Tasks 07-02 - Basic Probability

Section 07: Probability & Statistics

Problem 1: Sample Spaces (x)

Define the sample space for each experiment:

  1. Rolling a six-sided die
  2. Flipping two coins
  3. Drawing a card from a standard deck and noting its suit
  4. A customer rating satisfaction on a scale of 1-5

  1. \(S = \{1, 2, 3, 4, 5, 6\}\)

  2. \(S = \{HH, HT, TH, TT\}\)

  3. \(S = \{\text{Hearts, Diamonds, Clubs, Spades}\}\)

  4. \(S = \{1, 2, 3, 4, 5\}\)

Problem 2: Basic Probability Calculations (x)

A fair six-sided die is rolled. Find:

  1. \(P(\text{rolling a 4})\)
  2. \(P(\text{rolling an even number})\)
  3. \(P(\text{rolling greater than 4})\)
  4. \(P(\text{rolling a 7})\)
  5. \(P(\text{not rolling a 6})\)

  1. \(P(4) = \frac{1}{6}\)

  2. \(P(\text{even}) = P(\{2,4,6\}) = \frac{3}{6} = \frac{1}{2}\)

  3. \(P(>4) = P(\{5,6\}) = \frac{2}{6} = \frac{1}{3}\)

  4. \(P(7) = 0\) (impossible event)

  5. \(P(\text{not 6}) = 1 - P(6) = 1 - \frac{1}{6} = \frac{5}{6}\)

Problem 3: Addition Rule (x)

In a class of 100 students:

  • 45 study German
  • 35 study French
  • 15 study both German and French
  1. Find \(P(\text{German})\)
  2. Find \(P(\text{German or French})\)
  3. Find \(P(\text{neither German nor French})\)
  4. Find \(P(\text{German only})\)

  1. \(P(G) = \frac{45}{100} = 0.45\)

  2. \(P(G \cup F) = P(G) + P(F) - P(G \cap F) = 0.45 + 0.35 - 0.15 = 0.65\)

  3. \(P(\text{neither}) = 1 - P(G \cup F) = 1 - 0.65 = 0.35\)

  4. \(P(\text{G only}) = P(G) - P(G \cap F) = 0.45 - 0.15 = 0.30\)

Problem 4: Independence (xx)

Two machines operate independently. Machine A works 95% of the time, Machine B works 90% of the time.

  1. Find \(P(\text{both work})\)
  2. Find \(P(\text{neither works})\)
  3. Find \(P(\text{at least one works})\)
  4. Find \(P(\text{exactly one works})\)

  1. \(P(A \cap B) = P(A) \cdot P(B) = 0.95 \times 0.90 = 0.855\)

  2. \(P(A' \cap B') = P(A') \cdot P(B') = 0.05 \times 0.10 = 0.005\)

  3. \(P(\text{at least one}) = 1 - P(\text{neither}) = 1 - 0.005 = 0.995\)

  4. \(P(\text{exactly one}) = P(A)P(B') + P(A')P(B) = 0.95 \times 0.10 + 0.05 \times 0.90 = 0.095 + 0.045 = 0.140\)

Problem 5: Cards (xx)

A card is drawn from a standard 52-card deck. Find:

  1. \(P(\text{Ace})\)
  2. \(P(\text{Heart})\)
  3. \(P(\text{Ace or Heart})\)
  4. \(P(\text{Face card})\) (Jack, Queen, King)
  5. Are “Ace” and “Heart” mutually exclusive? Independent?

  1. \(P(\text{Ace}) = \frac{4}{52} = \frac{1}{13}\)

  2. \(P(\text{Heart}) = \frac{13}{52} = \frac{1}{4}\)

  3. \(P(\text{Ace} \cup \text{Heart}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}\) (There’s 1 Ace of Hearts, counted twice if we just add)

  4. \(P(\text{Face}) = \frac{12}{52} = \frac{3}{13}\) (4 Jacks + 4 Queens + 4 Kings)

  5. Not mutually exclusive (Ace of Hearts exists). Check independence: \(P(\text{Ace}) \times P(\text{Heart}) = \frac{1}{13} \times \frac{1}{4} = \frac{1}{52}\) \(P(\text{Ace} \cap \text{Heart}) = \frac{1}{52}\) ✓ Yes, they ARE independent!

Problem 6: Business Application (xx)

A company surveyed 500 customers:

  • 320 are satisfied with the product
  • 280 are repeat customers
  • 200 are satisfied AND repeat customers
  1. Find \(P(\text{Satisfied})\)
  2. Find \(P(\text{Satisfied or Repeat})\)
  3. Find \(P(\text{Satisfied but not Repeat})\)
  4. Are satisfaction and repeat status independent?

  1. \(P(S) = \frac{320}{500} = 0.64\)

  2. \(P(S \cup R) = \frac{320}{500} + \frac{280}{500} - \frac{200}{500} = \frac{400}{500} = 0.80\)

  3. \(P(S \cap R') = P(S) - P(S \cap R) = \frac{320 - 200}{500} = \frac{120}{500} = 0.24\)

  4. Check: \(P(S) \times P(R) = 0.64 \times 0.56 = 0.358\) \(P(S \cap R) = \frac{200}{500} = 0.40\) \(0.358 \neq 0.40\), so NOT independent.

Problem 7: Quality Control (xxx)

A factory produces items with a 3% defect rate. A sample of 5 items is randomly selected.

  1. Are these selections independent? Why or why not?
  2. Find \(P(\text{all 5 are good})\)
  3. Find \(P(\text{at least one is defective})\)
  4. Find \(P(\text{exactly one is defective})\)

  1. If the population is large enough (much larger than 5), we can treat selections as approximately independent.

  2. \(P(\text{all good}) = (0.97)^5 \approx 0.859\)

  3. \(P(\text{at least 1 defective}) = 1 - P(\text{all good}) = 1 - 0.859 = 0.141\)

  4. \(P(\text{exactly 1 defective}) = \binom{5}{1}(0.03)^1(0.97)^4 = 5 \times 0.03 \times 0.885 = 0.133\)