Slide 1

  • Probability - Problems on conditional probability

Slide 2

  • Recap: Basic probability concepts
    • Sample space
    • Event
    • Probability of an event
    • Rule of sum
    • Rule of product

Slide 3

  • Conditional probability
    • Definition: Probability of an event given that another event has already occurred
    • Notation: P(A|B)
    • Formula: P(A|B) = P(A ∩ B) / P(B)

Slide 4

  • Solving problems on conditional probability
    • Step 1: Identify the given information and events involved
    • Step 2: Use the formula P(A|B) = P(A ∩ B) / P(B) to calculate the conditional probability
    • Step 3: Simplify and solve the equation
    • Step 4: Interpret the result in the context of the problem

Slide 5

  • Example 1:
    • Problem: A box contains 4 red balls and 6 blue balls. Two balls are drawn at random without replacement. What is the probability of drawing a blue ball on the second draw given that the first ball drawn was red?
    • Solution:
      • Event A: Drawing a blue ball on the second draw
      • Event B: Drawing a red ball on the first draw
      • P(A|B) = ?
      • P(A|B) = P(A ∩ B) / P(B) = (4/9) / (4/10) = 2/3

Slide 6

  • Example 2:
    • Problem: A bag contains 3 red balls and 5 green balls. Two balls are drawn at random with replacement. What is the probability of getting a red ball on the second draw, given that the first ball drawn was red?
    • Solution:
      • Event A: Getting a red ball on the second draw
      • Event B: Getting a red ball on the first draw
      • P(A|B) = ?
      • P(A|B) = P(A ∩ B) / P(B) = (3/8) / (3/8) = 1

Slide 7

  • Example 3:
    • Problem: A box contains 10 balls, of which 4 are red and 6 are blue. Two balls are drawn at random without replacement. What is the probability of drawing a red ball on the second draw, given that the first ball drawn was blue?
    • Solution:
      • Event A: Drawing a red ball on the second draw
      • Event B: Drawing a blue ball on the first draw
      • P(A|B) = ?
      • P(A|B) = P(A ∩ B) / P(B) = (4/9) / (6/10) = 20/54

Slide 8

  • Properties of conditional probability:
    1. P(A|B) + P(A¯|B) = 1
    2. P(A|B) = P(A ∩ B) / P(B)
    3. P(B|A) = P(B ∩ A) / P(A)
    4. P(A ∪ B|C) = [P(A|C) * P(B|C)] + P(A¯ ∩ B¯|C)

Slide 9

  • Independence of events
    • Definition: Two events A and B are independent if and only if P(A ∩ B) = P(A) * P(B)
    • If A and B are independent, then P(A|B) = P(A) and P(B|A) = P(B)
    • Relevance to conditional probability: If A and B are independent, then P(A|B) = P(A)

Slide 10

  • Example 4:
    • Problem: Two dice are rolled. Find the probability of getting a sum of 7 on the second roll, given that the sum of the two rolls is at least 8.
    • Solution:
      • Event A: Getting a sum of 7 on the second roll
      • Event B: Getting a sum of at least 8 on the two rolls
      • P(A|B) = ?
      • P(A|B) = P(A ∩ B) / P(B) = (1/6) / (5/36) = 6/30 = 1/5

Slide 11

  • Example 5:
    • Problem: A deck of 52 cards contains 4 aces. Two cards are drawn at random without replacement. Find the probability of drawing an ace on the second draw, given that the first card drawn was an ace.
    • Solution:
      • Event A: Drawing an ace on the second draw
      • Event B: Drawing an ace on the first draw
      • P(A|B) = ?
      • P(A|B) = P(A ∩ B) / P(B) = (3/51) / (4/52) = 1/17

Slide 12

  • Example 6:
    • Problem: In a box, there are 8 red balls and 12 green balls. Two balls are drawn at random with replacement. What is the probability of getting two green balls, given that the first ball drawn was green?
    • Solution:
      • Event A: Getting two green balls
      • Event B: Getting a green ball on the first draw
      • P(A|B) = ?
      • P(A|B) = P(A ∩ B) / P(B) = (12/20) / (12/20) = 1

Slide 13

  • Example 7:
    • Problem: A jar contains 5 red balls and 3 blue balls. Three balls are drawn at random without replacement. Find the probability of drawing two red balls and one blue ball on the second draw, given that the first ball drawn was red.
    • Solution:
      • Event A: Drawing two red balls and one blue ball on the second draw
      • Event B: Drawing a red ball on the first draw
      • P(A|B) = ?
      • P(A|B) = P(A ∩ B) / P(B) = (4/7 * 3/6) / (5/8) = 2/7

Slide 14

  • Bayes’ Theorem
    • The probability of an event A given an event B can also be calculated using Bayes’ Theorem.
    • Bayes’ Theorem: P(A|B) = (P(B|A) * P(A)) / P(B)
    • Useful for reversing conditional probabilities when only the reverse is known.

Slide 15

  • Example 8:
    • Problem: A factory has three machines A, B, and C. Machine A produces 40% of the total output, machine B produces 30%, and machine C produces 30% of the total output. The percentages of defective items produced by machines A, B, and C are 2%, 3%, and 5% respectively. If a randomly chosen item is found to be defective, what is the probability that it was produced by machine B?
    • Solution:
      • Event A: The item is produced by machine B
      • Event B: The item is defective
      • P(A|B) = ?
      • P(A|B) = (P(B|A) * P(A)) / P(B) = (0.3 * 0.3) / (0.4 * 0.02 + 0.3 * 0.03 + 0.3 * 0.05) = 0.27 / 0.027 = 0.1

Slide 16

  • Example 9:
    • Problem: A bag contains 5 red balls and 3 green balls. Two balls are drawn at random without replacement. Find the probability that the second ball drawn is red, given that the first ball drawn is green.
    • Solution:
      • Event A: The second ball drawn is red
      • Event B: The first ball drawn is green
      • P(A|B) = ?
      • P(A|B) = P(A ∩ B) / P(B) = (5/7) / (3/8) = 20/21

Slide 17

  • Example 10:
    • Problem: A bag contains 10 balls, of which 6 are white and 4 are black. Two balls are drawn at random without replacement. Find the probability that both balls are black, given that the first ball drawn is white.
    • Solution:
      • Event A: Both balls are black
      • Event B: The first ball drawn is white
      • P(A|B) = ?
      • P(A|B) = P(A ∩ B) / P(B) = 0 / (6/10) = 0

Slide 18

  • Recap: Steps to solve problems on conditional probability
    1. Identify the given information and events involved.
    2. Use the formula P(A|B) = P(A ∩ B) / P(B) to calculate the conditional probability.
    3. Simplify and solve the equation.
    4. Interpret the result in the context of the problem.

Slide 19

  • Tips for solving conditional probability problems
    1. Draw a probability tree or create a probability distribution table to organize the information.
    2. Identify the events and calculate their individual probabilities.
    3. Calculate the conditional probabilities using the formula P(A|B) = P(A ∩ B) / P(B).
    4. Simplify and solve the equation to find the final answer.

Slide 20

  • Summary:
    • Conditional probability is the probability of an event given that another event has already occurred.
    • The formula for conditional probability is P(A|B) = P(A ∩ B) / P(B).
    • Bayes’ Theorem can be used to calculate conditional probabilities when only the reverse is known.
    • Solving conditional probability problems involves identifying the given information, using the formula, and interpreting the result.

Slide 21

  • Recap: Steps to solve problems on conditional probability
    • Identify the given information and events involved
    • Use the formula P(A|B) = P(A ∩ B) / P(B) to calculate the conditional probability
    • Simplify and solve the equation
    • Interpret the result in the context of the problem

Slide 22

  • Tips for solving conditional probability problems
    • Draw a probability tree or create a probability distribution table to organize the information
    • Identify the events and calculate their individual probabilities
    • Calculate the conditional probabilities using the formula P(A|B) = P(A ∩ B) / P(B)
    • Simplify and solve the equation to find the final answer

Slide 23

  • Summary:
    • Conditional probability is the probability of an event given that another event has already occurred
    • The formula for conditional probability is P(A|B) = P(A ∩ B) / P(B)
    • Bayes’ Theorem can be used to calculate conditional probabilities when only the reverse is known
    • Solving conditional probability problems involves identifying the given information, using the formula, and interpreting the result

Slide 24

  • Example 1:
    • Problem: A box contains 4 red balls and 6 blue balls. Two balls are drawn at random without replacement. What is the probability of drawing a blue ball on the second draw given that the first ball drawn was red?
    • Solution:
      • Event A: Drawing a blue ball on the second draw
      • Event B: Drawing a red ball on the first draw
      • P(A|B) = P(A ∩ B) / P(B) = (4/9) / (4/10) = 2/3

Slide 25

  • Example 2:
    • Problem: A bag contains 3 red balls and 5 green balls. Two balls are drawn at random with replacement. What is the probability of getting a red ball on the second draw, given that the first ball drawn was red?
    • Solution:
      • Event A: Getting a red ball on the second draw
      • Event B: Getting a red ball on the first draw
      • P(A|B) = P(A ∩ B) / P(B) = (3/8) / (3/8) = 1

Slide 26

  • Example 3:
    • Problem: A box contains 10 balls, of which 4 are red and 6 are blue. Two balls are drawn at random without replacement. What is the probability of drawing a red ball on the second draw, given that the first ball drawn was blue?
    • Solution:
      • Event A: Drawing a red ball on the second draw
      • Event B: Drawing a blue ball on the first draw
      • P(A|B) = P(A ∩ B) / P(B) = (4/9) / (6/10) = 20/54

Slide 27

  • Example 4:
    • Problem: Two dice are rolled. Find the probability of getting a sum of 7 on the second roll, given that the sum of the two rolls is at least 8.
    • Solution:
      • Event A: Getting a sum of 7 on the second roll
      • Event B: Getting a sum of at least 8 on the two rolls
      • P(A|B) = P(A ∩ B) / P(B) = (1/6) / (5/36) = 6/30 = 1/5

Slide 28

  • Example 5:
    • Problem: A deck of 52 cards contains 4 aces. Two cards are drawn at random without replacement. Find the probability of drawing an ace on the second draw, given that the first card drawn was an ace.
    • Solution:
      • Event A: Drawing an ace on the second draw
      • Event B: Drawing an ace on the first draw
      • P(A|B) = P(A ∩ B) / P(B) = (3/51) / (4/52) = 1/17

Slide 29

  • Example 6:
    • Problem: A bag contains 5 red balls and 3 green balls. Two balls are drawn at random without replacement. Find the probability that the second ball drawn is red, given that the first ball drawn is green.
    • Solution:
      • Event A: The second ball drawn is red
      • Event B: The first ball drawn is green
      • P(A|B) = P(A ∩ B) / P(B) = (5/7) / (3/8) = 20/21

Slide 30

  • Example 7:
    • Problem: A bag contains 10 balls, of which 6 are white and 4 are black. Two balls are drawn at random without replacement. Find the probability that both balls are black, given that the first ball drawn is white.
    • Solution:
      • Event A: Both balls are black
      • Event B: The first ball drawn is white
      • P(A|B) = P(A ∩ B) / P(B) = 0 / (6/10) = 0