Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
45. a. Let M = event that a putt is made
P(M) = 983,764/1,613,234 = .6098
Note: The probability that a putt is missed is
P(MC)=1-P(M)=1-.6098 =.3902
b. Let A= event that a PGA Tour player has a par putt
( ) .640P A M =
( ) .203
C
P A M =
c. Let B = event that a PGA Tour player has a birdie putt
( ) .188P B M =
( ) .734
C
P B M =
d. These probabilities indicate that there is a much higher probability of making a par putt than making
a birdie putt. One reason is that par putts have an extra stroke than birdie putts. Because of the extra
stroke, par putts tend to be closer to the hole than birdie putts and there is thus a higher probability of
making a par putt. For example, the average length of all putts made is slightly less than 5 feet while
the average length of birdie putts made is 21 feet.
enough to have a very high probability of making a par.
46. a. 422 + 181 + 80 + 121 + 201 = 1005 respondents
b. Most frequent response a day or less; Probability = 422/1005 = .4199
47. a. (2)(2) = 4
b. Let S = successful
U = unsuccessful
c. O = {E1, E2}
M = {E1, E3}
48. a. There are a total of 1364 responses. Dividing each entry by 1364 provides the following joint
probability table.
A
B
Total
Female
.2896
.2133
.5029
Male
.2368
.2603
.4971
Total
.5264
.4736
1.0000
b. The marginal probability of a female is .5029 from above.
c. Let A = uses social media and other websites to voice opinions about television programs
F = female respondent
P(F)=.2896
.5029 =.5758
d.
P(FÇA)=.2896
from the joint probability table
P(F)P(A)=(.5029)(.5264) =.2647
Oi
l
Bond
s
S
U
S
U
S
U
E1
E2
E3
E4
P(FÇA)¹P(F)P(A)
49. Let I = treatment-caused injury
D = death from injury
N = injury caused by negligence
M = malpractice claim filed
$ = payment made in claim
We are given P(I) = .04, P(N | I) = .25, P(D | I) = 1/7, P(M | N) = 1/7.5 = .1333,
and P($ | M) = .50
50. a. Probability of the event = P(average) + P(above average) + P(excellent)
=
11
50
14
50
13
50
+ +
= .22 + .28 + .26 = .76
P(Under 25) = .2000
e. This is a conditional probability.
(Under 25 BD) .0378
(Under 25 BD) .1121
(BD) .3371
P
PP
= = =
f. No.
(100 or More HS) .1061P=
is not equal to P(100 or More) = .2508.
Household income is not independent of education level. Individuals with a high school degree have
More Than One
Age Group
Yes
No
Total
23 and Under
.1026
.0996
.2022
24 – 26
.1482
.1878
.3360
27 – 30
.0917
.1328
.2245
31 – 25
.0327
.0956
.1283
36 and Over
.0253
.0837
.1090
Total
.4005
.5995
1.0000
b. Marginal probability .2022
c. .2245 + .1283 + .1090 = .4618
d. Marginal probability .4005
53. a. P(24 to 26 | Yes) = .1482/.4005 = .3700
b. P(Yes | 36 and over) = .0253/.1090 = .2321
c. .1026 + .1482 + .1878 + .0917 + .0327 + .0253 = .5883
b.
P(30 -49)=.2273+.0907 =.3180
P(Okay 30-49) =P(OkayÇ30-49)
P(30-49) =.0907
.3180 =.2852
c.
P(50 +Not Okay) =P(50+ÇNot Okay)
P(Not Okay) =.4008
.7766 =.5161
P50+
( )
.4731 =.8472
PNot Okay|18-29
( )
=PNot OkayÇ18-29
( )
P18-29
( )
=.1485
.2089 =.7109
There is a higher probability the 50+ year olds will not be okay with this practice.
55. a.
(B ) .12
(B S) .30
(S) .40
PS
PP
= = =
We have P(B | S) > P(B).
Yes, continue the ad since it increases the probability of a purchase.
b. Estimate the company’s market share at 20%. Continuing the advertisement should increase the
market share since P(B | S) = .30.
c.
(B ) .10
(B S) .333
(S) .30
PS
PP
= = =
c. P(A B) = 10/800 = .0125
d. P(A | B) = P(A B)/P(B) = .0125/.125 = .10
Given: P(B) = .06, P(A) = .05, P(A | B) = .15
a. P(A B) = P(A | B)P(B) = .15(.06) = .009
b. P(A B) = P(A) + P(B) - P(A B) = .06 + .05 - .009 = .101 or 10.1%
M = male student
P(A1) = .095
P(A2) =
1-P(A
1)
= 1
-
.095 = .905
P(F | A1) = .60
P(F | A2) = .49
Pspam|shipping!
( )
=Pspam
( )
P shipping!|spam
( )
Pspam
( )
P shipping!|spam
( )
+Pham
( )
P shipping!|ham
( )
=.10
( )
.051
( )
.10
( )
.051
( )
+.90
( )
.0015
( )
=.7907
Pham|shipping!
( )
=Pham
( )
P shipping!|ham
( )
Pham
( )
P shipping!|ham
( )
+Pspam
( )
P shipping!|ham
( )
=.90
( )
.0015
( )
.90
( )
.0015
( )
+.10
( )
.051
( )
=.2093
.10
( )
.045
( )
+.90
( )
.0022
( )
Pspam|here!
( )
=Pspam
( )
P here!|spam
( )
Pspam
( )
P here!|spam
( )
+Pham
( )
P here!|ham
( )
=.10
( )
.034
( )
.10
( )
.034
( )
+.90
( )
.0022
( )
=.6320
A message that includes the word today! is more likely to be spam. This is because P(today!|spam)
is larger than P(here!|spam). Because today! occurs more often in unwanted messages (spam), it is
easier to distinguish spam from ham in messages that include today!.
c.
Pspam|available!
( )
=Pspam
( )
P available!|spam
( )
Pspam
( )
P available!|spam
( )
+Pham
( )
P available!|ham
( )
=.10
( )
.014
( )
.10
( )
.014
( )
+.90
( )
.0041
( )
=.2750
Pspam|fingertips!
( )
=Pspam
( )
P fingertips!|spam
( )
Pspam
( )
P fingertips!|spam
( )
+Pham
( )
P fingertips!|ham
( )
=.10
( )
.014
( )
.10
( )
.014
( )
+.90
( )
.0011
( )
=.5858
A message that includes the word fingertips! is more likely to be spam.
d. It is easier to distinguish spam from ham when a word occurs more often in unwanted messages
