INDEPENDENT EVENTS and the MULTIPLICATION RULE

Document technical information

Format pdf
Size 284.4 kB
First found May 22, 2018

Document content analysis

Category Also themed
Language
English
Type
not defined
Concepts
no text concepts found

Persons

Organizations

Places

Transcript

Last Name____________________ First Name _______________________Class Time________Chapter 3-1
Probability: Events and Probabilities
PROBABILITY: likelihood or chance that an outcome will happen; long-run relative frequency
A probability is a number between 0 and 1 , inclusive.
0 ≤ probability ≤ 1
Sample Space: Set of all possible simple outcomes
Event: Combination of Outcomes; (also called "compound event")
AND, OR , NOT, GIVEN THAT relate two or more events to each other to create new compound events.
EXAMPLE 1:
Assume that Mickey had a driving accident. But Mickey was shook up after the accident and wasn't
quite sure exactly where the accident occurred. He only knows that location of the accident was
related to Park and Bay, two streets which cross.
a. Shade in the relevant parts of the streets:
Park AND Bay streets
Park OR Bay streets
not on Park street
Describe in words where the accident occurred if it was:
b. on Park AND Bay streets.
c. on Park OR Bay streets.
d. The accident was not on Park.
EXAMPLE 2:
ROLLING 1 DIE:
Sample Space:
S={1,2,3,4,5,6}
Based on Example 3-2 Illowsky, B., & Dean, S. Collaborative Statistics. Connexions, http://cnx.org/content/col10522/1.29/
Odd
Even
Odd number
Even number Number less
Number greater
Event
number
number
larger than 2
less than 5
than 5
than 3
Event
A = {1, 3, 5}
B = {2, 4, 6}
C = {3, 5}
D = {2, 4}
E = {1, 2, 3, 4}
Probability
P(A) = 3
P(B) = 3
6
6
P(C) =
P(C) =
P(E) =
G={4, 5, 6}
P(G) =
Last Name____________________ First Name _______________________Class Time________Chapter 3-2
Terminology with Events:
AND: A and B means BOTH events A and B occur: Outcome that occurs satisfies both events A and B
Probability of Event C and E : P(C and E) = 1
a. Event C and E : { 3 }
6
b. Event B and G : {
}
Probability of Event B and G : P(B and G) = _______
OR: A or B means either event A occurs or event B occurs or both occur.
Outcome that occurs satisfies event A or event B or both
Probability of Event C or E : P(C or E) = 5
c. Event C or E = { 1,2,3,4,5 }
6
d. Event B or G = {
}
Probability of Event B or G : P(B or G) = _________
NOT: COMPLEMENT A means event A does NOT occur.
e.
A = {1, 3, 5}
f.
D = {2, 4}
A= { 2,4,6 }
D= {
COMPLEMENT RULE:
P(A) = 3
P(A) + P(A) = 3 + 3 = 1
P(D) =
P(D) + P(D) =
6
}
6
For any event A: P(A) + P(A) = 1
6
P(A) = 1  P(A)
EXAMPLE 3:
MUTUALLY EXCLUSIVE events are two events that can NOT both happen at the same time:
P(A and B) = 0
a. List all pairs of events above that are mutually exclusive. Justify algebraically with probabilities:
b. Are taking day classes and taking night classes at college are mutually exclusive? Why or why not?
c. Are being a full time student and being a part time student mutually exclusive? Why or why not?
Last Name____________________ First Name _______________________Class Time________Chapter 3-3
EXAMPLE 4:
CONDITIONAL PROBABILITY:
Probability that event A occurs GIVEN THAT (we know that) B has occurred
The outcome B that we know has occurred is called the condition.
The condition B reduces the sample space to be smaller by eliminating outcomes that did not occur.
Notation: P(A|B)
P(Event| Condition)
the vertical line
|
means
given that
R = the event of rain on one day in Cupertino.
M = the event being a day in March
The probability of rain on one day in Cupertino is 50/365 or 0.1370.
R|M is the event of rain on one day in Cupertino, GIVEN that the day is in March.
P(R|M) is the probability of rain on one day in Cupertino, GIVEN that the day is in March.
Circle the correct relation: P(R|M) < P(R)
P(R|M) =P(R)
P(R|M) > P(R)
Explain your answer.
EXAMPLE 5:
F is the event that the animal climbed a fence.
C is the event that the animal is a cat.
D is the event that the animal is a dog.
Insert one of the three symbols: <
a. P (F|C)_____P(F)
=
Event
> in the following:
b. P (F|D)_____P(F)
EXAMPLE 6:
ROLLING 1 DIE:
Event
A is the event that the driver caused an accident.
T is the event that the driver is texting.
Y is the event that the driver is 45 years old.
c. P (A)_____P(A|T)
Sample Space:
Odd
number
A = {1, 3, 5}
Even
number
B = {2, 4, 6}
Odd number
larger than 2
C = {3, 5}
S={1,2,3,4,5,6}
Even number
less than 5
D = {2, 4}
d. P (A)_____P(A|Y)
Number less
than 5
Number greater
than 3
E = {1, 2, 3, 4}
G={4, 5, 6}
Find the probability of rolling a 3 or 5 IF WE KNOW THAT the result was an odd number:
Knowing the result was odd gives us a smaller sample space ("reduces" the sample space).
The new reduced sample space is { 1 , 3 , 5 }
P(C|A) = P(3 or 5 | the result is odd – a 1, 3, or 5 ) = 2
{1,2,3,4,5, 6}
3
 GIVEN THAT we know that the outcome is odd,
the probability of rolling a 3 or a 5 out of the group 1, 3, or 5, is 2/3.
Last Name____________________ First Name _______________________Class Time________Chapter 3-4
EXAMPLE 7: Use probability symbols to represent the problem and write your final answer as a
fraction in unreduced form.
a. Event Given Condition:Find the probability rolling a number greater than 3 given that the outcome
is even.
Event Without Condition: P(G) =
b. Event Given Condition: Find the probability of rolling an odd number given that the outcome was
even.
Event Without Condition: P(A) =
c. Event Given Condition: Find the probability of rolling a number less than 5 given that the outcome
is odd.
Event Without Condition: P(E) =
EXAMPLE 8: CONDITIONAL PROBABILITY: TOSSING A FAIR COIN TWICE:
A fair coin has equal probability of landing on "Head" or "Tail"
a. Write the sample space of outcomes for tossing a coin twice:
S=
{
}
b. Find the probability of getting exactly ONE HEAD in two tosses.
Write your answer as an unreduced fraction.
c. Find the probability of getting exactly ONE HEAD in two tosses, GIVEN THAT AT LEAST ONE
HEAD was obtained.
Last Name____________________ First Name _______________________Class Time________Chapter 3-5
INDEPENDENT EVENTS: Two events are independent if and only if the probability of one event
occurring is not affected by whether the other event occurs or does not occur.


Events A and B are INDEPENDENT if and only if P(A | B) = P(A)
This also means that P(B | A) = P( B )
Independence means that the "condition B " does not change the probability of event A
EXAMPLE 9:
Even
number
Event
ROLLING ONE DIE
Event
Number less
than 5
B = {2, 4, 6} E = {1, 2, 3, 4}
Number greater
than 3
G={4, 5, 6}
The probability of rolling an number less than 5, given that the number is even: P(E | B) = ____
P(E | B) = 2/3
The probability of a number less than 5 : P(E) = ____
The probability of rolling an number greater than 3, given that the number is even: P(G | B) = ___
The probability of a number greater than 3 :
EXAMPLE 10:
a. Rolling two dice:
P(G) = ____
State whether the following describe independent or dependent events? Why?
_____________________
b. Selecting 2 cards consecutively from a deck of 52 cards, without replacement _______________
c. Selecting 2 cards from a deck of cards, with replacement ________________
Last Name____________________ First Name _______________________Class Time________Chapter 3-6
EXAMPLE 11:
Statistics from the FHDA (Foothill/De Anza) district website for a recent quarter:
 15% of students registered at De Anza College were Hispanic
 15 % of students registered at Foothill College were Hispanic
 15% of all students registered at FHDA District were Hispanic
H = event that a student in the FHDA district was Hispanic
D = event that a student attended De Anza
F = event that a student attended Foothill
Were the events "student is Hispanic" and "student attends De Anza College" independent?
Justify your answer using appropriate probabilities in a clear, logical presentation.
EXAMPLE 12: http://www.censusindia.gov.in/2011-prov-results/indiaatglance.html
http://www.censusindia.gov.in/Census_Data_2001/India_at_glance/literates1.aspx
Adult (15+ years) literacy rates are 82.1% for men and 65.5% for women.
The overall literacy rate is estimated as approximately 74%.
Note: The literacy rates have improved since 2001 when the rates were:
Overall 64.8%
Male: 75.3%
Female: 53.7%
Consider the population of residents of India age 15 and over:
F = event that resident of India is female
M = event that resident of India is male
L = event that a randomly selected resident of India is literate
Is the literacy rate in India independent of gender?
Justify your answer using appropriate probabilities in a clear, logical presentation.
TWO-WAY CONTINGENCY TABLES
A contingency table with frequencies displays data for two variables; the number of individuals or
items in each category is shown.
We can use the data in the table to find probabilities as fractions and create a corresponding
contingency table with probabilities.
Fractions for all probabilities EXCEPT conditional probabilities have the grand total in the
denominator of the fraction. The fraction may be changed to a decimal.
Conditional Probabilities: The condition limits you to the particular row or column in the table
specified by the reduced sample space.
The denominator of the fraction will be the total for the row or column in the table that
corresponds to the condition.
Last Name____________________ First Name _______________________Class Time________Chapter 3-7
EXAMPLE 13: A large car dealership examined a sample of vehicles sold or leased in the past year.
The vehicles were classified by two variables:
 whether the vehicle was a car, SUV, van, or truck
 whether the transaction was a new car sale, a used car sale or a vehicle lease
Car (C) SUV (S) Van (V) Truck(T) Total
New vehicle sale(N)
86
25
21
38
170
Used vehicle sale (U)
39
13
4
22
78
Vehicle Lease (L)
34
12
6
0
52
Total
159
50
31
60
300
Suppose a vehicle in the sample is randomly selected to review its sales or lease papers. For each of the
probability questions that follow:
 Write the symbolic notation of the probability.
 Find the probability as an unreduced fraction
 Find the probability as a decimal, estimated to 4 places.
a. Find the probability that the vehicle was leased.
b. Find the probability that a vehicle is a truck.
c. Find the probability that a vehicle is NOT a truck.
d. Find the probability that the vehicle was a car AND was leased.
e. Find the probability that the vehicle was used GIVEN THAT it was a van.
f. Find the probability that the vehicle was a van GIVEN THAT it was used.
g. Find the probability that the vehicle was used AND was a van.
Addition Rule for OR Events:
P(A or B) = P(A) + P(B) – P(A and B)
h. Find the probability that the vehicle was used OR was a van.
i.
Find the probability that the vehicle was leased OR was a truck.
Last Name____________________ First Name _______________________Class Time________Chapter 3-8
INDEPENDENCE in CONTINGENCY TABLES


Let A be one row event.
Let B be one column event.
Check if two events are independent with any of the following three equations.
Use the equation with probabilities that are easiest to find.
1.
2.
3.



P(A) = P(A | B)
P(B) = P(B | A)
P(A and B) = P(A)•P(B)
If any equation is true, the two events are independent.
If any equation is false, the two events are not independent. They are dependent.
All three equations are either all true or all false.
A and B are INDEPENDENT EVENTS if and only if these probability statements are true.
A and B are DEPENDENT EVENTS if and only if these probability statements are true.
EXAMPLE 14: Using the same data from EXAMPLE 12, previous page,
a. Are the events N and V independent?
Car (C) SUV (S) Van (V) Truck(T) Total
New vehicle sale(N)
86
25
21
38
170
Used vehicle sale (U)
39
13
4
22
78
Vehicle Lease (L)
34
12
6
0
52
Total
159
50
31
60
300
Show your work in a CLEAR AND LOGICAL way to justify your answer.
Show three different ways to arrive at your conclusion.
1. P(N| V) = 21/31 = 0.6774 ≠ P(N) = 170/300 = 0.5667,
so N and V are not independent.
2. P(V| N) = 21/170 = 0.1235 ≠ P(V) = 31/300 =0.1033,
so N and V are not independent.
3. P(V and N) = 21/300 = 0.07 ≠ P(V)•P(N) = (31/300)(170/300) = 0.0569,
so N and V are not independent.
Last Name____________________ First Name _______________________Class Time________Chapter 3-9
b. Are the events S and U independent?
Car (C) SUV (S) Van (V) Truck(T) Total
New vehicle sale(N)
86
25
21
38
170
Used vehicle sale (U)
39
13
4
22
78
Vehicle Lease (L)
34
12
6
0
52
Total
159
50
31
60
300
Show your work in a CLEAR AND LOGICAL way to justify your answer.
Show three different ways to arrive at your conclusion.
PROBABILITY RULES
Complement Rule:
P(A) = 1  P(A)
Conditional Probability Rule:
P(A|B) =P(A and B)
P(B)
or
P(B|A) =P(A and B)
P(A)
Multiplication Rule
for AND Events:
General Multiplication Rule:
P(A and B) = P(A|B) P(B) or P(A and B) = P(B|A) P(A)
IF AND ONLY IF events are INDEPENDENT:
P(A and B) = P(A) P(B)
Addition Rule for OR Events:
General Addition Rule:
P(A or B) = P(A) + P(B) – P(A and B)
IF AND ONLY IF events are MUTUALLY EXCLUSIVE: P(A or B) = P(A) + P(B)
Using the Addition Rule to solve for P(A and B):
P(A and B) = P(A) + P(B) – P(A or B)
Last Name____________________ First Name _______________________Class Time________Chapter 3-10
For Examples 15 - 23:
 Write the symbolic notation of the probability.
 Find the probability as an unreduced fraction, when possible.
 Find the probability as a decimal, estimated to 4 places.
 Show all work in a NEAT, logical, clear manner.
EXAMPLE 15 Conditional Probability Rule:
At a large company:
34% of employees live over 30 miles away from the company's offices,
52% of employees sometimes work from home,
28% of employees sometimes work from home AND live over 30 miles away from the company's offices.
Let H = the event that the employee sometimes works at home.
Let F = be the event that the employee lives over 30 miles away from the company's offices.
Find the probability that an employee sometimes works at home, given that the employee lives over
30 miles away from the company's offices.
EXAMPLE 16: Conditional Probability Rule:
In a certain neighborhood, 65 % of residents subscribe to the Mercury News and 30% of residents
subscribe to the SF Chronicle. These figures include the fact that 20% of residents subscribe to both.
M = event that a person subscribes to the Mercury News
C = event that a person subscribes to the Chronicle
Find the probability that a person subscribes to Mercury News given that the person subscribes to the
SF Chronicle.
EXAMPLE 17: Addition Rule for OR Events:
In a certain neighborhood, 65 % of residents subscribe to the Mercury News and 30% of residents
subscribe to the SF Chronicle. These figures include the fact that 20% of residents subscribe to both.
Find the probability that a person subscribes to Mercury News OR the SF Chronicle
M = event that a person subscribes to the Mercury News ; C = event that a person subscribes to the
Chronicle
Last Name____________________ First Name _______________________Class Time________Chapter 3-11
EXAMPLE 18: Addition Rules for OR Events
a. When rolling two dice the probability of rolling a sum of 6
is 5/36 and the probability of rolling a double is 6/36.
The probability of rolling a 6 by rolling a 3 on each die is 1/36.
Find the probability of rolling a sum of 6 or a double.
Sample Space for sum when rolling 2 dice
36 possible outcomes
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
b. When rolling two dice the probability of rolling a sum of 3 is 2/36 and the probability of
rolling a double is 6/36. Find the probability of rolling a sum of 3 or a double.
EXAMPLE 19: Addition Rules for OR Events (Harder)
In a city, 50% of residents watch the 6PM news and 30% of residents watch the 11PM news.
35% of residents do not watch either the 6PM news or the 11PM news.
S = event that a person watches the 6 pm news ; E = event that a person watches the 11 pm news
(a) Find the probability that a person watches the 6PM news or the 11PM news (or both).
(b) Find the probability that a person watches both the 6PM news and 11 PM news.
EXAMPLE 20: Consider the experiment of rolling one fair die and tossing a fair coin.
Find the probability of rolling 6 on the die AND tossing "Head" on the coin.
EXAMPLE 21: Consider the experiment of tossing a fair coin twice.
Find the probability of getting Head on both tosses.
Last Name____________________ First Name _______________________Class Time________Chapter 3-12
EXAMPLE 22:
In a class there are male and female students, and students with long or short hair
60% of students in a class are female
40% of students are male.
50% of students have long hair.
45% of students are female and have long hair.
Events: F = student is female
Of the male students, 12.5% have long hair.
M = student is male
L = student has long hair
a. Identify each of the events and probabilities as stated in the problem. Use correct symbols:

60% of the students in a class are female.
_____________________

40% of the students are male.
_____________________

50% of the students have long hair.
_____________________

45% of the students are female and have long hair.

Of the male students, 12.5% have long hair.
_____________________
_____________________
b. Fill in the probability contingency table with the appropriate probabilities:
Hair Length by Gender
for a class of students
Female (F)
Long Hair
(L)
Short Hair (or not long hair)
(L')
Total
Male (M)
Total
For parts b through f, show how to use an appropriate probability rule to calculate probabilities.
Write the probability statement using proper notation and show the work for each calculation.
a. Of the female students, 75% have long hair. ________________________________________ Rule
b. 5% of the students are male and have long hair.______________________________________ Rule
c. 85% of the students are male or have long hair. _____________________________________ Rule
d. 15% of students are female and have short hair. ____________________________________ Rule
(This one needs a little bit of creative thinking!)
Last Name____________________ First Name _______________________Class Time________Chapter 3-13
e. 90% of students who have long hair are female. ____________________________________ Rule
f. Are being MALE and having LONG hair independent? Justify your answer with probabilities.
EXAMPLE 23: http://www.censusindia.gov.in/2011-prov-results/indiaatglance.html
India's population is 48.5% female and 51.5% male
Adult (15+ years) literacy rates are 82.1% for men and 65.5% for women
The overall adult literacy rate is estimated as approximately 74%.
Consider the population of residents of India age 15 and over. Assume that the population age 15
and over follows the same gender distribution: 48.5% female and 51.5% male
F = event that resident of India is female
M = event that resident of India is male
L = event that a randomly selected resident of India is literate
a. Find the probability that a randomly selected person in India (age 15+) is both literate and female.
b. Find the probability that a randomly selected person in India (age 15+) is literate and male
c. Use the Conditional Probability Rule to find the probability that a randomly selected
resident of India age 15 or over is female, given that the person is literate:
Last Name____________________ First Name _______________________Class Time________Chapter 3-14
TESTING WHETHER 2 EVENTS ARE INDEPENDENT:
 Two events are independent if and only if P(A|B) = P(A)
This also means that P(B|A) = P(B)
 IF AND ONLY IF events A and B are independent, P(A and B) = P(A) P(B)
You can use this equation to test for independence IF you already know P(A and B)
3 ways to test for independence

"Conditional" probability:
Compare P(A|B) to P(A) .
If P(A|B)=P(A), the events are independent

"Conditional" probability:
Compare P(B|A) to P(B) .
If P(B|A)=P(B), the events are independent

"And" probability:
Compare P(A and B) to P(A) P(B).
If P(A and B) = P(A) P(B), the events are independent
Do one method only. All three tests for independence always give the same result (if done correctly)
Do whichever is easier with the information you know for the problem.
EXAMPLE 24: Test for Independence
A class is 65% female. 35 % of all students in the class have long hair.
25% of the students are female and have long hair.
55% of the female students have long hair.
Event F = student is female
Event L = student has long hair
Are the events being female and having long hair independent?
Test with Conditional Probabilities:
P(L | F) =_______
P(L) =_______
Does P(L | F ) = P(L)? _______
Test using "AND" probabilities:
P(F and L) =_____ P(F) P(L) =___________________
Does P(F and L) = P(F) P(L)? _______
Conclusion:
EXAMPLE 25: Test for Independence
60% of students take a math class and 50% of students take a history class.
30% of students take math class and a history class
Event M = student takes a math class
Event H = student takes a history class
Are events M and H independent?
Test using "AND" probabilities:
P(M and H) =_____
Conclusion:
P(M) P(H) =________________
Does P(M and H) = P(M) P(H) ? ____
Last Name____________________ First Name _______________________Class Time________Chapter 3-15
TREE DIAGRAMS
Tree diagrams are a useful tool in solving probability problems.
Each complete path through the tree represents a separate and mutually exclusive outcome in the
sample space.
 Draw a tree representing the possible mutually exclusive outcomes.
 Label the end of each branch with the event.
 Write the probability both symbolically and numerically along each branch of the tree.
 Multiply probabilities along each complete path through the tree to find probabilities of each
"AND" outcomes in the sample space.
 Note: To find the probability of a compound OR event, add probabilities for the appropriate paths
of a tree.
EXAMPLE 26:
A box of chocolates contains 25 pieces. There are 10 milk chocolates (M) and 15 dark chocolates (D).
Homer selects two pieces of chocolate at random from the box.
Homer selects 2 pieces of chocolate WITH REPLACEMENT:
a. Draw the tree diagram. Show the events and probabilities for each branch and each complete path
of the tree. What should the probabilities for each branch add up to
For parts b through g,
 write the problem in symbolic form
 show your final answer in both unreduced fraction format and as a decimal, estimated to
4 places.
Find the probability that:
b. Home chose a milk chocolate on his first selection.
Last Name____________________ First Name _______________________Class Time________Chapter 3-16
c. Homer chose exactly one milk chocolate.
d. Homer chose exactly one milk chocolate, given his first selection was a dark chocolate.
e. Homer chose at least one dark chocolate.
f. Homer chose two chocolates of the same type, either both milk or both dark chocolate.
g. Are M1 and D1 mutually exclusive events? Justify your answer BOTH numerically with
calculations and with logic using a complete sentence.
h. Are M1 and M2 independent events? Justify your answer with calculations.
Last Name____________________ First Name _______________________Class Time________Chapter 3-17
EXAMPLE 27:
A box of chocolates contains 25 pieces. There are 10 milk chocolates (M) and 15 dark chocolates (D).
Homer selects two pieces of chocolate at random from the box.
Homer selects 2 pieces of chocolate WITHOUT REPLACEMENT:
a. Draw the tree diagram. Show the events and probabilities for each branch and each complete path
of the tree.
What should the probabilities for each branch add up to?
For parts b through g,
 write the problem in symbolic form
 show your final answer in both unreduced fraction format and as a decimal, estimated to
4 places.
Find the probability that:
b. Home chose a milk chocolate on his first selection.
c. Homer chose exactly one milk chocolate.
d. Homer chose exactly one milk chocolate, given his first selection was a dark chocolate.
e. Homer chose at least one dark chocolate.
Last Name____________________ First Name _______________________Class Time________Chapter 3-18
f. Homer chose two chocolates of the same type, either both milk or both dark chocolate.
g. Are M1 and D1 mutually exclusive events? Justify your answer BOTH numerically with
calculations and with logic using a complete sentence.
h. Are M1 and M2 independent events? Justify your answer with calculations.
HELPFUL FORMULAS to solve for joint and conditional probabilities:

P(A or B) = P(A) + P(B) – P(A and B)
o If A and B are MUTUALLY EXCLUSIVE, then P(A or B) = P(A) + P(B)

P(A and B) = P(A) + P(B) – P(A or B)
o If A and B are MUTUALLY EXCLUSIVE, then P(A and B) = 0

P(A and B) = P(A | B) P(B)
o If A and B are INDEPENDENT, then P(A and B) = P(A) P(B)

P(A and B) = P(B | A) P(A)
o If A and B are INDEPENDENT, then P(A and B) = P(A) P(B)
Last Name____________________ First Name _______________________Class Time________Chapter 3-19
PRACTICE PROBLEMS: USING PROBABILITY RULES
Practice Example A: Suppose H and J are events. P(H) = 0.2 P(J) = 0.4 P(J and H) = 0.12
(a) Are H and J independent?
(b) Find P(J | H)
(c) Find P(H | J)
(d) Find P(H or J)
Practice Example B:
Suppose R and S are events. P(R) =0.2 P(S) = 0.5 P(R | S) = 0.3
(a) Are R and S independent?
(b) Find P(R and S)
(c) Find P(S | R)
(d) Find P(R or S)
Practice Example C: Suppose F and G are events. P(F) = 0.4, P(G) = 0.3 P(F or G) = 0.6
(a) Find P(G and F)
(b) Find P(G|F)
(c) Find P(F|G)
(d) Are F and G mutually exclusive?
(e) Are F and G independent?
Practice Example D:
Redo example C using P(F or G) = 0.58 instead of 0.60
P(F) = 0.4, P(G) = 0.3 P(F or G) = 0.58
(a) Find P(G and F)
(b) Find P(G|F)
(c) Find P(F|G)
(d) Are F and G mutually exclusive?
(e) Are F and G independent?
Last Name____________________ First Name _______________________Class Time________Chapter 3-20
PRACTICE PROBLEMS FOR TREES
For both examples:




Draw a tree representing the possible mutually exclusive outcomes.
Label the end of each branch with the event.
Write the probability both symbolically and numerically along each branch of the tree.
Write symbols and values of probabilities along each complete path through the tree. Check that the
sum of your probabilities at the end of the paths is 1!
PRACTICE EXAMPLE E: TREE Source: http://www.censusindia.gov.in/2011-prov-results/indiaatglance.html
India's population is 48.5% female and 51.5% male
Adult (15+ years) literacy rates are 82.1% for men and 65.5% for women
Consider the population of residents of India age 15 and over. Assume that the population age 15 and
over follows the same gender distribution: 48.5% female and 51.5% male
F = event that resident of India is female
M = event that resident of India is male
L = event that a randomly selected resident of India is literate
Use the tree to find the probability that an adult in India is literate.
PRACTICE EXAMPLE F: TREE
Two statistics professors in Poland had their students do an experiment that showed that the Belgian one Euro
coin is not a fair coin. It is a biased coin. In their experiment, it showed a 56% chance of landing on a "Head".
Suppose we do an experiment of tossing the Belgian one Euro coin twice in a row.
Use the tree to find the probability of getting at least one head.

Similar documents

×

Report this document