Section 6.7
This section is a matter of using the data that you are given to find a
probability…and a probability is simply the ratio of number of whatever answers
the question divided by the total.
Problem 173
“…probability of a randomly chosen visitor would like to return…” is found by
dividing how many like to return (in this case, “…887 say they would like to
return…”) by the total (in this case, “Of 947 visitors polled…”):
Answer: 887/947 or 0.9366 (as for what method…it is called relative frequency)
Problem 174
“…probability that a randomly chosen can of nuts will have no cashews.“ is found
by dividing how many have no cashews (in this case, “…420 can have any
cashews…”, so 500-420 = 80 cans have no cashews) by the total (in this case, “…a
sample of 500 cans…”):
Answer: 80/500 = 4/25 or 0.16
Problem 175
“…probability that a randomly chosen new model California car will not be able to
pass emission control standards.“ is found by dividing how many will not pass (in
this case, “…65 that would not meet emission…”) by the total (in this case, “…a
sample of 186 new model cars…”):
Answer: 65/186 or 0.3495
Problem 176
“…probability that the next on-schedule departure...“ is found by dividing how
many were on-schedule (in this case, “…1,789 of 2,154 buses arrived on time…”)
by the total (in this case, “…of 2,154…”):
Answer: 1789/2154 or 0.8305 (As for whether you would feel safe with the
answer, that is up to you…for me, 83% is a good percentage to feel safe with)
Problem 177
“…probability that the next the next machine of this type is defective...“ is found
by dividing how many were defective (in this case, “…14 were “D.O.A.” – dead
(defective)…”) by the total (in this case, “A sample of 88…”):
Answer: 14/88 = 7/44 or 0.1591
Problem 178 and 179
Let’s answer both of these at the same time by using a Venn Diagram.
Whenever we deal with percentages the total for the Venn Diagram is 1.00 or
100%; the other information is in our text: “…20% … have bad eyesight and 12%
have poor hearing.” So the diagram looks like:
H
E
20
12
This cannot be completed until we have the intersection, so the question must be
what is the smallest and largest numbers that we could place in the intersection?
The smallest is easy (in this case) ZERO (since probabilities have to be between 0
and 1); now the largest will be the smaller of the 2 totals given (in this case, 12),
because anything larger would require a negative number in what of the sections,
which is not allowed. With those intersections you can complete the Venn
Diagram:
E
H
12
12
0
20
0
12
8
20
68
80
Problem 178
“…largest percentage…bad eyesight or poor hearing…”, the “or” being a union of
the two circles…
So the answer is the largest group of numbers in the two circles: 12+0+20 = 32%
Problem 179
“…smallest percentage…bad eyesight or poor hearing…”…the “or” being a union
of the two circles…
So the answer is the smallest group of numbers in the two circles: 0+12+8 = 20%
Problems 180-183
Will use the above Venn Diagram but with a specific intersection of 6% because it
says “…6%…bad eyesight and poor hearing…”, the “and” being an intersection of
the two circles…giving us:
E
H
12
6
6
20
14
74
Problem 180
“…poor hearing or bad eyesight …”, the “or” being a union of the two circles…
Answer: 6+6+14 = 26%
Problem 181
“…bad eyesight but good hearing …”, the section that is in the circle (E) for bad
eyesight, but is outside the circle (H) for poor hearing (since it is good hearing that
we want)…
Answer: 14%
Problem 182
“…poor hearing but acceptable eyesight …”, the section that is in the circle (H) for
poor hearing, but is outside the circle (E) for bad eyesight (since it is acceptable
eyesight that we want)…
Answer: 6%
Problem 182
“…one or the other…but not both …”, this is within both circles, but does not
include the intersection (because the intersection includes BOTH)…
Answer: 6+14 = 20%
Problems 184-186
The mortality table is based off a starting total and decreases with time:
Age
20
30
40
#Living
92,857
87,641
81,239
Age
50
60
70
#Living
73,921
57,917
45,674
Age
80
90
100
#Living
18,783
985
0
Problem 184
“…probability that a 20-year –old will still be alive at age 60.”, at age 60 there are
57,917 alive and this must be divided by the total (or starting point), which in this
case is age 20, so 92,857…
Answer: 57917/92857 = 0.6237
Problem 185
“…probability that a 10-year –old will reach 80.”, at age 80 there are 18,783 alive
and this must be divided by the total (or starting point), which in this case is age
10, so 100,000…
Answer: 18783/100000 = 0.1878
Problem 186
“…probability that a 40-year –old will reach 70.”, at age 70 there are 45,674 alive
and this must be divided by the total (or starting point), which in this case is age
40, so 81,239…
Answer: 45674/81239 = 0.5622
Problems 187-191
To solve these it is best to start with solving the Venn Diagram or Table (we will
do both), with B for Basketball and F for Football:
Venn Diagram
F
Table
B
In B
Out B Total
In B
Out B Total
In F
Out F
Total
“A survey of 100…” is the Total:
Venn Diagram
F
Table
B
In F
Out F
Total
100
“…55 watch football…” is everything in the F circle:
100
Venn Diagram
55
Table
In B
B
F
In F
Out F
Total
Out B Total
55
100
100
“…41 watch basketball…” is everything in the B circle:
Venn Diagram
55
Table
In B
B
F
41
In F
Out F
Total
41
Out B Total
55
100
100
“…27 watch both…” is intersection (key word “both”):
Venn Diagram
55
Table
B
F
41
27
In F
Out F
Total
In B
27
41
Out B Total
55
100
100
Now to solve each…
The Table is solved by solving any row or column that has two of the three values
(order is up to you)… I would start with the Total column and row: 100-55 = 45
and 100-41 =59; then do the Football row and Basketball column: 55-27 = 28 and
41-27 = 14 and finally the “Out F” row: 45-14 = 31.
The Venn Diagram is solved by completing the numbers in each of the circles: for
F: 55-27 = 28 and for B” 41-27 = 14; then adding up all the segments and solving
so they add up to the total: 100 = 27+14+28+x = 69+x, so x = 31
This gives us the following to solve the following problems:
Venn Diagram
55
Table
B
F
41
28
27
14
31
In F
Out F
Total
In B
27
14
41
Out B
28
31
59
Total
55
45
100
100
Problem 187
“…only basketball.”, is the segment in the basketball (B) circle but does not
include the part where F overlaps it (because it is ONLY basketball, not both)…and
this must be divided by the total to get a probability:
Answer: 14/100 = 7/50 or 0.14
Problem 188
“…only football.”, is the segment in the football (F) circle but does not include the
part where B overlaps it (because it is ONLY football, not both)…and this must be
divided by the total to get a probability:
Answer: 28/100 = 7/25 or 0.28
Problem 189
“…basketball or football.”, is the union of the two circles because “or” stands for
union, so all the numbers in both circles…and this must be divided by the total to
get a probability:
Answer: (28+27+14)/100 = 69/100 or 0.69
Problem 190
“…basketball or football but not both.”, is the union of the two circles because
“or” stands for union, but excludes the intersection (“..not both…”, so all the
numbers in both circles, less the intersection…and this must be divided by the
total to get a probability:
Answer: (28+14)/100 = 42/100 = 21/50 or 0.42
Problem 191
“…at least one of the two sports.”; “at least one: means one or more, so in this
case, 1 or 2 sports, so this is the union of the two circles , so all the numbers in
both circles…and this must be divided by the total to get a probability:
Answer: (28+27+14)/100 = 69/100 or 0.69
Problems 192-195
To solve these it is best add a Totals column/row to the table:
Class Standing
Favor 4 Days
Do Not Favor Total
Freshmen
120
40
160
Sophomores
130
20
150
Juniors
95
65
160
Seniors
60
70
130
Total
405
195
600
And it is nice when your total matches “All 600 Anemia Tech undergraduates…”
Problem 192
“…freshman.”; total freshmen is 160…and this must be divided by the total (600)
to get a probability:
Answer: 160/600 = 4/15 or 0.2667
Problem 193
“…favors a four-day week.”; total favoring a 4-day week is 405…and this must be
divided by the total (600) to get a probability:
Answer: 405/600 = 27/40 or 0.6750
Problem 194
“…sophomore or does not favor...”; since “or” designates a union it is all the
sophomores and all those that do not favor a 4-day week (not counting the
intersection of those twice)…and this must be divided by the total (600) to get a
probability:
Answer: (130+20+40+65+70)/600 = 325/600 = 13/24 or 0.5417
Problem 195
“…junior and favors...”; since “and” designates an intersection, the intersection of
junior and favors a 4-day week is 95…and this must be divided by the total (600)
to get a probability:
Answer: 95/600 = 19/120 or 0.1583