3.14 Two fair dice are tossed, and the up face on each die is MI recorded. a. List the 36 sample points contained in the sample space. b. Assign probabilities to the sample points. c. Find the...

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3.14 Two fair dice are tossed, and the up face on each die is MI recorded.
a. List the 36 sample points contained in the sample space. b. Assign probabilities to the sample points. c. Find the probability of observing each of the following
3.21 Fungi in beech forest trees. Bcechwood forests in East Cen-tral Europe are being threatened by dynamic changes in land ownership and economic upheaval. The current status of the beech tree species in this area was evaluated by Hun-garian university professors in Applied Ecology and Envi-ronmental Research (Vol. 1, 2003). Of 188 beech trees surveyed, 49 had been damaged by fungi. Depending on the species of fungus, damage will occur on either the trunk, branches, or leaves of the tree. In the damaged trees, the trunk was affected 85% of the time, the leaves W% of the time, and the branches 5% of the time. a. Give a reasonable estimate of the probability of a beech tree in East Central Europe being damaged by fungi. b. A fungus-damaged beech tree is selected at random, and the area (trunk, leaf, or branch) affected is observed. List /aka
the sample points for this experiment, and assign a rea-sonable probability to each one.
3.30 Matching socks. Consider the following question posed to Marilyn vos Savant in her weekly newspaper column, "Ask Marilyn":
1 have two pairs of argyle socks, and they look nearly identi-cal—one navy blue and the other black. [When doing the laundry] my wife matches the socks incorrectly much more often than she does correctly.... 1 f all four socks are in front Of her, it seems to me that her chances are 50% for a wrong match and 50% for a right match. What do you think? Source: Parade Magazine, Feb. 27,1994.
3.40 Suppose P(A) = .4, P(B) = .7 and P(A fl B) = .3. Find the following probabilities: a. P(B") b. P(A`) c. P(AU B)
144 Consider the following Venn diagram, where /1( = .10, P(E2) = .05, P(E3) = P(E4) = .2, = .06, P(E6) = .3, P(E7) = .06, and Pa:0 = .03:
Federal civil trial appeals. Lie Journal of the American taw and tconontieN Association (Vol. 1. 2()01) p111,11,,hcd the results of a study of appeals of federal tis it I i h?k. I he accompanying t able, extracted twin the article, gives a breakilimn of 2,143 civil cases that \N ere appealed by ei-ther the plaintiff or the delendant.1 he outcome of the ap-peal, as well as the type of trial (judge or jury), was determined for each civil case. Suppose one of the 2,143 cases is selected at random and both the outcome of the appeal and the type of trial are observed.
Plaintiff trial win —Jury Judge Totals reversed 194 71 265 Plaintiff trial win —affirmed/dismissed 429 240 669 Defendant trial win—reversed 111 68 179 Defendant trial win—affirmed/ dismissed 731 299 1,030 Totals 1,465 678 2,143
a. List the sample points for this experiment. b. Find P(A), where A = {jury trial}. c. Find P(B), where B = {plaintiff trial win is reversed}.
d. Are A and B mutually exclusive events? e. Find P(A`). 1. Find P(A U B). g. Find I'(A fl B).
1.
Learning the Mechanics 3.63 For two events A and B, P(A) = .4, P(B) = .2, 1111 P(A n B) = .1. a. Find P(AIB). b. Fiid P(BIA). c. Are A and B independent events? 3.64 For two events A and B, P(A) = .4, P(B) = /'(AfB)=.6. a. Lind P(A B). b. find P(BIA).
3.66
An experiment results in one of three mutually exclusive events A, B, and C. It is known that P(A) = .30, P( B) = .55, and P(C) = .15, Find each of the following probabilities: a. P(A U B) b. P(A n B)) c. P(AIB) d. P(B U C)) e. Are B and C independent events? Explain.
3.-4 National Firearms Sur?e%. Refer to the Harvard School Of IlcaIth studx of pm atek held firearm stock in the United Stales. presented in Exercise 2.? (p. 34). Recall that in a rePresentatne household telephone survey of 2.770 adults. ...!0% reported that the owned at least one gun. (Injury Prevention. Jan. 2007.) The accompanying pie chart summarizes the types of firearms owned by those who own at least one gun. Suppose I of the 2.770 adults surveyed is randonth selected. a. What is the probability that the adult owns at least one gun? b. Gixen that the adult does own at least one gun. what is the prohahilit that the adult owns a revolver? c. What is the prohabilit? that the adult owns at least one gun and the gun is a handgun?
3.84 Cigar smoking and cancer. The Journal of the National Can-cer Institute (Feb. 16, 2000) published the results of a study that investigated the association between cigar smoking and death from tobacco-related cancers. Data were obtained for a national sample of 137,243 American men. The results are summarized in the accompanying table. Each male in the study was classified according to his cigar-smoking status and whether or not he died from a tobacco-related cancer. a. Find the probability that a randomly selected man never smoked cigars and died from cancer. b. Find the probability that a randomly selected man was a former cigar smoker and died from cancer. c. Find the probability that a randomly selected man was a current cigar smoker and died from cancer. d. Given that a male was a current cigar smoker, find the probability that he died from cancer. e. Given that a male never smoked cigars, find the proba-bility that he died from cancer.
Died from Cancer
Cigars Yes No Totals Never Smoked 782 120,747 121,529 Former Smoker 91 7,757 7,848 Current Smoker 141 7,725 7,866 Totals 1,014 136,229 137,243
Source': Shapiro. J. A.. Jacobs. E. J.. and Thum M. J. "Cigar smoking in men and risk of death from tobacco-related cancers." Journal of the National Cancer Institute, Vol. 92. No. 4. Feb. lb. 2(XX) (Table 2.1.
events: A: {A 3 appears on each of the two dice.} B: {The sum of the numbers is even.} C: The sum of the numbers is equal to 7. } D: {A 5 appears on at least one of the dice. } E: {The sum of the numbers is 10 or more.}
Answered Same DayDec 22, 2021

Answer To: 3.14 Two fair dice are tossed, and the up face on each die is MI recorded. a. List the 36 sample...

David answered on Dec 22 2021
116 Votes
3.14)
a.) The 36 sample points are given below:
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)
(3,1),(3,2),(3,3),(3,4),(3,5),(3
,6)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
b.) Note that, for any pair (x,y) {x=1,2,3,4,5,6 and y=1,2,3,4,5,6}
P(x appears)=1/6 and P(y appears)=1/6
Hence P((x,y) appears)=P(x appears) P(y appears) [ since they are independent]
=1/36
Thus we probability of each of the pairs occurring is 1/36.
c.)
(A) The only favorable case is (3,3) whose probability is 1/36.
(B) The favorable cases are
(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(4,6),(5,1),(5,3),(5,5),(6,2),(6,4),(6,6)
Each has probability 1/36 and hence total probability is the sum of them which is 18/36=1/2.
(C) The favorable cases are
(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)
Each has probability 1/36 and hence total probability is the sum of them which is 6/36=1/6.
(D) The favorable cases are
(1,5),(2,5),(3,5),(4,5),(5,1), (5,2),(5,3),(5,4),(5,5),(5,6),(6,5)
Each has probability 1/36 and hence total probability is the sum of them which is 11/36.
(E) The favorable cases are
(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)
Each has probability 1/36 and hence total probability is the sum of them which is 6/36=1/6.
3.21)
a.) Out of 188 beech trees surveyed, 49 had been damaged by fungi.
Hence the probability of a beech tree in East Central Europe being damaged by fungi
=49/188=0.26
b.) The Probability that trunk of the selected tree would be damaged by fungi =
...
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