Slide 1

Math 1520, Lecture 20

The concepts of conditional probability and independent events are considered in this lecture. This is a very important and useful idea in probability.

Slide 2

Conditional Probability

As an example of conditional probability consider the following: Two cards are drawn in sequence (without replacement) from a well shuffled deck of cards. 1. What is the probability that the first card drawn is a King? 2. What is the probability that the second card drawn is a King, given the the first card drawn was not a King? 3. What is the probability that the second card drawn is a King, given the the first card drawn was a King? The items in parts 2 and 3 are called conditional probability since their probability depends on a previous event.

Slide 3

Definition of Conditional Probability

Definition: In general, suppose A and B are events. The probability that event B happens given that the event A has already occurred, is called the conditional probability of B given A and is written using the notation: P (B|A) For example, suppose as on the previous slide that two cards are drawn in sequence (without replacement) from a well shuffled deck of cards. • Let A be the event that the first card drawn is a King. • Let B be the event that the second card drawn is a King. • The probability of B given A is written P (B|A) and means the probability of drawing a King, given that the first card drawn is a King.

Slide 4

iClicker

Question Two cards are drawn in sequence (without replacement) from a well shuffled deck of cards. • Let A be the event that the first card drawn is a King. • Let B be the event that the second card drawn is a King. Which of the following is the correct notation for the probability that the second card drawn is a King given that the first card drawn is not a King. A. P (B|A) B. P (B|AC ) C. P (A|B) D. P (AC |B) E. None of the above

Answer to Question Two cards are drawn in sequence (without replacement) from a well shuffled deck of cards. • Let A be the event that the first card drawn is a King. • Let B be the event that the second card drawn is a King. Which of the following is the correct notation for the probability that the second card drawn is a King given that the first card drawn is not a King. A. P (B|A) B. P (B|AC )

is the correct answer.

C. P (A|B) D. P (AC |B) E. None of the above

Slide 5

Formula for Conditional Probability of an Event

If A and B are events in an experiment and P (A) 6= 0 then the conditional probability that the event B will occur given that the event A has already occurred is P (A ∩ B) P (B|A) = P (A) 1. Use the formula to P (B|A) and P (A|B) if P (A) = .4, P (B) = .6, and P (A ∩ B) = .3 2. Justify the formula using a Venn Diagram.

Slide 6

iClicker

Question A pair of fair dice is rolled. What is the probability that one of the numbers is 4 given that the sum of the numbers is 7? A. 11/36 B. 4/7 C. 1/6 D. 1/3 E. None of the above

Answer to Question A pair of fair dice is rolled. What is the probability that one of the numbers is 4 given that the sum of the numbers is 7? A. 11/36 B. 4/7 C. 1/6 D. 1/3

is the correct answer.

E. None of the above

Slide 7

Product Rule

The Product Rule states that P (A ∩ B) = P (A) · P (B|A) and follows from the rule for conditional probability. Here is a simple example of what the rule means: Two cards are drawn from a deck. What is the probability that the first card drawn is a two and the second card drawn is a face card? • Let A denote the event that the first card drawn is a two. P (A) = 1/13 • Let B denote the probability that the second card drawn is a face card. P (B|A) = 12/51 • The probability we want is P (A ∩ B) which by the product rule is 1 · 12 P (A ∩ B) = P (A) · P (B|A) = 13 51 Draw a tree diagram to illustrate this situation.

Slide 8

Tree Diagrams

Tree diagrams are very useful for analyzing what happens to probabilities over several events. Consider the following example: Figures indicate that of all widgets shipped, there were 60% bought new and 40% bought used. Of those bought new, 1% were defective on arrival, 4% were slightly damaged on arrival and 95% were in good condition on arrival. Of those bought used, 10% were defective on arrival, 20% were slightly damaged on arrival and 70% were in good condition on arrival. 1. Draw a tree diagram representing these data. 2. Suppose you buy a widget that you know is new. What is the probability that it will be slightly damaged or defective? 3. Suppose you buy a widget, but you don’t know whether it is new or not. What is the probability that it will be new and in good condition? 4. Suppose you buy a widget, but you don’t know whether it is new or not. What is the probability that it will be in good condition?

Slide 9

iClicker

Question A box contains eight batteries of which two are known to be defective. The batteries are selected one at a time without replacement and tested until a non-defective one is found. What is the probability that 3 batteries have to be selected before a non-defective one is found? A. 3/4 B. 3/14 C. 3/28 D. 1/28 E. None of the above

Answer to Question A box contains eight batteries of which two are known to be defective. The batteries are selected one at a time without replacement and tested until a non-defective one is found. What is the probability that 3 batteries have to be selected before a non-defective one is found? A. 3/4 B. 3/14 C. 3/28

is the correct answer.

D. 1/28 E. None of the above

Slide 10

Independent Events

Two events A and B are independent events if the outcome of one does not effect the other, and vice-versa. This means P (B|A) = P (B) and P (A|B) = P (A) Test for Independent Events Two events A and B are independent events if and only if P (A ∩ B) = P (A) · P (B) 1. Give an argument about why this test works. 2. Given that P (A) = .5, P (B) = .8, P (A ∩ B) = .2 are A and B independent? 3. Given that P (AC ) = .3, P (B C ) = .4, P (A ∩ B) = .42 are A and B independent?