Question 9 to 10

(ns assignments.hw2.q9-10
  (:require
    [assignments.hw2.utils :refer :all]
    [scicloj.hanamicloth.v1.api :as haclo]
    [tablecloth.api :as tc]))

Question 9

9) Events \(A1\) and \(A2\) are such that \(A1 ∪ A2 = S\) (the sample space), and \(A1 ∩ A2 = ∅\) (the empty set). Find \(p1\) if \(P(A1) = p1\) , \(P(A2) = p2\), and \(3p_1 − p_2 = 0.5\).

Given:

\[A1 ∪ A2 = S\]

\[A1 ∩ A2 = ∅\]

\[P(A1) = p1\]

\[P(A2) = p_2\]

\[3p_1 − p_2 = 0.5\]

Steps:

Since A1 and A2 are mutually exclusive and cover the entire sample space:

\[P(A1) + P(A2) = 1\]

\[p_1 + p_2 = 1\]

We now have a system of two equations:

\[p_1 + p_2 = 1\]

\[3p_1 - p_2 = 0.5\]

We can solve this system using linear algebra:

\[A = \begin{bmatrix} 1 & 1 \\ 3 & -1 \end{bmatrix}\]

\[x = \begin{bmatrix} p1 \\ p2 \end{bmatrix}\]

\[b = \begin{bmatrix} 1 \\ 0.5 \end{bmatrix}\]

We’ll solve this using LU decomposition and forward/backward substitution. Similar to Question 1:

(comment
  (defn solve-probabilities [b1 b2]
    (let [A (dge 2 2 [1 1
                      3 -1]
                 {:layout :row})
          b (dv [b1 b2])
          LU (trf! A)
          x (mv (tri! LU) b)]
      x)))

(let [solution (solve-probabilities 1 0.5)
      p1 (first solution)
      p2 (second solution)]
  (str [(str "p1 = " p1)
        (str "p2 = " p2)]))

"[\"p1 = 0.375\" \"p2 = 0.625\"]"

Answer:

(let [solution (solve-probabilities 1 0.5)
      p1 (first solution)]
  (answer (str "p1 = " p1)))

p1 = 0.375

Verification:

Let’s verify our solution satisfies both equations:

(let [solution (solve-probabilities 1 0.5)
      p1 (first solution)
      p2 (second solution)]
  (answer (str "p1 + p2 = " (+ p1 p2)
               ",   \n"
               "3p1 - p2 = " (- (* 3 p1) p2))))

p1 + p2 = 1.0,
3p1 - p2 = 0.5

Interpretation: 1. \(p_1\) represents the probability of event \(A1\) occurring. 2. \(p_2\) represents the probability of event \(A2\) occurring. 3. The sum of \(p_1\) and \(p_2\) equals 1, confirming that \(A1\) and \(A2\) cover the entire sample space. 4. The equation \(3p_1 - p_2 = 0.5\) is satisfied by our solution.

Conclusion: We have successfully found \(p_1\), which satisfies all the given conditions. This solution demonstrates how mutually exclusive events that cover the entire sample space relate to each other probabilistically.

Question 10

10) Suppose that two fair dice are tossed. What is the probability that the sum equals ten given that it exceeds eight?

Given:

Two fair dice are tossed
We need to find \(P(Sum = 10 | Sum > 8)\)

Steps:

1. Sample space for sums exceeding 8:

(def sums-exceeding-8 [9 10 11 12])

2. Count the number of ways to get each sum:

(defn count-ways [sum]
  (count (for [d1 (range 1 7)
               d2 (range 1 7)
               :when (= (+ d1 d2) sum)]
           [d1 d2])))

(let [sums (range 2 13)
      frequencies (map count-ways sums)
      data (tc/dataset {:sum       sums
                        :frequency frequencies})]
  (haclo/layer-bar data
                   {:=x :sum :=x-title "Sum of Dice" :=mark-size 25
                    :=y :frequency :=y-title "Frequency"}))

(def ways-for-sums
  (map #(vector % (count-ways %)) sums-exceeding-8))

Number of ways for each sum:

(into [] (map (fn [[sum ways]]
                {:sum sum :ways ways}) ways-for-sums))

[{:sum 9, :ways 4}
 {:sum 10, :ways 3}
 {:sum 11, :ways 2}
 {:sum 12, :ways 1}]

3. Calculate the probability of sum exceeding 8:

(def total-ways-exceeding-8 (reduce + (map second ways-for-sums)))

(def p-sum-exceeding-8 (/ total-ways-exceeding-8 36))

P(Sum > 8) = 10/36 = 0.2777778

4. Calculate the probability of sum equals 10:

(def ways-sum-10 (count-ways 10))

(def p-sum-10 (/ ways-sum-10 36))

P(Sum = 10) = 3/36 = 0.083333336

5. Calculate the conditional probability:

(def p-sum-10-given-exceeding-8 (/ p-sum-10 p-sum-exceeding-8))

Answer:

(answer (str "P(Sum = 10 | Sum > 8) = " (float p-sum-10-given-exceeding-8)))

P(Sum = 10 | Sum > 8) = 0.3

Verification:

We can verify this result by calculating the probability directly from our counted ways:

(def direct-probability (/ (count-ways 10) total-ways-exceeding-8))

(answer (str "Direct calculation: " (float direct-probability)))

Direct calculation: 0.3

Interpretation:

The probability of getting a sum of 10, given that the sum exceeds 8, is about 0.4.
This means that if we know the sum is greater than 8, there’s a 40% chance it’s exactly 10.
This probability is higher than the unconditional probability of getting a sum of 10 (which is about 0.0833 or 8.33%) because we’ve eliminated all the outcomes where the sum is 8 or less.

Conclusion:

We’ve successfully calculated the conditional probability using the definition of conditional probability: P(A|B) = P(A ∩ B) / P(B). This problem demonstrates how additional information (in this case, knowing the sum exceeds 8) can significantly change the probability of an event

source: src/assignments/hw2/q9_10.clj