stats chap 5
hat does it mean for an event to be unusual? Why should the cutoff for identifying unusual events not always be 0.05?
An event is unusual if it has a low probability of occurring. The choice of a cutoff should consider the context of the problem.
Describe what an unusual event is. Should the same cutoff always be used to identify unusual events?
An event is unusual if it has a low probability of occurring. The same cutoff should not always be used to identify unusual events. Selecting a cutoff is subjective and should take into account the consequences of incorrectly identifying an event as unusual.
E: A randomly selected person finding cheese revolting. F: A different randomly selected person finding cheese delicious.
E cannot affect F and vice versa because the people were randomly selected, so the events are independent.
In probability, a(n) experiment is any process that can be repeated in which the results are uncertain.
In probability, an experiment is any process with uncertain results that can be repeated. The result of any single trial of the experiment is not known ahead of time. However, the results of the experiment over many trials produce regular patterns that enable one to predict with remarkable accuracy.
If events E and F are disjoint and the events F and G are disjoint, must the events E and G necessarily be disjoint?
No, events E and G are not necessarily disjoint. For example, E={0,1,2}, F={3,4,5}, and Gequals{2,6,7} show that E and F are disjoint events, F and G are disjoint events, and E and G are events that are not disjoint.
When a probability is based on an empirical experiment, a probability of zero does not mean that the event cannot occur. The probability of an event E is approximately the number of times event E is observed divided by the number of repetitions of the experiment, as shown below. Just because the event is not observed, does not mean that the event is impossible.
P(E)= relative freq of E = Freq of E/ # of trails of experiment
A survey of 500 randomly selected high school students determined that 458 play organized sports. (a) What is the probability that a randomly selected high school student plays organized sports? (b) Interpret this probability.
STEPS 1= empirical method 2. p(e)= F(e) / # of trails of experiemnt 3. F ( E) = 458 number of trails = 500 4. 458/500= 0.916 part 2. If 1,000 high school students were sampled, it would be expected that about 916 of them play organized sports 0.916 * 1000= 916
(c) E: The consumer demand for synthetic diamonds. F: The amount of research funding for diamond synthesis.
The consumer demand for synthetic diamonds could affect the amount of research funding for diamond synthesis, so E and F are dependent.
Bob is asked to construct a probability model for rolling a pair of fair dice. He lists the outcomes as 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. Because there are 11 outcomes, he reasoned, the probability of rolling a two must be 1/11 . What is wrong with Bob's reasoning?
The experiment does not have equally likely outcomes.
Which of the following numbers could be the probability of an event? 0.26, 1.53, -0.5, 0.05, 1, 0
The numbers that could be a probability of an event are 0.26, , 0.05, 1, 0 .... 1.53 is not the answer becuase its greater then 1 and -0.5 is less then 0
What is the probability of obtaining two heads in a row when flipping a coin? Interpret this probability.
The probability of obtaining two heads in a row when flipping a coin is ? steps = 1 prob of geting a head p(s) = total sum is heads and tails which is 2 so 1/2 * 2 heads in a row = .25 Consider the event of a coin being flipped two times. If that event is repeated ten thousand different times, it is expected that the event would result in two heads about ? multipy given answer .25 * 10000 = 2500
According to a certain country's department of education, 42.3% of 3-year-olds are enrolled in day care. What is the probability that a randomly selected 3-year-old is enrolled in day care?
The probability that a randomly selected 3-year-old is enrolled in day care is ? STEP 1: 42.3 devide by 100 answer is 0.423
5.2/13 A golf ball is selected at random from a golf bag. If the golf bag contains 8 type A balls, 7 type B balls, and 6 type C balls, find the probability that the golf ball is not a type A ball.
The probability that the golf ball is not a type A ball is ? if its not type A is P(E^c) =1- P P=n/m n=8 type A m= 21 sum 8/21=.381 P(E^c) is 1-.381=.619
If a person spins a six dash space spinner and then flips a coin, describe the sample space of possible outcomes using 1, 2 , 3 ,4 , 5 ,6 for the spinner outcomes and Upper H comma Upper T for the coin outcomes. (Make sure your answers reflect the order stated.)
The sample space is S= __ ? 1. The sample space of the given probability experiment is the collection of all possible outcomes. 2. The possible outcomes are each possible spinner outcome followed by each possible coin outcome because every possible combination could be an outcome. 3. the sample space is all possible outocmes so = H1, H2, H3, H4, H5, H6, T1,T2,T3, T4, T5, T6
Let the sample space be Upper S = 1, 2 ,3 ,4 , 5 ,6 ,7 ,8 ,9 , 10 EndSet. Suppose the outcomes are equally likely. Compute the probability of the event ="an odd number less than 8."
p(e)= steps= use classical method formula = p(e) = n(e) / n(s) N(e) = odd number less then 8 so ( 1,3,5,7) so n(e) = 4 digits n(s) = 10 digits all possible outocmes SO P(E) = 0.4
Suppose you toss a coin 100 times and get 76 heads and 24 tails. Based on these results, what is the probability that the next flip results in a head?
steps for this Q. 1. when a probability experiment is run, probabilities are approximated using the empirical approach. 2. P(E)= relatiev FREQ of E= freq of e/ # of trails of experiment 3. Let event E be the result of flipping a head. What is the frequency of E? head is 76 4. How many trials of the experiment were run? 100 5. 76/ 100 = 0.76
Why is the following not a probability model?
This is not a probability model because at least one probability is less than 0.
Probability is a measure of the likelihood of a random phenomenon or chance behavior.
True
True or False: In a probability model, the sum of the probabilities of all outcomes must equal 1.
True. In a probability model, the sum of the probabilities of all outcomes must equal 1.
Two events E and F are
Two events E and F are independent if the occurrence of event E in a probability experiment does not affect the probability of event F.
Explain the Law of Large Numbers. How does this law apply to gambling casinos
As the number of repetitions of a probability experiment increases, the proportion with which a certain outcome is observed gets closer to the probability of the outcome. This applies to casinos because they are able to make a profit in the long run because they have a small statistical advantage in each game.
When two events are disjoint, they are also independent.
False- Two events are disjoint if they have no outcomes in common. In other words, the events are disjoint if, knowing that one of the events occurs, we know the other event did not occur. Independence means that one event occurring does not affect the probability of the other event occurring. Therefore, knowing two events are disjoint means that the events are not independent.
Describe the difference between classical and empirical probability.
The empirical method obtains an approximate probability of an even by conducting a probability experiment. The probability is approximate because different runs of the probability experiment lead to different outcomes, and, therefore, different estimates of the probability. The classical method of computing probabilities relies on counting techniques, and requires equally likely outcomes. An experiment has equally likely outcomes when each outcome has the same probability of occurring.
5.2/7 Find the probability of the indicated event if P(E)=0.30 and P(F)= 0.50, P(E and F)= 0.20.
formula = P( E or F)= P(E) + P(F) - P(E &F) 0.30+0.50-.20= .60
5.2/11 If P(E)=0.60, P(E or F)=0.65, P(E and F)=0.20, find P(F).
formula = P(e or f) = P(e) + P(f)- P(e and F) 0.65= 0.60 + p(f) - .20 0.65 -0.60+0.20= P(f) =.25
5.2/10 Find the probability P(E c) if P(E)=0.29
formula is P(E^c) = 1 - 0.29 =.71
Event
A(n) event is any collection of outcomes from a probability experiment.
What is the probability of an event that is impossible?
0
The word (and) in probability implies that we use the
Multiplication rule P(E & F & G &..) = P(E) * P(F) * P( G)=
5.2- # 5 List the outcomes in E and G. Choose the correct answer below.
Sample space= (7,8,9,10,11,12,13,14,15,16,17,18) E=( 8,9,10,11) G= (13,14,15,16) since E and G are disjointed and #12 is missing. answer is A= E and G = ( ) and E an G is mutually exculisve becuase. Yes, because the events E and G have no outcomes in common.
The word or in probability implies
The word or in probability implies that we use the Addition Rule.
In a national survey college students were asked, "How often do you wear a seat belt when riding in a car driven by someone else?" The response frequencies appear in the table to the right. (a) Construct a probability model for seat-belt use by a passenger. (b) Would you consider it unusual to find a college student who never wears a seat belt when riding in a car driven by someone else?
step 1. add total of freq to get = 4737 step 2. divide response by total freq "never" 143/4737 step 3 round each to third number b. to find if unsual look at first answer ".030" a probality cant be less then 5% or 0.05 so yes its unsual.
According to a center for disease control, the probability that a randomly selected person has hearing problems is 0.157. The probability that a randomly selected person has vision problems is 0.091. 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.
5.2- #6 List the outcomes in Upper E(c) ? S( 10,11,12,13,14,15,16,17,18,19,20,21) e(12,13,14,15)
E ( c) = all numbers in S that arent in E 10,11,16,17,18,19,20,21 p(E c) = number of E(c) / number of S 8/12= 0.667
(a) E: A person living at least 70 years. F: The same person regularly handling venomous snakes.
E and F are dependent because regularly handling venomous snakes can affect the probability of a person living at least 70 years.
5.2/8 P(E)=0.35 and P(F)=0.40 and P(E or F) =0.60
Formula= P(E & F)= P (E)+P (F)-P(EorF) 0.30+0.40- 0.60=.15
If E and F are disjoint events, then
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)
If the addition rule for disjoint events P(E or F)= P(E) +P(f) is used to compute the probability, the outcomes that E and F have in common will be double counted, and so P(Eor F) must be if E and F are not disjoint. Hence, the general addition rule is used, which states that for any two events E and F, P(E or F ) = P(E) + P(F) - P (E and F)
E & F are independt so P(e)= 0.7 P(f) = 0.9 what is P(e & F)
P (E &F) = 0.7 * 0.9= 0.63 MULTIPLY E AND F
Let the sample space be Upper S equals Start Set = 1, 2, 3 , 4, 5 , 6 , 7 , 8 , 9, 10 EndSet. Suppose the outcomes are equally likely. Compute the probability of the event Upper E= 3, 5 , 9.
P( E) = steps 1. use Classical method 2. p(e)= # of ways E occurs / # of possible outcomes = M/N 3. p(E) = N(E) / N(S) N( s) = all outcomes which is 10 n( E) = NUMBER OF OUTCOMES OF E ( 3,5,9 ) so 3 digits 3/10 = 0.3
In a certain card game, the probability that a player is dealt a particular hand is 0.34. Explain what this probability means. If you play this card game 100 times, will you be dealt this hand exactly 34 times? Why or why not?
The probability 0.34 means that approximately 34 out of every 100 dealt hands will be that particular hand. No, you will not be dealt this hand exactly 34 times since the probability refers to what is expected in the long-term, not short-term.
5.2/9 Find the probability P(E or F) if E and F are mutually exclusive, P(E)=0.25, and P(F)=0.45
if E and F are mutally exculsive it will be 0 so formuala is P (e or f) = P(e)+P(f)-P(e and f) .25+.45-0=.7
Suppose that a probability is approximated to be zero based on empirical results. Does this mean that the event is impossible?
no
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. Complete parts a) through c) below.
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. -ss comma SS 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? Interpret this probability. The probability is 0.25. This means that there is a 25% 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)? The probability .50 so 50% chane of offsprinng being a carrier
In a certain game of chance, a wheel consists of 58 slots numbered 00, 0, 1, 2,..., 56. To play the game, a metal ball is spun around the wheel and is allowed to fall into one of the numbered slots. Complete parts (a) through (c) below.
a) Determine the sample space. Choose the correct answer below. A. The sample space is {00, 0, 1, 2,..., 56}. b. The number of outcomes in E is N(E)equals= 1 N(s) = 58 1/58 =0.0172 c. If the wheel is spun 1,000 times, it is expected that about 17 of those times result in the ball landing in slot 7. 1000* 0.0172 d. Find the number of outcomes in E, N(E), that is, the number of odd slots. Note that 0 and 00 are not odd slots. The number of outcomes in E is N(E)= 28 N(s)= 58 28/58=0.4828 e. Interpret the probability 0.4828. Select the correct choice below and fill in the answer box within your choice. (Type a whole number.) A. If the wheel is spun 100 times, it is expected that about 48 of those times result in the ball landing on an odd number.
A bag of 100 tulip bulbs purchased from a nursery contains 20 red tulip bulbs, 40 yellow tulip bulbs, and 40 purple tulip bulbs. (a) What is the probability that a randomly selected tulip bulb is red? (b) What is the probability that a randomly selected tulip bulb is purple? (c) Interpret these two probabilities.
a. 20 red tulips divide by 100= 0.2 b. 40 purple divide 100= 0.4 c. If 100 tulip bulbs were sampled with replacement, one would expect about 20 of the bulbs to be red and about 40 of the bulbs to be purple. basically solve by 0.2 * 100 & 0.4 *100 to find answer c
A golf ball is selected at random from a golf bag. If the golf bag contains 7 orange balls, 4 red balls, and 13 yellow balls, find the probability of the following event.
a. The probability that the golf ball is orange or red formula= p( Red +orange)= P(ORNAGE) + P(RED) total sum of balls = 24 P(red) = 4/24 = .167 p(orange)= 7/24= .292 .167+.292= 0.459
