From 82889f6a0652927bd0e182bb8e468a46b6321831 Mon Sep 17 00:00:00 2001 From: andyeisenberg <125768484+andyeisenberg@users.noreply.github.com> Date: Tue, 30 Dec 2025 16:59:11 +0000 Subject: [PATCH] Figured out the list error --- source/ch-Probability.ptx | 284 +++++++++++++++++++++++--------------- 1 file changed, 172 insertions(+), 112 deletions(-) diff --git a/source/ch-Probability.ptx b/source/ch-Probability.ptx index 4eea216..e2b1646 100644 --- a/source/ch-Probability.ptx +++ b/source/ch-Probability.ptx @@ -62,49 +62,48 @@

Consider sets A and B, each contained inside \Omega. - We can combine sets in a variety of ways: + We can combine sets in a variety of ways:

+
  • + Union + +

    + The union of A and B is the set A \cup B = \{x \mid x \in A \text{ or } x \in B\}. +

    +
    • + +
  • + Intersection + +

    + The intersection of A and B is the set A \cap B = \{x \mid x \in A \text{ and } x \in B\}. +

    +
    • + +
  • + Difference + +

    + The set difference A-B is the set A - B = \{x \mid x \in A \text{ and } x \notin B\}. +

    +
    • + +
  • + Complement + +

    + The complement of A is the set A^c = \{x \in \Omega \mid x \notin A\}. +

    +
    • + +
  • + Empty Set + +

    + The empty set, usually written \emptyset or \{\}, is the set which contains no elements. +

    +
    • +

    -
    -
  • - Union - -

    - The union of A and B is the set A \cup B = \{x \mid x \in A \text{ or } x \in B\}. -

    -
    • - -
  • - Intersection - -

    - The intersection of A and B is the set A \cap B = \{x \mid x \in A \text{ and } x \in B\}. -

    -
    • - -
  • - Difference - -

    - The set difference A-B is the set A - B = \{x \mid x \in A \text{ and } x \notin B\}. -

    -
    • - -
  • - Complement - -

    - The complement of A is the set A^c = \{x \in \Omega \mid x \notin A\}. -

    -
    • - -
  • - Empty Set - -

    - The empty set, usually written \emptyset or \{\}, is the set which contains no elements. -

    -
    • -
    @@ -220,30 +219,93 @@

    A probability distribution on a sample space \Omega assigns probabilities to every event, satisfying the following conditions: +

      +
    1. +

      + \Pr(\Omega) = 1. +

      +
      2. + +
    2. +

      + 0 \leq \Pr(A) \leq 1 for any event A. +

      +
      3. + +
    3. +

      + If A \cap B = \emptyset, then \Pr(A\cup B) = \Pr(A) + \Pr(B). +

      +
      4. +

    - -
      -
    1. -

      - \Pr(\Omega) = 1. -

      -
      2. - -
    2. -

      - 0 \leq \Pr(A) \leq 1 for any event A. -

      -
      3. - -
    3. -

      - If A \cap B = \emptyset, then \Pr(A\cup B) = \Pr(A) + \Pr(B). -

      -
      4. -
    +

    + For small probability spaces (i.e., with finitely many outcomes in the sample space), we'll usually assign probabilities to each individual outcome, and perhaps list them in a table. + Then, to find the probability of any event, simply add together the probabilities of each outcome in that event. +

    + + + +

    + An experiment consists of rolling a standard 6-sided die. + The sample space is \Omega = \{1, 2, 3, 4, 5, 6\}. + The probability distribution (assuming a fair die) is shown below. +

    + + + Distribution for a fair die + + + + x + \Pr(x) + + + + 1 + 1/6 + + + + 2 + 1/6 + + + + 3 + 1/6 + + + + 4 + 1/6 + + + + 5 + 1/6 + + + + 6 + 1/6 + + +
    + +

    + One possible event is A = \{2, 4, 6\}, i.e., the event that the result of the roll is even. + The probability of A is: + + \Pr(A) = \Pr(2) + \Pr(4) + \Pr(6) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} + +

    +     +     + @@ -255,31 +317,55 @@ - - - x - 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 + + + x + \Pr(x) - \Pr(x) - 0.1 - 0.05 - 0.2 - 0.15 - 0.15 - 0.1 - 0.05 - 0.1 - 0.1 + 1 + 0.1 + + + + 2 + 0.05 + + + + 3 + 0.2 + + + + 4 + 0.15 + + + + 5 + 0.15 + + + + 6 + 0.1 + + + + 7 + 0.05 + + + + 8 + 0.1 + + + + 9 + 0.1
    @@ -511,35 +597,9 @@

    - Let A = \{1, 2, 3\} and B = \{3, 4, 5\}. - Fill in the following probability distribution table so that A, B are independent. + Let A = \{1, 2, 3\} and B = \{3, 4, 5\} be events in the sample space \Omega = \{1, 2, 3, 4, 5, 6\}. + Create a probability distribution for \Omega so that A, B are independent.

    - - - - - - - x - 1 - 2 - 3 - 4 - 5 - 6 - - - - \Pr(x) - - - - - - - - -