Download Math 165-01 Chapter13HWAnswer

MATH 165 - Ch. 13 – Homework
Name (Print)
Student ID #
Independent and the Multiplication Rule
(1) Lost Internet sites Internet sites often vanish or move, so that references to them
can’t be followed. In fact, 13 % of Internet sites referenced in major scientific journals are
lost within two years after publication. If a paper contains seven Internet references,
what is the probability that all seven are still good two years later? What specific
assumptions did you make to calculate this probability?
P(all seven are still good two years later)= [P(one still good two years later)]7
= (1 − 0.13)7
= (0.87)7
= 0.377254794
≈ 0.3773
The General Addition Rule
(2) Photo and Video Sharing Photos and videos have become an important part of the
online social experience, with more than half of Internet users posting photos or videos
online that they have taken themselves. Let A be the event an Internet user posts photos
that they have taken themselves, and B be the event an Internet user posts videos that
they have taken themselves. Pew Research Center finds that P(A) = 0.52, P(B) = 0.26,
and P(A or B) = 0.54.
(a) Make a Venn diagram similar to Figure 13.4 showing the events {A and B},
{A and not B}, {B and not A}, and {neither A nor B}.
(b) Describe each of these events in words.
(c) Find the probabilities of all four events and add the probabilities to your Venn diagram.
P(A or B) = P(A) + P(B) − P(A and B)
= 0.52 + 0.26 − P(A and B)
= 0.54
P(A and B) = P(A) + P(B) − P(A or B)
P(A and B) = 0.52 + 0.26 − 0.54 = P(a user who posts both)
P(A ∩ B) = 0.24
P(A and not B) = 0.52 − 0.24 = P(a user who posts photos but not videos)
P(A ∩ not B) = 0.28
P(B and not A) = 0.26 − 0.24 = P(a user who posts videos but not photos)
P(B ∩ not A) = 0.02
P(neither A nor B) = 1 − 0.54 = P(a user who posts neither photos nor videos)
P(neither A nor B) = 0.46
Online
Videos
No Videos
Total
—————————————————————————–
Photos
0.24
0.28
0.52
No Photos
0.02
0.46
0.48
—————————————————————————Total
0.26
0.74
1
Conditional Probability
(3) Photo and Video Sharing In the setting of the question # 2, what is the conditional
probability that an Internet user posts photos that they have taken themselves, given
that they post video that they have taken themselves?
P (A ∩ B) = 0.24
P (a user who posts phot| a user who posts video) =
=
0.24
0.26
= 0.923076923
≈ 0.9231
P (a user who posts both)
P (a user who posts video)
(4) Computer games Here is the distribution of computer games sold by type of game:
Game type
Probability
——————————————————————————————————————–
Strategy
0.354
Role playing
0.139
Family Entertain.
Shooters
Children’s
0.127
0.109
0.057
Other
0.214
What is the conditional probability that a computer game is a role-playing game,
given that it is not a strategy game?
P (B | A) =
P (A and B)
P (A)
P (role playing game | not strategy game) =
=
P (not strategy and role playing)
P (not strategy)
0.139
1 − 0.354
= 0.215170278
≈ 0.2152
Check Your Skills
(5) An instant lottery game gives you probability 0.02 of winning on any one play. Plays
are independent of each other. If you play three times, the probability that you win on
none of your plays is about
P(no winning on three lotteries) = [P(no winning on each lottery)]3
= (1 − 0.02)3
= (0.98)3
= 0.941192
≈ 0.9412
(6) The probability that you win on one or more of your three plays of the game in previous
exercise is about
P(winning on one or more of three lotteries) = 1− P(no winning on three lotteries)
= 1−0.941192
(from the previous exercise)
= 0.058808
≈ 0.0588
(7) An athlete suspected of having used steroids is given two tests that operate independently of each other. Test A has probability 0.9 of being positive if steroids have been
used. Test B has probability 0.8 of being positive if steroids have been used. What is
the probability that at least one test is positive if steroids have been used?
P(at least one of them is positive) = 1− P(both negative)
P(both negative) = (1 − 0.9) · (1 − 0.8) = 0.02
P(at least one of them is positive) = 1− P(both negative)
= 1 − 0.02
= 0.98
(8) What is the distribution of doctorates conferred by field and sex? Here are the counts
from the most popular fields in 2012. The physical sciences include mathematics and
computer and information sciences; the life sciences include agricultural sciences/natural
resources, biological/biomedical sciences, and health sciences; and the social sciences
include psychology.
Degree
Male
Female
Total
——————————————————————————————————————–
Engineering
6, 527
1, 883
8, 410
Physical sciences
6, 393
2, 551
8, 944
Life sciences
Social sciences
Education
5, 331
3, 488
1, 501
6, 698
4, 861
3, 297
12, 029
8, 349
4, 798
Humanities
2, 654
2, 847
5, 501
Other
1, 496
1, 425
2, 921
———————————————————————————————–
Total
27, 390
23, 562
50, 952
(a) Choose a doctoral recipient at random from this group. The probability that
the recipient is male is about
P(the recipient is male) =
27, 390
= 0.537564766 ≈ 0.5376
50, 952
(b) The conditional probability that the recipient is male, given that the degree is
in education, is about
P(the recipient is male | in Education) =
=
P (Eduation and male)
P (Eduation)
1, 501
= 0.312838682 ≈ 0.3128
4, 798
(c) The conditional probability that the degree is in education, given that the recipient is
female, is about
P(the recipient is in Education | female) =
=
P (f emale and Eduation)
P (f emale)
3, 297
= 0.139928698 ≈ 0.1399
23, 562
(d) Let A be the event that the degree is in engineering and B the event that the recipient
is a female. The proportion of engineering doctorates conferred on females is expressed
in probability notation as
A: The degree is in Engineering
B: The recipient is a female
P (The proportion of engineering doctorates conferred on female)
=
P (Engineering and female)
P (Engineering)
=
P (A ∩ B)
P (A)
= P (B|A)
Answer is (c).
(9) Choose an American adult at random. The probability that you choose a woman is 0.52.
The probability that the person you choose has never married is 0.25. The probability
that you choose a woman who has never married is 0.11. The probability that the person
you choose is either a woman or never married (or both) is therefore about
Adult
Never Married
Married
Total
—————————————————————————–
Woman
0.11
0.41
0.52
Man
0.14
0.34
0.48
—————————————————————————Total
0.25
0.75
1
P (woman or never married)
= P (woman) + P (never married ) − P (woman and never married)
= 0.52 + 0.25 − 0.11
= 0.66
(10) Universal blood donors People with type O-negative blood are referred to as universal donors, although if you give type O-negative blood to any patient you run the
risk of a transfusion reaction due to certain antibodies present in the blood. However,
any patient can receive a transfusion of O-negative red blood cells. Only 7.2 % of the
American population have O-negative blood. If 10 people appear at random to give
blood, what is the probability that at least one of them is a universal donor?
P(at least one of them is a universal donor) = 1− P(All 10 people are not universal donor)
P(the person is not universal donor) = 1− P(the person is a universal donor)
= 1 − 0.072 = 0.928
P(at least one of them is a universal donor) = 1− P(All 10 people are not universal donor)
= 1 − (0.928)10
= 1 − 0.473674234
= 0.526325765
≈ 0.5263