section 5.3 - the addition rule and disjoint events
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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?. Section 5.3 - The Addition Rule and Disjoint Events. - PowerPoint PPT PresentationTRANSCRIPT
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)
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and 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
Section 5.3 - The Addition Rule and 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)If A and B are mutually exclusive, P(A and B) =0.P(A or B) =P(A) + P(B)
Mutually exclusive
Section 5.3 - The Addition Rule and Disjoint Events
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?
Section 5.3 - The Addition Rule and Disjoint Events
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?
Section 5.3 - The Addition Rule and Disjoint Events
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?
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.
Section 5.3 - The Addition Rule and Disjoint Events
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.
b. Suppose you randomly select a person from among those who fish. Can you find the probability that the person fishes in salt water?
Section 5.3 - The Addition Rule and Disjoint Events
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.
b. Suppose you randomly select a person from among those who fish. Can you find the probability that the person fishes in salt water?
Yes:
P( fish in salt water) =905134071
=0.266
Section 5.3 - The Addition Rule and Disjoint Events
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.
c. Can you find the probability that the person fishes in fresh water?
Section 5.3 - The Addition Rule and Disjoint Events
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.
c. Can you find the probability that the person fishes in fresh water?
You can’t find the probability that the person fishes in fresh water, since the “Great Lakes” and “other fresh water” categories overlap.
Section 5.3 - The Addition Rule and Disjoint Events
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.
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?
Section 5.3 - The Addition Rule and Disjoint Events
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.
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 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
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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
Not supposed to receive payments 992
Section 5.3 - The Addition Rule and Disjoint Events
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)
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
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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
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
Section 5.3 - The Addition Rule and Disjoint Events
P16. If you roll two dice, are these pairs of events mutually exclusive?
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
Section 5.3 - The Addition Rule and Disjoint Events
P16. If you roll two dice, are these pairs of events mutually exclusive?
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
Section 5.3 - The Addition Rule and Disjoint Events
P16. If you roll two dice, are these pairs of events mutually exclusive?
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
Section 5.3 - The Addition Rule and Disjoint Events
P16. If you roll two dice, are these pairs of events mutually exclusive?
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
Section 5.3 - The Addition Rule and Disjoint Events
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.
Section 5.3 - The Addition Rule and Disjoint Events
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?
Section 5.3 - The Addition Rule and Disjoint Events
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?
Yes. A student can’t be a junior and a senior at the same time.
Section 5.3 - The Addition Rule and Disjoint Events
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.
b. Find the probability that the student will be a freshman or a sophomore.
Section 5.3 - The Addition Rule and Disjoint Events
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.
b. Find the probability that the student will be a freshman or a sophomore.
P(F or S) =P(F ) + P(S)=0.33+ 0.27 =0.60
Section 5.3 - The Addition Rule and Disjoint Events
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?
Section 5.3 - The Addition Rule and Disjoint Events
P18. A tetrahedral die has the numbers 1, 2, 3, and 4 on its four faces. Suppose you roll a pair of tetrahedral dice.
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
Section 5.3 - The Addition Rule and Disjoint Events
P18. A tetrahedral die has the numbers 1, 2, 3, and 4 on its four faces. Suppose you roll a pair of tetrahedral dice.
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
Section 5.3 - The Addition Rule and Disjoint Events
P18. A tetrahedral die has the numbers 1, 2, 3, and 4 on its four faces. Suppose you roll a pair of tetrahedral dice.
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
Section 5.3 - The Addition Rule and Disjoint Events
P18. A tetrahedral die has the numbers 1, 2, 3, and 4 on its four faces. Suppose you roll a pair of tetrahedral dice.
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
Section 5.3 - The Addition Rule and Disjoint Events
P18. A tetrahedral die has the numbers 1, 2, 3, and 4 on its four faces. Suppose you roll a pair of tetrahedral dice.
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
Section 5.3 - The Addition Rule and Disjoint Events
P18. A tetrahedral die has the numbers 1, 2, 3, and 4 on its four faces. Suppose you roll a pair of tetrahedral dice.
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
Section 5.3 - The Addition Rule and Disjoint Events
P18. A tetrahedral die has the numbers 1, 2, 3, and 4 on its four faces. Suppose you roll a pair of tetrahedral dice.
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)
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
P19. Display 5.34 gives information about all reportable crashes on state-maintained roads in North Carolina in a recent year.
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
Section 5.3 - The Addition Rule and Disjoint Events
P19. Display 5.34 gives information about all reportable crashes on state-maintained roads in North Carolina in a recent year.
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
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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.
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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.
Section 5.3 - The Addition Rule and Disjoint Events
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)}
Section 5.3 - The Addition Rule and Disjoint Events
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)}
P((no doubles) or (8))
=P(no doubles) + P(8) - P(no doubles and (8))
=3036
+536
−436
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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
P(F) =932116638
=0.56
P(F ∩30+) =238316638
=0.143
P(F ∪30+) =9321+ 3960 −2383
16638=0.655
Section 5.3 - The Addition Rule and Disjoint Events
What percentage of college students are male? Male and under age 30? Male or under age 30?
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
Section 5.3 - The Addition Rule and Disjoint Events
What percentage of college students are male? Male and under age 30? Male or under age 30?
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
P(M ) =731716638
=0.44
P(M∩< 30) =(7317 −1577)
16638=0.345
P(M∪< 30) =7317 + (16638 −3960)−(7317 −1577)
16638=0.857
Section 5.3 - The Addition Rule and Disjoint Events
What proportion of female college students are age 30 or older?
What proportion of college students age 30 or older are female?
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
Section 5.3 - The Addition Rule and Disjoint Events
What proportion of female college students are age 30 or older?
What proportion of college students age 30 or older are female?
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
P(30+ |F ) =23839431
=0.256
P(F |30+) =23833960
=0.602
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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 521 316 837
No Fatality 144,767 8,303 153,070
Total 145,288 8,619 153,907
Section 5.3 - The Addition Rule and Disjoint Events
E32. Display 5.38 classifies crashes in a recent year in Virginia.
b. What proportion of crashes involved a fatality and a traffic law violation?
Driver violated a traffic law
Driver didn’t violate a traffic law
Total
Fatality 521 316 837
No Fatality 144,767 8,303 153,070
Total 145,288 8,619 153,907
Section 5.3 - The Addition Rule and Disjoint Events
E32. Display 5.38 classifies crashes in a recent year in Virginia.
b. What proportion of crashes involved a fatality and a traffic law violation?
Driver violated a traffic law
Driver didn’t violate a traffic law
Total
Fatality 521 316 837
No Fatality 144,767 8,303 153,070
Total 145,288 8,619 153,907
P( fatality and violation) =521
153907=0.0034
Section 5.3 - The Addition Rule and Disjoint Events
E32. Display 5.38 classifies crashes in a recent year in Virginia.
c. What proportion of crashes involved a fatality or a traffic law violation?
Driver violated a traffic law
Driver didn’t violate a traffic law
Total
Fatality 521 316 837
No Fatality 144,767 8,303 153,070
Total 145,288 8,619 153,907
Section 5.3 - The Addition Rule and Disjoint Events
E32. Display 5.38 classifies crashes in a recent year in Virginia.
c. What proportion of crashes involved a fatality or a traffic law violation?
Driver violated a traffic law
Driver didn’t violate a traffic law
Total
Fatality 521 316 837
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
Section 5.3 - The Addition Rule and Disjoint Events
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.)
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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.)
Wallet
Yes No Total
Backpack Yes 40
No 30
Total 100
Section 5.3 - The Addition Rule and Disjoint Events
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.)
Wallet
Yes No Total
Backpack Yes 10 40
No 30
Total 100
Section 5.3 - The Addition Rule and Disjoint Events
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.)
Wallet
Yes No Total
Backpack Yes 10 40 50
No 30 20 50
Total 40 60 100
Section 5.3 - The Addition Rule and Disjoint Events
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.)
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
Section 5.3 - The Addition Rule and Disjoint Events
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.
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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?
Republican Democrat Total
Approve 449 937
Disapprove
Total 640 860 1500
Section 5.3 - The Addition Rule and Disjoint Events
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?
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
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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)
Section 5.3 - The Addition Rule and Disjoint Events
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
Section 5.3 - The Addition Rule and Disjoint Events
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)