A survey asked students about their weight and displayed results by sex. Of 129 female students, 87 thought they were about right weight, 39 overweight, and 3 underweight. Of 83 male students, 64 thought they were about right, 3 overweight, and 16 underweight. The probabilities of various outcomes are calculated from this data, and it is determined that ideas about weight and sex are not independent.

1. SexSex About rightAbout right OverweightOverweight UnderweightUnderweight TotalTotal
FemaleFemale 87 39 3 129
MaleMale 64 3 16 83
TotalTotal 151 42 19 212
In a class survey students were asked, “Regarding your weight, do you think youIn a class survey students were asked, “Regarding your weight, do you think you
are About right? Overweight? Underweight?” The following table displays theare About right? Overweight? Underweight?” The following table displays the
results by sex.results by sex.
In a class survey students were asked, “Regarding your weight, do you think youIn a class survey students were asked, “Regarding your weight, do you think you
are About right? Overweight? Underweight?” The following table displays theare About right? Overweight? Underweight?” The following table displays the
results by sex.results by sex.
If a person from this class is chosen at random, what is the probability that
the person:
A.thinks they are overweight?
B.is a female?
C.is a female who thinks she is overweight?
D.is a male who thinks he is overweight?
E.Thinks she is overweight if we know the chosen person is a female?
F.is a female or thinks they are overweight?
G.Are ideas about weight and sex independent?

2. SexSex About rightAbout right OverweightOverweight UnderweightUnderweight TotalTotal
FemaleFemale 87 39 3 129
MaleMale 64 3 16 83
TotalTotal 151 42 19 212
In a class survey students were asked, “Regarding your weight, do you think youIn a class survey students were asked, “Regarding your weight, do you think you
are About right? Overweight? Underweight?” The following table displays theare About right? Overweight? Underweight?” The following table displays the
results by sex.results by sex.
In a class survey students were asked, “Regarding your weight, do you think youIn a class survey students were asked, “Regarding your weight, do you think you
are About right? Overweight? Underweight?” The following table displays theare About right? Overweight? Underweight?” The following table displays the
results by sex.results by sex.
If a person from this class is chosen at random, what is the probability that the
person:
A.thinks they are overweight? P(O)=
B.is a female? P(F)=
C.is a female who thinks she is overweight? P(F∩O)=
D.is a male who thinks he is overweight? P(M∩O)=
E.Thinks she is overweight if we know the chosen person is a female? P(O/F)=
F.is a female or thinks they are overweight? P(F∪O)=
G.Are ideas about weight and sex independent?
18.0
212
39
=
20.0
212
42
=
61.0
212
129
=
01.0
212
3
=
30.0
129
39
=
62.0
212
132
=
No because P(F∩O)≠P(F)·P(O)

3. A population of children were classified according to sex and left-handednessA population of children were classified according to sex and left-handedness
A child was chosen at random from this population. Use the table to findA child was chosen at random from this population. Use the table to find
the following probabilitiesthe following probabilities
P(L)P(L)
P(R)P(R)
P(M)P(M)
P(F)P(F)
P(L|M)P(L|M)
P(L|F)P(L|F)
P(L and F)P(L and F)
Are sex and left handedness dependent?Are sex and left handedness dependent?
A child was chosen at random from this population. Use the table to findA child was chosen at random from this population. Use the table to find
the following probabilitiesthe following probabilities
P(L)P(L)
P(R)P(R)
P(M)P(M)
P(F)P(F)
P(L|M)P(L|M)
P(L|F)P(L|F)
P(L and F)P(L and F)
Are sex and left handedness dependent?Are sex and left handedness dependent?
Male (M)Male (M) Female (F)Female (F) TotalTotal
Left-handed (L)Left-handed (L) 200200 100100 300300
Right-handed (R)Right-handed (R) 18001800 900900 27002700
TotalTotal 20002000 10001000 30003000

4. A population of children were classified according to sex and left-handednessA population of children were classified according to sex and left-handedness
A child was chosen at random from this population. Use the table to findA child was chosen at random from this population. Use the table to find
the following probabilitiesthe following probabilities
P(L) = 0.1P(L) = 0.1
P(R) = 0.9P(R) = 0.9
P(M) = 0.67P(M) = 0.67
P(F)= 0.33P(F)= 0.33
P(L|M)= 0.1P(L|M)= 0.1
P(L|F)= 0.1P(L|F)= 0.1
P(L and F) = 0.03P(L and F) = 0.03
Are sex and left handedness dependent? Yes! Since P(L)=P(L/M)Are sex and left handedness dependent? Yes! Since P(L)=P(L/M)
A child was chosen at random from this population. Use the table to findA child was chosen at random from this population. Use the table to find
the following probabilitiesthe following probabilities
P(L) = 0.1P(L) = 0.1
P(R) = 0.9P(R) = 0.9
P(M) = 0.67P(M) = 0.67
P(F)= 0.33P(F)= 0.33
P(L|M)= 0.1P(L|M)= 0.1
P(L|F)= 0.1P(L|F)= 0.1
P(L and F) = 0.03P(L and F) = 0.03
Are sex and left handedness dependent? Yes! Since P(L)=P(L/M)Are sex and left handedness dependent? Yes! Since P(L)=P(L/M)
Male (M)Male (M) Female (F)Female (F) TotalTotal
Left-handed (L)Left-handed (L) 200200 100100 300300
Right-handed (R)Right-handed (R) 18001800 900900 27002700
TotalTotal 20002000 10001000 30003000

5. The following table shows the employment status of differentThe following table shows the employment status of different
tradesmen in a branch of a Building Trades Uniontradesmen in a branch of a Building Trades Union
The following table shows the employment status of differentThe following table shows the employment status of different
tradesmen in a branch of a Building Trades Uniontradesmen in a branch of a Building Trades Union
A tradesman is chosen at random from this branch. Find theA tradesman is chosen at random from this branch. Find the
probability that he is:probability that he is:
A.employed
B.a bricklayer
C.an employed bricklayer
D.an unemployed plumber
E.employed, given that he is a carpenter
F.a plumber, given that he is unemployed
G.a carpenter or a bricklayer
H.a plumber or unemployed
I.Decide whether trade and employment status are independent
A tradesman is chosen at random from this branch. Find theA tradesman is chosen at random from this branch. Find the
probability that he is:probability that he is:
A.employed
B.a bricklayer
C.an employed bricklayer
D.an unemployed plumber
E.employed, given that he is a carpenter
F.a plumber, given that he is unemployed
G.a carpenter or a bricklayer
H.a plumber or unemployed
I.Decide whether trade and employment status are independent
CarpenterCarpenter PlumberPlumber BricklayerBricklayer TotalTotal
EmployedEmployed 81 48 34 163
UnemployedUnemployed 56 42 49 147
TotalTotal 137 90 83 310

6. The following table shows the employment status of differentThe following table shows the employment status of different
tradesmen in a branch of a Building Trades Uniontradesmen in a branch of a Building Trades Union
The following table shows the employment status of differentThe following table shows the employment status of different
tradesmen in a branch of a Building Trades Uniontradesmen in a branch of a Building Trades Union
A tradesman is chosen at random from this branch. Find the probabilityA tradesman is chosen at random from this branch. Find the probability
that he is:that he is:
A.Employed P(E)=
B.a bricklayer P(B)=
C.an employed bricklayer P(E∩B)=
D.an unemployed plumber P(U∩P)=
E.employed, given that he is a carpenter P(E/C)=
F.a plumber, given that he is unemployed P(P/U)=
G.a carpenter or a bricklayer P(C∪B) =
H.a plumber or unemployed P(P∪U)=
I.Decide whether trade and employment status are independent. Dependent since P(E)≠P(E/C)
A tradesman is chosen at random from this branch. Find the probabilityA tradesman is chosen at random from this branch. Find the probability
that he is:that he is:
A.Employed P(E)=
B.a bricklayer P(B)=
C.an employed bricklayer P(E∩B)=
D.an unemployed plumber P(U∩P)=
E.employed, given that he is a carpenter P(E/C)=
F.a plumber, given that he is unemployed P(P/U)=
G.a carpenter or a bricklayer P(C∪B) =
H.a plumber or unemployed P(P∪U)=
I.Decide whether trade and employment status are independent. Dependent since P(E)≠P(E/C)
CarpenterCarpenter PlumberPlumber BricklayerBricklayer TotalTotal
EmployedEmployed 81 48 34 163
UnemployedUnemployed 56 42 49 147
TotalTotal 137 90 83 310
53.0
310
163
=
27.0
310
83
=
11.0
310
34
=
14.0
310
42
=
59.0
137
81
=
29.0
147
42
=
71.0
310
220
=
63.0
310
195
=

7. The emergency room of a hospital has two back-up generators, either of which canThe emergency room of a hospital has two back-up generators, either of which can
supply enough electricity for basic hospital operations. If we define events A and B assupply enough electricity for basic hospital operations. If we define events A and B as
follows:follows:
Event A:Event A: Generator 1 works properlyGenerator 1 works properly
Event B:Event B: Generator 2 works properlyGenerator 2 works properly
Describe the following events in words:
a) Ᾱ
b) B/A
c) A or B
If A and B are independent events and P(A) = 0.25 and P(B) = 0.60, find:If A and B are independent events and P(A) = 0.25 and P(B) = 0.60, find:
a. P(A ∩B)
b. P(A|B)
c. P(A ∪B)
d. P(Ᾱ ∩B)

8. The emergency room of a hospital has two back-up generators, either of which canThe emergency room of a hospital has two back-up generators, either of which can
supply enough electricity for basic hospital operations. If we define events A and B assupply enough electricity for basic hospital operations. If we define events A and B as
follows:follows:
Event A:Event A: Generator 1 works properlyGenerator 1 works properly
Event B:Event B: Generator 2 works properlyGenerator 2 works properly
Describe the following events in words:
a) Ᾱ Generator 1 does not work
b) B/A Generator 2 works, as we know that generator 1 works
c) A or B At least one of the two back-up generators works properly
If A and B are independent events and P(A) = 0.25 and P(B) = 0.60, find:If A and B are independent events and P(A) = 0.25 and P(B) = 0.60, find:
a. P(A ∩B) =0.15
b. P(A|B) = 0.25
c. P(A ∪B) =0.7
d. P(Ᾱ ∩B) = 0.3