section 5.3 - the addition rule and disjoint events

Section 5.3 - The Addition Rule and Disjoint Events

D16. The diagrams at the bottom of the slide are called Venn diagrams.

How do these diagrams justify the two forms of the Addition Rule?

BA BA

P(A or B) =P(A) + P(B)P(A or B) =P(A) + P(B)−P(A and B)


D16. How do these Venn diagrams justify the two forms of the Addition Rule?

BA BA

Disjoint events:

P(A or B) =P(A) + P(B)Non-disjoint events :P(A or B) =P(A) + P(B)−P(A and B)

Disjoint eventsNon-disjoint events


D17. What happens to the general form of the Addition Rule in a situation where A and B are mutually exclusive?

BA BA

P(A or B) =P(A) + P(B)−P(A and B)

Mutually exclusive


D17. What happens to the general form of the Addition Rule in a situation where A and B are mutually exclusive?

BA BA

P(A or B) =P(A) + P(B)−P(A and B)If A and B are mutually exclusive, P(A and B) =0.P(A or B) =P(A) + P(B)

Mutually exclusive


P14. Of the 34,071,000 people in the U.S. who fish, 1,847,000 fish in the Great Lakes, 27,913,000 fish in other fresh water, and 9,051,000 fish in salt water.

a. In categorizing people who fish, are these three categories disjoint? Are they complete?

b. Suppose you randomly select a person from among those who fish. Can you find the probability that the person fishes in salt water?

c. Can you find the probability that the person fishes in fresh water?

d. The number of people who fish in fresh water is 28,439,000. How many people fish in both salt water and fresh water?




The three categories are not disjoint:

1847000 + 27913000 + 9051000 > 34071000

The categories are complete since people who fish must belong to at least one of these three categories.




Yes:

P( fish in salt water) =905134071

=0.266




You can’t find the probability that the person fishes in fresh water, since the “Great Lakes” and “other fresh water” categories overlap.




The number of people who fish in fresh water is 28,439,000. The number of people who fish in both salt water and fresh water is

fresh or salt = fresh + salt - (fresh and salt)

fresh and salt = fresh + salt - (fresh or salt)

fresh and salt = 28,439,000 + 9,051,000 - 34,071,000 = 3,419,000


P15. Display 5.33 categorizes the child support received by custodial parents with children under age 21.

Revise the table so that the categories are complete and disjoint. One category was not included: With agreement or award but not supposed to receive payment in 2001

Child Support Status Number (thousands)

With agreement or award 7,916

Supposed to receive payments 6,924

Actually received payments 5,119

Received full amount 3,099

Received partial payments 2,020

Did not receive payments 1,805

Child support not awarded 5,467

Total 13,383





With agreement or award 7,916

Supposed to receive payments 6,924

Not supposed to receive payments 992





Not supposed to receive payments 992

Received full amount 3,099

Received partial payments 2,020

Did not receive payments 1,805

Child support not awarded 5,467

Total 13,383


P16. If you roll two dice, are these pairs of events mutually exclusive?

a. Doubles; sum is 8

b. Doubles; sum is odd

c. A 3 on one die; sum is 10

d. A 3 on one die; doubles



a. Doubles; sum is 8

b. Doubles; sum is odd

c. A 3 on one die; sum is 10

d. A 3 on one die; doubles

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12



a. Doubles; sum is 8: Not mutually exclusive:

Doubles = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Sum is 8 = {(2,6),(6,2),(3,5),(5,3),(4,4)}

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12



b. Doubles; sum is odd: Mutually exclusive:

Doubles = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Sum is odd = {(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(3,6),(4,1),

(4,3),(4,5),(5,2),(5,4),(5,6),(6,1),(6,3),(6,5)}

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12



c. A 3 on one die; sum is 10: Mutually exclusive:

A 3 on one die = {(1,3),(2,3),(3,3),(4,3),(5,3),(6,3),(3,1),(3,2),(3,4),(3,5),(3,6)}

Sum is 10 = {(4,6),(5,5),(6,4)}

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12



d. A 3 on one die; doubles: Not mutually exclusive:

A 3 on one die = {(1,3),(2,3),(3,3),(4,3),(5,3),(6,3),(3,1),(3,2),(3,4),(3,5),(3,6)}

Doubles = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12


P17. A researcher will select a student at random from a school population where 33% of the students are freshmen, 27% are sophomores, 25% are juniors, and 15% are seniors.

a. Is it appropriate to use the Addition Rule for Disjoint Events to find the probability that the student will be a junior or a senior? Why or why not?

b. Find the probability that the student will be a freshman or a sophomore.




Yes. A student can’t be a junior and a senior at the same time.




P(F or S) =P(F ) + P(S)=0.33+ 0.27 =0.60


P18. A tetrahedral die has the numbers 1, 2, 3, and 4 on its four faces. Suppose you roll a pair of tetrahedral dice.

a. Make a table of all 16 possible outcomes.

b. Use the Addition Rule for Disjoint Events to find the probability that you get a sum of 6 or a sum of 7.

c. Use the Addition Rule for Disjoint Events to find the probability that you get doubles or a sum of 7.

d. Why can’t you use the Addition Rule for Disjoint Events to find the probability that you get doubles or a sum of 6?



Make a table of all 16 possible outcomes.

sum 1 2 3 4

1 2 3 4 5

2 3 4 5 6

3 4 5 6 7

4 5 6 7 8



Use the Addition Rule for Disjoint Events to find the probability that you get a sum of 6 or a sum of 7.

sum 1 2 3 4

1 2 3 4 5

2 3 4 5 6

3 4 5 6 7

4 5 6 7 8



Use the Addition Rule for Disjoint Events to find the probability that you get a sum of 6 or a sum of 7.

sum 1 2 3 4

1 2 3 4 5

2 3 4 5 6

3 4 5 6 7

4 5 6 7 8

P(6 or 7) =P(6) + P(7) =316

+216

=516



Use the Addition Rule for Disjoint Events to find the probability that you get doubles or a sum of 7.

sum 1 2 3 4

1 2 3 4 5

2 3 4 5 6

3 4 5 6 7

4 5 6 7 8



Use the Addition Rule for Disjoint Events to find the probability that you get doubles or a sum of 7.

sum 1 2 3 4

1 2 3 4 5

2 3 4 5 6

3 4 5 6 7

4 5 6 7 8

P(doubles or 7) =P(doubles) + P(7) =416

+216

=616



Why can’t you use the Addition Rule for Disjoint Events to find the probability that you get doubles or a sum of 6?

sum 1 2 3 4

1 2 3 4 5

2 3 4 5 6

3 4 5 6 7

4 5 6 7 8



Why can’t you use the Addition Rule for Disjoint Events to find the probability that you get doubles or a sum of 6?

sum 1 2 3 4

1 2 3 4 5

2 3 4 5 6

3 4 5 6 7

4 5 6 7 8

The events "Doubles" and "Sum of 6" are not disjoint (3,3)


P19. Display 5.34 gives information about all reportable crashes on state-maintained roads in North Carolina in a recent year.

No Teen Driver(s)

Teen Driver(s)

Total

Not Speed Related

120,082 24,291 144,373

Speed Related

67,331 19,755 87,086

Total 187,413 44,046 231,459



Are the events crash involved a teen driver and crash was speed related mutually exclusive? How can you tell?

No Teen Driver(s)

Teen Driver(s)

Total

Not Speed Related

120,082 24,291 144,373

Speed Related

67,331 19,755 87,086

Total 187,413 44,046 231,459



Are the events crash involved a teen driver and crash was speed related mutually exclusive? How can you tell?

Not mutually exclusive. Look at the yellow cell.

No Teen Driver(s)

Teen Driver(s)

Total

Not Speed Related

120,082 24,291 144,373

Speed Related

67,331 19,755 87,086

Total 187,413 44,046 231,459


Use numbers from the cells of the table to compute the probability that a randomly selected crash involved a teen driver or was speed related.

No Teen Driver(s)

Teen Driver(s)

Total

Not Speed Related

120,082 24,291 144,373

Speed Related

67,331 19,755 87,086

Total 187,413 44,046 231,459


Use numbers from the cells of the table to compute the probability that a randomly selected crash involved a teen driver or was speed related.

No Teen Driver(s)

Teen Driver(s)

Total

Not Speed Related

120,082 24,291 144,373

Speed Related

67,331 19,755 87,086

Total 187,413 44,046 231,459

P(teen or speed) =24,291+19, 755 + 67, 331

231, 459=111, 377231, 459

≈0.4812


Now use two of the marginal totals and one number from a cell of the table to compute the probability that a randomly selected crash involved a teen driver or was speed related.

No Teen Driver(s)

Teen Driver(s)

Total

Not Speed Related

120,082 24,291 144,373

Speed Related

67,331 19,755 87,086

Total 187,413 44,046 231,459


Now use two of the marginal totals and one number from a cell of the table to compute the probability that a randomly selected crash involved a teen driver or was speed related.

No Teen Driver(s)

Teen Driver(s)

Total

Not Speed Related

120,082 24,291 144,373

Speed Related

67,331 19,755 87,086

Total 187,413 44,046 231,459

P(teen or speed) =44,046 + 87,086 −19, 775

231, 459=111, 377231, 459

≈0.4812


P20. Use the Addition Rule to compute the probability that if you roll two six-sided dice,

a. You get doubles or a sum of 4

b. You get doubles or a sum of 7

c. You get s 5 on the first die or you get a 5 on the second die


P(doubles or 4) =P(doubles) + P(sum of 4)−P((2,2))

=636

+336

−136

=836

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12


P(doubles or 7) =P(doubles) + P(sum of 7)

=636

+636

=1236

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12


P(5 on first or 5 on second) =P((5,x)) + P((x, 5))−P(5,5)

=636

+636

−136

=1136

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12


P21. Use the Addition Rule to compute the probability that if you flip two fair coins, you get heads on the first coin or you get heads on the second coin.


P21. Use the Addition Rule to compute the probability that if you flip two fair coins, you get heads on the first coin or you get heads on the second coin.

P((H , x) or (x,H )) =P((H ,x)) + P((x,H )) - P((H ,H ))

=12+12−14=34


P22. Use the Addition Rule to find the probability that if you roll a pair of dice, you do not get doubles or you get a sum of 8.



Doubles: {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Sum of 8: {(2,6),(3,5),(4,4),(5,3),(6,2)}



Doubles: {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Sum of 8: {(2,6),(3,5),(4,4),(5,3),(6,2)}

P((no doubles) or (8))

=P(no doubles) + P(8) - P(no doubles and (8))

=3036

+536

−436


E30. Display 5.36 shows the U.S. college population in 2006 categorized by age and sex.

Age Male EnrollmentFemale

EnrollmentTotal

19 and under 1,629 2,033 3,662

20 to 24 3,129 3,724 6,853

25 to 29 982 1,181 2,163

30 and older 1,577 2,383 3,960

Total 7,317 9,321 16,638


What percentage of college students are female? Female and age 30 or older? Female or age 30 or older?

Age Male Enrollment Female Enrollment Total

19 and under 1,629 2.033 3,662

20 to 24 3,129 3,724 6,853

25 to 29 982 1,181 2,163

30 and older 1,577 2,383 3,960

Total 7,317 9,321 16,638


What percentage of college students are female? Female and age 30 or older? Female or age 30 or older?


19 and under 1,629 2.033 3,662

20 to 24 3,129 3,724 6,853

25 to 29 982 1,181 2,163

30 and older 1,577 2,383 3,960

Total 7,317 9,321 16,638

P(F) =932116638

=0.56

P(F ∩30+) =238316638

=0.143

P(F ∪30+) =9321+ 3960 −2383

16638=0.655


What percentage of college students are male? Male and under age 30? Male or under age 30?


19 and under 1,629 2.033 3,662

20 to 24 3,129 3,724 6,853

25 to 29 982 1,181 2,163

30 and older 1,577 2,383 3,960

Total 7,317 9,321 16,638


What percentage of college students are male? Male and under age 30? Male or under age 30?


19 and under 1,629 2.033 3,662

20 to 24 3,129 3,724 6,853

25 to 29 982 1,181 2,163

30 and older 1,577 2,383 3,960

Total 7,317 9,321 16,638

P(M ) =731716638

=0.44

P(M∩< 30) =(7317 −1577)

16638=0.345

P(M∪< 30) =7317 + (16638 −3960)−(7317 −1577)

16638=0.857


What proportion of female college students are age 30 or older?

What proportion of college students age 30 or older are female?


19 and under 1,629 2.033 3,662

20 to 24 3,129 3,724 6,853

25 to 29 982 1,181 2,163

30 and older 1,577 2,383 3,960

Total 7,317 9,321 16,638


What proportion of female college students are age 30 or older?

What proportion of college students age 30 or older are female?


19 and under 1,629 2.033 3,662

20 to 24 3,129 3,724 6,853

25 to 29 982 1,181 2,163

30 and older 1,577 2,383 3,960

Total 7,317 9,321 16,638

P(30+ |F ) =23839431

=0.256

P(F |30+) =23833960

=0.602


E32. Display 5.38 classifies crashes in a recent year in Virginia.

a. Fill in the missing cells

Driver violated a traffic law

Driver didn’t violate a traffic law

Total

Fatality 521 837

No Fatality

Total 145,288 153,907



a. Fill in the missing cells



Total

Fatality 521 316 837

No Fatality 144,767 8,303 153,070

Total 145,288 8,619 153,907



b. What proportion of crashes involved a fatality and a traffic law violation?



Total


No Fatality 144,767 8,303 153,070

Total 145,288 8,619 153,907



b. What proportion of crashes involved a fatality and a traffic law violation?



Total


No Fatality 144,767 8,303 153,070

Total 145,288 8,619 153,907

P( fatality and violation) =521

153907=0.0034



c. What proportion of crashes involved a fatality or a traffic law violation?



Total


No Fatality 144,767 8,303 153,070

Total 145,288 8,619 153,907



c. What proportion of crashes involved a fatality or a traffic law violation?



Total


No Fatality 144,767 8,303 153,070

Total 145,288 8,619 153,907

P( fatality or violation) =837 +145288 −521

153907=0.946


E34. In a local school, 80% of the students carry a backpack (B) or a wallet (W). Forty percent carry only a backpack, and 30% carry only a wallet.

If a student is selected at random, find the probability that the student carries both a backpack and a wallet. (Note that the word only is important here.)




Yes No Total

Yes

No




Wallet

Yes No Total

Backpack Yes 40

No 30

Total 100




Wallet

Yes No Total

Backpack Yes 10 40

No 30

Total 100




Wallet

Yes No Total

Backpack Yes 10 40 50

No 30 20 50

Total 40 60 100




P(B or W ) =0.800P(B or W) =P(B) + P(W)−P(B and W)P(B or W) =P(B only) + P(W only) + P(B and W)

P(B and W) =P(B or W)−P(B only)−P(W only)=0.800 −0.400 −0.300=0.100


E36. Polls of registered voters often report the percentage of Democrats and the percentage of Republicans who approve of the job the president is doing. Suppose that in a poll of 1500 randomly selected voters 860 are Democrats and 640 are Republicans. Overall, 937 approve of the job the president is doing and 449 of these are Republicans.

Assuming the people in the poll are representative, what percentage of registered voters are Republicans or approve of the job the president is doing? First answer this question by using the addition rule. Then make a two-way table showing the situation.


E36. Polls of registered voters often report the percentage of Democrats and the percentage of Republicans who approve of the job the president is doing. Suppose that in a poll of 1500 randomly selected voters 860 are Democrats and 640 are Republicans. Overall, 937 approve of the job the

president is doing and 449 of these are Republicans.

Assuming the people in the poll are representative, what percentage of registered voters are Republicans or approve of the job the president is doing?

P(R or A) =P(R) + P(A)−P(R and A)

=6401500

+9371500

−4491500

=11281500

=0.752





Republican Democrat Total

Approve 449 937

Disapprove

Total 640 860 1500





Republican Democrat Total

Approve 449 488 937

Disapprove

191 372 563

Total 640 860 1500

P(A or R) =191+ 449 + 488

1500=0.752


E38. Jill computes the probability that she gets heads exactly once in two flips of a fair coin:

She defends her use of the addition rule because HT and TH are mutually exclusive. What would you say to her?

P(exactly one head) =P(tails on first and heads on second or heads on first and tails on second)=P(tails on first and heads on second) + P(heads on first and tails on second)

=14+14=12


E38. Jill computes the probability that she gets heads exactly once in two flips of a fair coin:

She defends her use of the addition rule because HT and TH are mutually exclusive. What would you say to her?

P(exactly one head) =P(tails on first and heads on second or heads on first and tails on second)=P(tails on first and heads on second) + P(heads on first and tails on second)

=14+14=12

{HH ,HT ,TH ,TT} ⇒ P(HT or TH ) =24=12


E40. Suppose events A, B, and C are three events where

P(A and B) ≠ 0, P(A and C) ≠ 0, P(B and C) ≠ 0, and

P(A and B and C) ≠ 0.

Draw a Venn diagram to illustrate this situation. Use the Venn diagram to write a rule for computing P(A or B or C)


E40. Suppose events A, B, and C are three events where

P(A and B) ≠ 0, P(A and C) ≠ 0, P(B and C) ≠ 0, and

P(A and B and C) ≠ 0.

Draw a Venn diagram to illustrate this situation. Use the Venn diagram to write a rule for computing P(A or B or C)

C

BA

C

BA

76 5

4 2

3

1


E40. Use the Venn diagram to write a rule for computing

P(A or B or C)

C

BA

76 5

4 2

3

1

P(A) =P(1) + P(4) + P(6) + P(7)P(B) =P(2) + P(4) + P(5) + P(7)P(C) =P(3) + P(5) + P(6) + P(7)P(A and B) =P(4) + P(7)P(A and C) =P(6) + P(7)P(B and C) =P(5) + P(7)P(A and B and C) =P(7)

P(A or B or C) =P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7)=P(A) + P(B) + P(C)−P(A and B)−P(A and C)−P(B and C)+P(A and B and C)

section 5.3 - the addition rule and disjoint events

Documents

addition rule

fresh water

salt water

disjoint eventsp14

disjoint eventsd16

disjoint eventsd17

categorizing people

great lakes