq = j$
  - they are exhaustive: $⋃_{i = 1}^{n} A_{i} = S$
- law of total probability
  - if $A_{1}, A_{2}, \dots A_{n}$ partitions the sample space and $P (A_{i}) > 0$
  - let B any event, then
  - $P (B) = P (B ∣ A_{1}) P (A_{1}) + \dots + P (B ∣ A_{n}) P (A_{n})$
    - proof:
      - because $⋃_{i = 1}^{n} A_{i} = S$
      - $B = B \cap S = B \cap (⋃_{i = 1}^{n} A_{i}) = ⋃_{i = 1}^{n} A \cap A_{i}$
      - because $A_{i} \cap A_{j} = \emptyset, \forall i \neq = j$
      - $⋃_{i = 1}^{n} B \cap A_{i} = \sum P (B A_{i}) = \sum P (B ∣ A_{i}) P (A_{i})$
Examples
- MCQ exam
  - a student: either knows the answer at probability $p$ , or guesses at probability $1 - p$
  - there are $m$ alternatives
  - what is the probability that he answered it correctly?
    - $K$ : student knows; $P (K) = p$
    - $G$ : student guesses: $P (G) = 1 - p$
    - $c$ : student answers correctly $P (C) = P (C ∣ K) P (K) + P (C ∣ K^{c}) P (K^{c}) = 1 \cdot p + \frac{1}{m} (1 - p) = p + \frac{1 - p}{m}$
  - what is the probability that the student knew the answer, given that he answered it correctly?
    - $p (k ∣ c) = \frac{P ( k c )}{P ( c )} = \frac{P ( k )}{P ( c )} = \frac{p}{p + \frac{1 - p}{m}}$
    - 👨‍🏫 aka: Bayes formula
- Insurance company: categorizes people into 2 groups: accident-prone and others
  - accident-prone person: will have accident in a fixed 1 yr time with probability of 0.4
    - 0.2 for others
  - if 30% of population is accident prone, what's the change that new policyholder will have an accident within an year?
    - $A$ : person being accident prone
    - $X$ : person having accident in 1 year $P (X) = P (X ∣ A) P (A) + P (X ∣ A^{c}) P (A^{c}) = 0.4 \cdot 0.3 + 0.2 \cdot 0.7 = 0.26$
- party support
  - 40% of people support party $A$ , 30% $B$ , 20% $C$ , 10% $D$
  - $Q$ : certain policy
    - 50% of $A$ supporter supports it
    - 40% of $B$ supporter supports it
    - 30% of $C$ supporter supports it
    - 100% of $D$ supporter supports it
  - what is the probability that a random citizen supports the policy?
    - $P (Q) = P (Q ∣ A) P (A) + P (Q ∣ B) P (B) + P (Q ∣ C) P (C) + P (Q ∣ D) P (D)$
    - $= 0.4 * 0.5 + 0.3 * 0.4 + 0.2 * 0.3 + 0.1 * 1 = 0.48$

Bayes' Theorem

Introduction
- if $A_{1}, A_{2}, \dots A_{n}$ partitions the sample space $S$
  - and $P (A_{i}) > 0$
  - then, for any $1 \leq i \leq n$ $P (A_{i} ∣ B) = \frac{P ( B ∣ A _{i} ) P ( A _{i} )}{P ( B ∣ A _{1} ) P ( A _{1} ) + \dots + P ( B ∣ A _{n} ) P ( A _{n} )}$
  - proof:
    - $P (A_{i} ∣ B) = \frac{P ( B A _{i} )}{P ( B )} = \frac{P ( B ∣ A _{i} ) P ( A _{i} )}{P ( B )}$
    - and expand the denominator

MATH 2421: Probability