Tasks 07-04 - Conditional Probability

Section 07: Probability & Statistics

Problem 1: Basic Conditional Probability (x)

Given \(P(A) = 0.6\), \(P(B) = 0.5\), and \(P(A \cap B) = 0.3\):

  1. Find \(P(A|B)\)
  2. Find \(P(B|A)\)
  3. Are A and B independent?

  1. \(P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.3}{0.5} = 0.6\)

  2. \(P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3}{0.6} = 0.5\)

  3. Check: \(P(A) \times P(B) = 0.6 \times 0.5 = 0.3 = P(A \cap B)\) ✓ Yes, A and B ARE independent. Note: \(P(A|B) = P(A) = 0.6\) confirms this.

Problem 2: Multiplication Rule (x)

A bag contains 6 red and 4 blue balls. Two balls are drawn without replacement.

  1. Find \(P(\text{both red})\)
  2. Find \(P(\text{first red, second blue})\)
  3. Find \(P(\text{different colors})\)

  1. \(P(R_1 \cap R_2) = P(R_1) \times P(R_2|R_1) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}\)

  2. \(P(R_1 \cap B_2) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90} = \frac{4}{15}\)

  3. \(P(\text{different}) = P(R_1 \cap B_2) + P(B_1 \cap R_2) = \frac{24}{90} + \frac{4}{10} \times \frac{6}{9} = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}\)

Problem 3: Tree Diagrams (xx)

A company has two production lines:

  • Line A produces 60% of items, with 5% defect rate
  • Line B produces 40% of items, with 8% defect rate
  1. Draw a tree diagram
  2. Find \(P(\text{defective})\)
  3. Find \(P(\text{Line A and defective})\)
  4. Given an item is defective, find \(P(\text{from Line A})\)

  1. Tree:

    • A (0.6) → D (0.05), G (0.95)
    • B (0.4) → D (0.08), G (0.92)

  2. \(P(D) = P(D|A)P(A) + P(D|B)P(B) = 0.05(0.6) + 0.08(0.4) = 0.03 + 0.032 = 0.062\)

  3. \(P(A \cap D) = P(D|A)P(A) = 0.05 \times 0.6 = 0.03\)

  4. \(P(A|D) = \frac{P(A \cap D)}{P(D)} = \frac{0.03}{0.062} \approx 0.484\)

Problem 4: Law of Total Probability (xx)

A store has three suppliers:

  • Supplier X: 50% of stock, 2% return rate
  • Supplier Y: 30% of stock, 4% return rate
  • Supplier Z: 20% of stock, 5% return rate
  1. What is the overall return rate?
  2. A returned item is selected. What’s the probability it came from Supplier Y?

  1. \(P(R) = P(R|X)P(X) + P(R|Y)P(Y) + P(R|Z)P(Z)\) \(= 0.02(0.5) + 0.04(0.3) + 0.05(0.2)\) \(= 0.01 + 0.012 + 0.01 = 0.032 = 3.2\%\)

  2. \(P(Y|R) = \frac{P(R|Y)P(Y)}{P(R)} = \frac{0.04 \times 0.3}{0.032} = \frac{0.012}{0.032} = 0.375\)

Problem 5: Sequential Selection (xx)

From a group of 8 men and 5 women, 3 people are selected randomly (without replacement).

  1. Find \(P(\text{all women})\)
  2. Find \(P(\text{all men})\)
  3. Find \(P(\text{at least one woman})\)
  4. Find \(P(\text{exactly 2 men})\)

  1. \(P(\text{all women}) = \frac{5}{13} \times \frac{4}{12} \times \frac{3}{11} = \frac{60}{1716} = \frac{5}{143}\)

  2. \(P(\text{all men}) = \frac{8}{13} \times \frac{7}{12} \times \frac{6}{11} = \frac{336}{1716} = \frac{28}{143}\)

  3. \(P(\text{at least 1 woman}) = 1 - P(\text{all men}) = 1 - \frac{28}{143} = \frac{115}{143}\)

  4. Using combinations: \(P = \frac{\binom{8}{2}\binom{5}{1}}{\binom{13}{3}} = \frac{28 \times 5}{286} = \frac{140}{286} = \frac{70}{143}\)

Problem 6: Independence Testing (xxx)

A survey of 400 employees collected data on job satisfaction (Satisfied/Not Satisfied) and work arrangement (Remote/Office):

Remote Office Total
Satisfied 90 150 240
Not Satisfied 60 100 160
Total 150 250 400
  1. Find \(P(\text{Satisfied})\)
  2. Find \(P(\text{Satisfied}|\text{Remote})\)
  3. Find \(P(\text{Satisfied}|\text{Office})\)
  4. Are satisfaction and work arrangement independent?

  1. \(P(S) = \frac{240}{400} = 0.60\)

  2. \(P(S|R) = \frac{90}{150} = 0.60\)

  3. \(P(S|O) = \frac{150}{250} = 0.60\)

  4. Since \(P(S|R) = P(S|O) = P(S) = 0.60\), the variables ARE independent!

Alternative check: \(P(S) \times P(R) = 0.60 \times 0.375 = 0.225\) \(P(S \cap R) = \frac{90}{400} = 0.225\) ✓

Problem 7: Exam-Style Problem (xxx)

At a university, 70% of students pass the statistics exam. Of those who pass, 80% studied more than 10 hours. Of those who fail, 30% studied more than 10 hours.

  1. Draw a complete tree diagram with all probabilities
  2. Find \(P(\text{studied more than 10 hours})\)
  3. A student studied more than 10 hours. What’s the probability they passed?
  4. Are “passing” and “studying more than 10 hours” independent?

  1. Tree diagram:

    • Pass (0.70) → Study>10 (0.80), Study≤10 (0.20)
    • Fail (0.30) → Study>10 (0.30), Study≤10 (0.70)

    Joint probabilities:

    • Pass ∩ Study>10: 0.70 × 0.80 = 0.56
    • Pass ∩ Study≤10: 0.70 × 0.20 = 0.14
    • Fail ∩ Study>10: 0.30 × 0.30 = 0.09
    • Fail ∩ Study≤10: 0.30 × 0.70 = 0.21

  2. \(P(\text{Study}>10) = 0.56 + 0.09 = 0.65\)

  3. \(P(\text{Pass}|\text{Study}>10) = \frac{0.56}{0.65} = \frac{56}{65} \approx 0.862\)

  4. Check: \(P(\text{Pass}) \times P(\text{Study}>10) = 0.70 \times 0.65 = 0.455\) \(P(\text{Pass} \cap \text{Study}>10) = 0.56\) \(0.455 \neq 0.56\), so NOT independent.