Question 7 to 8

(ns assignments.hw2.q7-8
  (:require
    [assignments.hw2.utils :refer :all]
    [fastmath.core :as m]))

Question 7

7) Suppose that three fair dice are tossed. Let \(A_i\) be the event that a 6 shows on the \(i^{th}\) die, i = 1, 2, 3. Does \(P(A_1 ∪ A_2 ∪ A_3) = 1/2\)? Explain.

To solve this problem, we’ll use the following approach:

Step 1: Probability of getting a 6 on a single die

\[P(A_i) = 1/6\]

Step 2: Probability of not getting a 6 on a single die

\[P(A_i^c) = 1 - P(A_i) = 5/6\]

Step 3: Probability of getting at least one 6 in three tosses, using the complement rule

\[P(A_1 ∪ A_2 ∪ A_3) = 1 - P(A_1^c ∩ A_2^c ∩ A_3^c)\]

\[P(A_1 ∪ A_2 ∪ A_3) = 1 - (5/6)^3\]

(defn probability-at-least-one-six [num-dice]
  (- 1 (m/pow (/ 5 6) num-dice)))

(let [result (probability-at-least-one-six 3)]
  (answer (str "$P(A_1 ∪ A_2 ∪ A_3) = "
               (m/approx result 4)
               "$")))

\(P(A_1 ∪ A_2 ∪ A_3) = 0.4213\)

Step 4: Compare the result with 1/2

The difference between the calculated probability and 1/2 is -0.0787

(answer "No, P(A_1 ∪ A_2 ∪ A_3) ≠ 1/2. The probability is slightly less than 1/2.")

No, P(A_1 ∪ A_2 ∪ A_3) ≠ 1/2. The probability is slightly less than 1/2.

Explanation: The probability of getting at least one 6 when tossing three fair dice is approximately 0.4213, which is not equal to 1/2 (0.5). The difference is about 0.0787 or 7.87%.

This result can be interpreted as follows:

The probability of getting at least one 6 in three tosses is higher than one might intuitively expect.
It’s more likely to get at least one 6 than to get no 6s at all in three tosses.
The probability is closer to 1/2 than it is to either 0 or 1, which might explain why someone might guess it to be exactly 1/2.

The probability is slightly less than 1/2 because:

The chance of getting at least one 6 increases with each additional die roll.
However, the increase is not linear, and it gets smaller with each additional die.
With three dice, we’re just short of the 50% mark, but adding more dice would eventually bring us closer to (but never exactly) 100%.

Question 8

8) A box contains 3 red balls and 2 white balls. Two balls are drawn at random without replacement. Let A be the event that the first ball drawn is red, and B the event that the second ball drawn is red. Are A and B independent? Explain.

To determine if events A and B are independent, we need to check if P(A ∩ B) = P(A) * P(B). Let’s calculate these probabilities:

Step 1: Calculate P(A)

\[P(A) = 3/5\]

Step 2: Calculate P(B)

\[P(B) = (2/4) * (3/5) + (3/4) * (2/5) = 3/5\]

Step 3: Calculate P(A ∩ B)

\[P(A ∩ B) = (3/5) * (2/4) = 3/10\]

Step 4: Check if P(A ∩ B) = P(A) * P(B)

\[P(A) * P(B) = (3/5) * (3/5) = 9/25\]

(defn are-events-independent? [p-a p-b p-a-and-b]
  (m/approx-eq p-a-and-b (* p-a p-b)))

(let [p-a (/ 3 5)
      p-b (/ 3 5)
      p-a-and-b (/ 3 10)
      independent? (are-events-independent? p-a p-b p-a-and-b)]
  (answer (str "A and B are "
               (if independent? "" "not ")
               "independent.")))

A and B are not independent.

Explanation:

\(P(A)\): The probability that the first ball is red is \(3/5\).
\(P(B)\): The probability that the second ball is red is also \(3/5\).
\(P(A ∩ B)\): The probability that both balls are red is \(3/10\).
Since \(P(A ∩ B) = 3/10\) and \(P(A) * P(B) = 9/25\), and these are not equal, events \(A\) and \(B\) are not independent.

Conclusion:

The outcome of the first draw affects the probability of the second draw.
Therefore, events \(A\) and \(B\) are dependent.

source: src/assignments/hw2/q7_8.clj