conditional probability and multiplication rule section 3.2

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Conditional Conditional Probability and Probability and Multiplication Rule Multiplication Rule Section 3.2 Section 3.2

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Page 1: Conditional Probability and Multiplication Rule Section 3.2

Conditional Probability Conditional Probability and Multiplication Ruleand Multiplication Rule

Section 3.2Section 3.2

Page 2: Conditional Probability and Multiplication Rule Section 3.2

Example 1Example 1

The table shows the estimated number of The table shows the estimated number of earned degrees conferred in the US in earned degrees conferred in the US in 2004 by level and gender. All numbers 2004 by level and gender. All numbers are in thousands.are in thousands.

GenderGender

malemale femalefemale TotalTotal

AssociateAssociate 231231 401401 632632

LevelLevel Bachelor’Bachelor’ss

553553 769769 13221322

OfOf Master’sMaster’s 197197 270270 467467

DegreDegreee

DoctoratDoctoratee

2525 2020 4545

TotalTotal 10061006 14601460 24662466

Page 3: Conditional Probability and Multiplication Rule Section 3.2

Find the probability of randomly Find the probability of randomly selecting someone who:selecting someone who:

Earned a bachelor’s degreeEarned a bachelor’s degree1322/2466=.5361322/2466=.536Earned a bachelor’s degree given Earned a bachelor’s degree given

that the person is a femalethat the person is a female769/1460=.527769/1460=.527 Is a female given the person earned Is a female given the person earned

a bachelor’s degreea bachelor’s degree769/1322=.582769/1322=.582

Page 4: Conditional Probability and Multiplication Rule Section 3.2

Classify as independent or Classify as independent or dependent eventsdependent events

P(A)=.2, P(B)=.3, P(A and B)=.06P(A)=.2, P(B)=.3, P(A and B)=.06.2*.3=.06 INDEPENDENT.2*.3=.06 INDEPENDENTP(A)=.5, P(B)=.2, P(A and B)=.12P(A)=.5, P(B)=.2, P(A and B)=.12.5*.2=.1 DEPENDENT.5*.2=.1 DEPENDENT

Page 5: Conditional Probability and Multiplication Rule Section 3.2

Example 3Example 3

You are dealt two cards successively You are dealt two cards successively without replacement from a standard without replacement from a standard deck of 52 playing cards. What is the deck of 52 playing cards. What is the probability that the first card is an probability that the first card is an ace and the second card is a jack?ace and the second card is a jack?

4/52*4/51=0.0064/52*4/51=0.006

Page 6: Conditional Probability and Multiplication Rule Section 3.2

Example 4Example 4

Find the probability of answering two Find the probability of answering two multiple choice questions correctly if multiple choice questions correctly if random guesses are made. Each random guesses are made. Each question has five choices. Only one question has five choices. Only one of the choices is correct.of the choices is correct.

1/5*1/5=0.041/5*1/5=0.04

Page 7: Conditional Probability and Multiplication Rule Section 3.2

At least one….At least one….

At least one means not none.At least one means not none.This means to find the probability of This means to find the probability of

at least one, you need to find the at least one, you need to find the complement of none.complement of none.

P(at least one)=1-P(none)P(at least one)=1-P(none)

Page 8: Conditional Probability and Multiplication Rule Section 3.2

Example 5Example 5

A true-false test has six questions. If A true-false test has six questions. If you randomly guess the answer to each you randomly guess the answer to each question, what is the probability you will question, what is the probability you will answer no questions correctly?answer no questions correctly?

½* ½* ½* ½* ½* ½=0.016½* ½* ½* ½* ½* ½=0.016What is the probability that you will What is the probability that you will

answer at least one question correctly?answer at least one question correctly?1-0.016=..9841-0.016=..984

Page 9: Conditional Probability and Multiplication Rule Section 3.2

ExercisesExercises

Pg 134-139 #10-17Pg 134-139 #10-17