Transcript Conditional Probability and Multiplication Rule Day 2

Conditional Probability and
Multiplication Rule
Section 3.2
Example 1

The table shows the estimated number of
earned degrees conferred in the US in 2004 by
level and gender. All numbers are in thousands.
Gender
male
Associate 231
Level
Bachelor’s 553
Of
Master’s
197
Degree Doctorate 25
Total
1006
female
401
769
270
20
1460
Total
632
1322
467
45
2466
Find the probability of randomly
selecting someone who:
Earned a bachelor’s degree
 1322/2466=.536
 Earned a bachelor’s degree given that the
person is a female
 769/1460=.527
 Is a female given the person earned a
bachelor’s degree
 769/1322=.582

Classify as independent or dependent
events
P(A)=.2, P(B)=.3, P(A and B)=.06
 .2*.3=.06 INDEPENDENT
 P(A)=.5, P(B)=.2, P(A and B)=.12
 .5*.2=.1 DEPENDENT

Example 3
You are dealt two cards successively
without replacement from a standard deck
of 52 playing cards. What is the
probability that the first card is an ace and
the second card is a jack?
 4/52*4/51=0.006

Example 4
Find the probability of answering two
multiple choice questions correctly if
random guesses are made. Each question
has five choices. Only one of the choices
is correct.
 1/5*1/5=0.04

At least one….
At least one means not none.
 This means to find the probability of at
least one, you need to find the
complement of none.
 P(at least one)=1-P(none)

Example 5
A true-false test has six questions. If you
randomly guess the answer to each
question, what is the probability you will
answer no questions correctly?
 ½* ½* ½* ½* ½* ½=0.016
 What is the probability that you will
answer at least one question correctly?
 1-0.016=..984

Exercises

Pg 134-139 #10-17