MATH 2421: Lecture 4

Date: 2024-09-11 12:02:33

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❓Questions

Notes

Sample Spaces Having Equally Likely Outcomes

Assuming Equally-Likely
- for many experience: natural to assume all outcomes in finite sample space: equally likely to occur
  - e.g. fair coin / die
- notation
  - $1 = P (S) = P (⋃_{i = 1}^{N} {s_{i}}) = \sum_{i = 1}^{n} P ({s_{i}}) = N c$
  - $c = 1/ N$ ; or $P ({s_{i}}) = \frac{1}{∣ S ∣}$
- then, follows from Axiom 3 that, for $E \subset S$
  - $P (E) = \frac{no. of outcomes in E}{no. of outcomes in S}$
Examples
- trump cards
  - assuming all 52 cards in a deck is equally likely to occur
  - and $A$ : drawing diamond; $B$ : drawing Jack
  - what is $P (A), P (B), P (A \cup B)$
  - $P (A) : \frac{13}{52} = \frac{1}{4}$
  - $P (B) : \frac{4}{52} = \frac{1}{13}$
  - $P (A \cup B) : \frac{1}{4} + \frac{1}{13} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$
- drawing random balls
  - if 3 balls are randomly drawn from set of 6 white and 5 black balls
  - what's probability that one of the drawn balls is white
    - and the other two black?
  - 3 ordering possible
    - $W BB : 6 \times 5 \times 4 = 120$
    - $B W B : 5 \times 6 \times 4 = 120$
    - $BB W : 5 \times 4 \times 6 = 120$
  - total ways: $(3 11) = 990$
  - $\frac{120 * 3}{990} = \frac{36}{99} = \frac{4}{11}$
  - alternatively
    - no. of choosing 1 white and 2 black:
    - $= (1 6) (2 5)$
    - $P (A) = \frac{( 1 6 ) ( 2 5 )}{( 3 11 )}$
- birthday problem 2
  - how large must the group be, so that there is a probability io greater than 0.5, that someone will have the same birthday as your do?
  - exclude Feb 29 for calculation; assume unifom distribution
  - $P (A) = 1 - P (A^{c}) = 1 - P (no one among n people w/ same BD as you)$
  - $= 1 - (\frac{364}{365})^{n}$
  - $n$ s.t. $P (A) > 0.5 ⟹ 1 - (\frac{364}{365})^{n} > 0.5$
  - $n ln \frac{364}{365} < ln 0.5$ // switch side
  - $n ln \frac{364}{365} < ln \frac{1}{2} = - ln 2$
  - $ln \frac{364}{365} < ln 364 - ln 365$
  - $n < - \frac{l n 2}{l n 364 - l n 365}$
- birthday problem 1
  - what is the probability that in a group of $n$ people, at least $2$ of the will have the same birthday?
  - $P (B) = 1 - P (B^{c}) = 1 - P (all having diff. birthday)$
  - $= 1 - \frac{365 \times 364 \times \dots \times ( 365 - n + 1 ) n terms}{365 ^{n}}$
  - $= 1 - \frac{365 !}{( 365 - n )! 365 ^{n}}$
  - $= 1 - \frac{\prod _{i = 0}^{n - 1} ( 365 - i )}{365 ^{n}}$
  - or: $= \prod_{i = 0}^{n - 1} (\frac{365 - i}{365}) = \prod_{i = 0}^{n - 1} (1 - \frac{i}{365})$
  - and as $1 + x \approx e^{x}$ when $∣ x ∣$ is small
    - $\approx \prod_{i = 0}^{n - 1} (e^{- \frac{i}{365}})$
    - $= e^{- \frac{1}{365} \frac{( n - 1 ) n}{2}}$
    - $= e^{- \frac{( n - 1 ) n}{365 * 2}}$
    - as $ln 0.5 \approx - 0.6931$
    - $- 0.6931 = - \frac{( n - 1 ) n}{365 * 2}$
    - $\dots n \approx 25$
  - $P (B) \approx 1 - e^{- \frac{( n - 1 ) n}{365 * 2}} = P_{n}$

Conditional Probability and Independence: Introduction

Why conditional probability
- We can leverage the information we have!

Conditional Probability

Definition
- $P (B ∣ A) = \frac{∣ A \cap B ∣}{∣ A ∣}$
  - $= \frac{\frac{∣ A \cap B ∣}{∣ S ∣}}{\frac{∣ A ∣}{∣ S ∣}} = \frac{P ( A \cap B )}{P ( A )}$
  - ⭐conditional probability: $P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )}$
    - read: $P (B)$ given that $A$ has occurred
  - if $A \subseteq B$ , then $P (B ∣ A) = 1$
- if $P (F) > 0$ , then
  - $P (B ∣ A) = \frac{P ( A B )}{P ( A )}$ (simple notation change)
- ⭐ by Kolmogorov's axioms, is $P (\cdot ∣ F)$ a valid probability?
  - check
    - $0 \leq P (E ∣ F) \leq 1$
      - true as $P (EF) < P (F)$
    - $P (S ∣ F) = 1$
      - true as $P (SF) = P (F)$ (as $F \subseteq S$ )
    - $P (⋃_{i = 1}^{\infty} A_{i} ∣ F) = \sum_{i = 1}^{\infty} P (A_{i} ∣ F)$ $P (i = 1 ⋃ \infty A_{i} ∣ F) = \frac{P (( ⋃ _{i = 1}^{\infty} A _{i} ) F )}{P ( F )} = \frac{P ( ⋃ _{i = 1}^{\infty} A _{i} F )}{P ( F )} = \frac{\sum _{i = 1}^{\infty} P ( A _{i} F )}{P ( F )} = i = 1 \sum \infty \frac{P ( A _{i} F )}{P ( F )} = i = 1 \sum \infty P (A_{i} ∣ F)$
    - holds as $A_{i} F$ are mutually exclusive to one another
  - thus, $P (\cdot ∣ F)$ is a valid probability
    - and therefore properties of probability holds for $P (\cdot ∣ F)$ , such as:
    - $P (E^{c} ∣ F) = 1 - P (E ∣ F)$
    - $P (A \cup B ∣ F) = P (A ∣ F) + P (B ∣ F) - P (A B ∣ F)$
Examples
- if we know that event $A = {1, 3, 5}$ has occurred
  - then $P (B = {1, 2, 3}) = \frac{2}{3}$ , not $\frac{1}{2}$ !
- suppose 2 fair dice are are rolled
  - and we observe that the first die is a $3$
  - given that, what is the probability that the sum of 2 dice equals 8?
    - i.e. second die = $5$ , so $\frac{1}{6}$
  - or: $P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )} = \frac{\frac{1}{36}}{\frac{1}{6}}$
    - $= \frac{1}{6}$
- student: taking 1-hr exam
  - suppose $P (A)$ : student will finish exam in less than $x$ hours: $\frac{x}{2}$ ( $\forall0 \leq x \leq 1$ )
  - given that student: still working after $0.75$ hrs ( $L_{0.75}^{c}$ )
  - what's the conditional probability that the full hour will be used? ( $L_{1}^{c}$ )
  - $P (L_{1}^{c} ∣ L_{0.75}^{c}) = \frac{P ( L _{0.75}^{c} \cap L _{1}^{c} )}{P ( L _{0.75}^{c} )} = \frac{P ( L _{1}^{c} )}{P ( L _{0.75}^{c} )} = \frac{1 - 1.0/2}{1 - 0.75/2} = \frac{1/2}{5/8} = \frac{4}{5}$
    - 80%

MATH 2421: Probability