Question 5 to 6

(ns assignments.hw2.q5-6
  (:require
    [assignments.hw2.utils :refer :all]
    [fastmath.core :as m]))

---

Question 5

---

5) Let A and B be two events such that \(P((A ∪ B)^c) = 0.6\) and \(P(A ∩ B) = 0.1\). Let E be the event that either A or B but not both occurs. Find \(P(E|A ∪ B)\).

This problem invites us to explore the intricacies of set theory and conditional probability. Let’s break it down step-by-step, much like we would approach a complex sociocultural phenomenon.

First, let’s define our events and their known probabilities:

\[P((A ∪ B)^c) = 0.6\]

\[P(A ∩ B) = 0.1\]

Now, let’s calculate \(P(A ∪ B)\) using the complement rule:

\[P(A ∪ B) = 1 - P((A ∪ B)^c) = 1 - 0.6 = 0.4\]

Event (E) is defined as (A ∪ B) setminus (A ∩ B), which is the symmetric difference or XOR of A and B.

We can calculate \(P(E)\) as follows:

\[P(E) = P(A ∪ B) - P(A ∩ B) = 0.4 - 0.1 = 0.3\]

Now, to find (P(E|A ∪ B)), we use the definition of conditional probability:

\[P(E|A ∪ B) = \frac{P(E ∩ (A ∪ B))}{P(A ∪ B)}\]

Since \(E | A ∪ B\), we have \(P(E ∩ (A ∪ B)) = P(E)\).

\[P(E|A ∪ B) = \frac{P(E)}{P(A ∪ B)} = \frac{0.3}{0.4} = 0.75\]

Note: E is a subset of A ∪ B because E represents the elements that are in A or B but not both, which is necessarily a subset of all elements in A or B.

(defn calculate-conditional-probability
  [p-a-union-b-complement p-a-intersect-b]
  (let [p-a-union-b (- 1 p-a-union-b-complement)
        p-e (- p-a-union-b p-a-intersect-b)
        p-e-given-a-union-b (/ p-e p-a-union-b)]
    p-e-given-a-union-b))

(let [result (calculate-conditional-probability 0.6 0.1)]
  (answer (str "$P(E|A ∪ B) = " (m/approx result 4) "$")))

\(P(E|A ∪ B) = 0.75\)

Question 6

---

6) Let A and B be two events on a sample space S. Suppose that \(P(A) = 0.4\), \(P(B) = 0.5\), and \(P(A ∩ B) = 0.1\). What is the probability that A or B but not both occur?

This problem is asking for the probability of the symmetric difference of events A and B, also known as the exclusive OR (XOR) of A and B. Let’s approach this step-by-step.

Given:

\[P(A) = 0.4\]

\[P(B) = 0.5\]

\[P(A ∩ B) = 0.1\]

The probability that A or B but not both occur can be calculated as:

\[P(A △ B) = P(A ∪ B) - P(A ∩ B)\]

First, let’s calculate P(A ∪ B) using the addition rule of probability:

\[P(A ∪ B) = P(A) + P(B) - P(A ∩ B)\]

\[P(A ∪ B) = 0.4 + 0.5 - 0.1 = 0.8\]

Now we can calculate the probability of A or B but not both:

\[P(A △ B) = P(A ∪ B) - P(A ∩ B) = 0.8 - 0.1 = 0.7\]

(defn calculate-symmetric-difference
  [p-a p-b p-a-intersect-b]
  (let [p-a-union-b (+ p-a p-b (- p-a-intersect-b))
        p-symmetric-difference (- p-a-union-b p-a-intersect-b)]
    p-symmetric-difference))

(let [result (calculate-symmetric-difference 0.4 0.5 0.1)]
  (answer (str "The probability that A or B but not both occur is " (m/approx result 4))))

The probability that A or B but not both occur is 0.7

source: src/assignments/hw2/q5_6.clj