chapter 5 questions
P(E and F) = .1 P(E) = .2 What is P(F I E)
(.1/.2) = .5
suppose events E and F are independent P(E)= .3 and P(F)=.6 what is the P(E and F)
(.3)(.6)=.18
when you want: at least 1 will not/negative/won't
1-(all will/positive)
A test to determine whether a certain antibody is present is 99.4% effective. This means that the test will accurately come back negative if the antibody is not present (in the test subject) 99.4% of the time. The probability of a test coming back positive when the antibody is not present (a false positive) is 0.006. Suppose the test is given to four randomly selected people who do not have the antibody. a) P(all 4 are negative) b)P(at least one positive)
a) (.994)^4 = .976 b) 1-(.994^4) =.0238
Died from it DIDN'T Never smoked 859 116171 Former smoker 72 5019 Current smoker 141 8394 a) P(died from cancer) b) P(current cigar smoker) c) P(died from cancer and current cigar smoker) d) P(died from cancer or current cigar smoker)
a) (1072/130656) = .008 b) (.065/130656) = .065 c) (141/130656) = .001 d) (1072/130656) + (8394/130656) = .072
A flush in the card game of poker occurs if a player gets five cards that are all the same suit (clubs, diamonds, hearts, or spades). Complete parts (a) and (b) to obtain the probability of being dealt a flush in five cards. a) Initially concentrate on one suit, say clubs. There are 13 clubs in a deck. Compute P(five clubs)=P(first card is clubs and second card is clubs and third card is clubs and fourth card is clubs and fifth card is clubs). b) A flush can occur if a player receives five clubs or five diamonds or five hearts or five spades. Compute P(five clubs or five diamonds or five hearts or five spades). Note that the events are mutually exclusive.
a) (13/52)(12/51)(11/50)(10/49)(9/48) = .000495 b) (.000495*4)=.00198 1-.00198=.002
A baseball player hit 54 home runs in a season. Of the 54 home runs, 18 went to right field, 18 went to right center field, 8 went to center field, 8 went to left center field, and 2 went to left field. (a)What is the probability that a randomly selected home run was hit to right field? (b)What is the probability that a randomly selected home run was hit to left field? (c) Was it unusual for this player to hit a home run to left field? Explain.
a) (18/54) = .333 b) (2/54) = .037 c) yes because P(left field) < .05
Fresh Soph Junior Sr Total Satisfied 55 55 62 58 230 Neutral 28 10 19 12 69 Not Sat 22 21 19 24 86 Total 105 86 100 94 385 a) P(satisfied) b) P(junior) c) P(satisfied and junior) d) P(satisfied or junior)
a) (230/385) = .597 b) (100/385) = .260 c) (62/385) = .161 d) (230/385)+(100/385)-(62/385) = .696
Suppose that a single card is selected from a standard 52-card deck. a) What is the probability that the card drawn is a queen? b) Now suppose that a single card is drawn from a standard 52-card deck, but it is told that the card is court (jack, queen, or king). What is the probability that the card drawn is a queen?
a) (4/52) = .077 b) (4/12) =.333
a) What is the probability that the first card is a king and the second card is a king if the sampling is done without replacement? (b) What is the probability that the first card is a king and the second card is a king if the sampling is done with replacement?
a) (4/54)(3/53) =.005 b) (4/52)(4/52) = .006
a bag of 100 tulip bulbs purchased from a nursery contains 25 red tulip bulbs, 25 yellow tulip bulbs, and 50 purple tulip bulbs. a) probability that a selected tulip is red? b) probability that a selected tulip is purple? c) interpret these two probabilities
a) .25 b) .5 c) if 100 bulbs were sampled with replacement, one would expect about 25 of the bulbs to be red and about 50 of the bulbs to be purple
a survey of 500 randomly selected high school students determined that 253 play organized sports. a) What is the probability that a randomly selected high school student plays organized sports? b) interpret this probability.
a) 253/500 = .506 b) if 1,000 high school students were sampled, it would be expected that about 506 of them play organized sports
A probability experiment is conducted in which the sample space of the experiment is S={5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}, event E={6, 7, 8, 9} and event G={12, 13, 14, 15}. Assume that each outcome is equally likely. (A) List the outcomes in E and G. (B) Are E and G mutually exclusive?
a) E and G = { } b) Yes, because the events E and G have no outcomes in common
A probability experiment is conducted in which the sample space of the experiment is S={5,6,7,8,9,10,11,12,13,14,15,16}. Let event E={7,8,9,10}. Assume each outcome is equally likely. a) List the outcomes in Ec. b) Find P(Ec).
a) Ec= {5,6,11,12,13,14,15,16} b) (8/12) = .667
According to a center for disease control, the probability that a randomly selected person has hearing problems is 0.148. The probability that a randomly selected person has vision problems is 0.089. Can we compute the probability of randomly selecting a person who has hearing problems or vision problems by adding these probabilities? Why or why not?
No, because hearing and vision problems are not mutually exclusive. So, some people have both hearing and vision problems. These people would be included twice in the probability
P(E and F) = ? P(E) = .8 P(F I E) = .4
(.4 * .8) =.32
Find the probability of the indicated event if P(E)=0.35 and P(F)=0.45. Find P(E or F) if P(E and F)=0.15
(.45+.35)-.15 =.65
The probability of obtaining ten tails in a row when flipping a coin is
(.5)^10 = .00098
A golf ball is selected at random from a golf bag. If the golf bag contains 7 type A balls, 8 type B balls, and 9 type C balls, find the probability that the golf ball is NOT a type A ball.
(17/24) = .708
A golf ball is selected at random from a golf bag. If the golf bag contains 6 orange balls, 9 green balls, and 6 black balls, find the probability of the following event the ball is orange or green.
(6/21) + (9/21) = .714
let the sample space be S=(1,2,3,4,5,6,7,8,9,10) compute the probability of the event (3,5,7,8)
4/10 = .4
P(FIE) =
P(E and F)/E
If E and F are disjoint events, then P(E or F)=
P(E) + P(F)
If E and F are not disjoint events, then P(E or F)=
P(E) + P(F) - P(E AND F)
probability of rolling a dice S={1,2,3,4,5,6} E={2,4,6} P(E)?
P(E)= 1/2
If P(E)=0.40, P(E or F)=0.50, and P(E and F)=0.15, find P(F)
P(F)= .25 .5 = .4 + x - .15
probability is a measure of the likelihood of a random phenomenon or chance behavior true or false
true
the sum of the probabilities of all outcomes must equal 1 true or false
true
was x was y =
y/x
A bag of 30 tulip bulbs contains 11 red tulip bulbs, 12 yellow tulip bulbs, and 7 purple tulip bulbs. Suppose two tulip bulbs are randomly selected without replacement from the bag. prop one is red and one is yellow (no order)
.152 * 2 =.304
In a recent survey, it was found that the median income of families in country A was $57,500. What is the probability that a randomly selected family has an income greater than $57,500?
.5 (because it is the median)
when you want: at least 1 will be postive/will
1-(all will not/negative/won't)
notion P(F I E) means the probability of event ? given event ?
F given event E
For the month of June in a certain city, 48% of the days are cloudy. Also in the month of June in the same city, 33% of the days are cloudy and snowy. What is the probability that a randomly selected day in June will be snowy if it is cloudy?
F=cloudy (.48) F and E = both .33 EIF=? .33/.48 =.688
Never 135 Rarely 348 Sometimes 501 Most of the time 1167 Always 2322 a) make probability table b) would you consider it to be unusual to find a student who never wears a seatbelt?
Never .030 Rarely .078 Sometimes .112 Most of the time .261 Always .519 b) Yes because P(never) < .05
About 14% of the population of a large country is math phobic. a) If two people are randomly selected, what is the probability both are math phobic? b) What is the probability at least one is math phobic? Assume the events are independent.
a) (.14)(.14) = .0196 b) 1-(.86^2)=.2604
Among 43-to 48-year-olds, 25% say they have operated heavy machinery while under the influence of alcohol. Suppose six 43-to 48-year-olds are selected at random. a) all six have operated b) at least one has NOT c) NONE have d) at least one HAS
a) (.25^6)= .0002 b) 1-.002= .9998 c) (.75^6)= .1780 d) 1-.1780= .822
Players in sports are said to have "hot streaks" and "cold streaks." For example, a batter in baseball might be considered to be in a slump, or cold streak, if that player has made 10 outs in 10 consecutive at-bats. Suppose that a hitter successfully reaches base 38% of the time he comes to the plate. The hitter makes an out 62% of the time. a) P(hitter makes 10 consecutive outs)
a) (.62)^10 = .00839
The probability that a randomly selected 1-year-old male garter snake will live to be 2 years old is 0.98787. (a) What is the probability that two randomly selected 1-year-old male garter snakes will live to be 2 years old? (b) What is the probability that eight randomly selected 1-year-old male garter snakes will live to be 2 years old? (c) What is the probability that at least one of eight randomly selected 1-year-old male garter snakes will not live to be 2 years old? Would it be unusual if at least one of eight randomly selected 1-year-old male garter snakes did not live to be 2 years old?
a) (.98787)^2 = .97589 b) (.98787)^8 = .90698 c) 1-.90689 = .09302 no because it is greater than .05
A gene is composed of two alleles. An allele can be either dominant or recessive. Suppose that a husband and wife, who are both carriers of the sickle-cell anemia allele but do not have the disease, decide to have a child. Because both parents are carriers of the disease, each has one dominant normal-cell allele (S) and one recessive sickle-cell allele (s). Therefore, the genotype of each parent is Ss. Each parent contributes one allele to his or her offspring with each allele being equally likely. a) Genes are always written with the dominant gene first. Therefore, there are two instances the offspring could have genotype Ss (one if the mother contributes the dominant allele and the father contributes the non-dominant allele; and one if the father contributes the dominant allele and the mother contributes the non-dominant allele). List the other two possible genotypes of the offspring. b) What is the probability that the offspring will have sickle-cell anemia? In other words, what is the probability that the offspring will have genotype ss? c) What is the probability that the offspring will not have sickle-cell anemia but will be a carrier (one normal-cell allele and one sickle-cell allele)?
a) SS, ss b) (1/4) = .25 c) (2/4) =.5
Red .35 Green .1 Blue .15 Brown .1 Yellow 0 Orange .25 is this a probability model? what do we call the outcome yellow?
a) no because the probabilities do not sum to 1 b) impossible event
classical, empirical, or subjunctive On the basis of a survey of 1000 families with six children, the probability of a family having six girls is 0.0064
empirical
subjunctive
based on personal judgement
Suppose you just purchased a digital music player and have put 12 tracks on it. After listening to them you decide that you like 4 of the songs. With the random feature on your player, each of the 12 songs is played once in random order. Find the probability that among the first two songs played Can not be replayed a) like both b) like neither c) like exactly 1 can be replayed a) like both b) like neither c) like exactly 1
cannot be replayed a) (4/12)(3/11)= .091 b) (8/12)(7/11)= .424 c) 1-(.091)-(.424)=.485 can be replayed a) (4/12)(4/12) =.111 b) (8/12)(8/12)= .444 c) 1-(.111)-(.444) =.445
classical, empirical, or subjunctive The probability of having six girls in an six-child family is 0.015625
classical
classical, empirical, or subjunctive On the basis of clinical trials, the probability of efficacy of a new drug is 0.85.
empirical
unusal event =
less than .05
classical
number of ways E can occur divided by number of possible outcomes
empirical
probability of an event E is the approx number of times it is observed
classical, empirical, or subjunctive According to a sports analyst, the probability that a football team will win the next game is 0.49
subjunctive