2. Fundamentals

36

Consider the sample space Ω of pure outcomes for two dice rolls. How many elements are in Ω?

C; The numerator equals n(n−1)...(2)(1) while the denominator equals (n−k)(n−k−1)...(2)(1), causing all the lower terms through n−k to be cancelled out, leaving n(n−1)...(n−k+1).

For positive integers n, k with n>k, which of the following also expresses n!/(n−k)!? A) n^k B) n(n−1)...(n−k) C) n(n−1)...(n−k+1) D) n^(n−k)

8, 0

suppose X∼B(n,p), with n = 8 and p=1. What is the mean/standard deviation of this variable?

A, B, As discussed previously, (nk)0.5k(1−0.5)n−k is the probability that k of the tosses are heads and the remaining n−k are tails. This can be simplified in this case to (nk)0.5n.

Consider an unbiased coin with P(Heads)=0.5. What is the probability of exactly k heads in n tosses? (Select all that apply) A. (n \choose k)0.5^k(1−0.5)^{n−k} B. (n \choose k)0.5^n C. 0.5^n D. 0.5^k

D; In order to compute the PDF from a CDF of a continuous random variable, you simply take the derivative of the CDF.

Given the CDF of a continuous random variable, which of the following processes allows you to get the PDF of that random variable? A. You cannot recover the PDF knowing only the CDF B. Integrate from 0 to 1 C. Integrate over the relevant region D. Take the derivative of the CDF

1/36

If the die are fair, then each of the outcomes in Ω has what associated probability?

False; If A and B were disjoint, then we will have that P(A∪B)=P(A)+P(B). Here, we have that P(A)=1/3 and P(B)=1−P(BC)=1−1/4=3/4. Using this information, we have that P(A)+P(B)=1/3+3/4=13/12>1. Thus, the statement is false.

State whether the following statement is True or False: IfP(A)=1/3andP(BC)=1/4, A and B can be disjoint.

.8 ; To compute this probability we find the area of the specified region, and since this is a uniform distribution this corresponds to dividing the length of the specified range of values by the total length: 7−38−3 = 0.8.

Assume that you have a continuous random variable which is uniformly distributed in the range: [3,8]. What is the probability that the random variable takes on a value less than or equal to 7?

B; X−1(x) is the set of outcomes s in which X(s)=x. This means that the two sets must be disjoint, since if s was in both sets we would have x=X(s)=y, but x,y are distinct. They are not necessarily exhaustive, because there could be other possible values of X(s) other than x or y. They are only identical if they are both empty, but this is not always true.

Consider a discrete random variable X, and the set of outcomes in Ω. Then X−1(x) is the inverse function of X where X−1(x) is the set of outcomes s in which X(s)=x. Which of the following are always true about the two sets X−1(x) and X−1(y), both subsets of Ω, when x≠y? A. They are exhaustive B. They are disjoint C. They are identical

(n(n+1))/2 The number of pieces with different numbers is equal to n choose 2. Then, the total pieces with different numbers is n(n−1)2. There are n additional pieces with the same number. Thus, the total is: n+n(n−1)2=(n+1)n2.

In the game of dominoes, each piece is marked with two numbers and each piece is unique. The pieces are symmetrical so that the numbered pair is not ordered: this means that (2,6) = (6,2) and there is only one of such tile. The piece may have identical numbers as well, such as (1,1), (2,2), or (3,3). How many different pieces can be formed using the numbers 1, 2, ..., n?

B; The probability that the first toss is a head is 12, as is the probability that the second toss is a head. The probability that they are both heads are (12)2=14. Following the inclusion exclusion principle resolves the probability of at least one heads as 12+12−14=34

For this simple sample space problem, which of these probabilities captures the inclusion-exclusion principle statemen: tP(at least one head)=P(first toss is a head)+P(second toss is a head)−P(both coin tosses are heads) A. 2/3=1/2+1/2−1/3 B. 3/4=1/2+1/2−1/4 C. 1/2=1/4+1/4

A is correct, where the probability distribution depicted shows that rolling each of 1, 2, 3, 4, 5, or 6 are equally likely.

Suppose that you roll one six-sided, fair die once. Which of the following diagrams represents the associated probability function of observing each of the faces (1-2-3-4-5-6)?

False; This is false. The "success" and "failure" outcomes are not required to be equally likely. Note also that the Bernoulli trials must be independent for the distribution to be a binomial.

True or false: The binomial distribution describes the number of successes in n Bernoulli (binary outcome) trials, with the additional constraint that in each trial the probability of success has to be equal to the probability of failure.

Syphilis

A research experiment regarding which disorder prompted increased federal protection for human subjects?

B; All valid probability distribution functions integrate to 1 over their entire domain.

In the region selected in the above question what is true of the double integral of the joint distribution? A. It equals 0 B. It equals 1 C. It is undefined

A, C; The first statement is the so-called multiplicative criterion that defines independent events. The third statement can be derived from Bayes's law, P(A|B)=P(A∩B)/P(B) and substituting the multiplicative criterion.

Select all the true statements about independent events A and B. A. P(A∩B)=P(A)P(B) B. P(A∩B)=P(A)+P(B) C. P(A|B)=P(A) D. P(A|B)=P(B)

C correctly represents that there are several equally likely combinations that would each add up to the middle values. For example, a total of 7 could be achieved from combinations of 1+6, 2+5, or 4+3. In contrast, there are few combinations that would add up to the high and low values. For example, achieving a 2 requires both die to roll a 1 and achieving a 12 requires both die to roll 6.

Suppose that you will roll two six-sided, fair dice one at a time, and then add up the two values rolled. Which of the following diagrams approximately represents the associated probability of the die adding up to each of 1,2,3...12?

C; The joint distribution of (X,Y) lies in the cross product of the two random variables' dimensions (you can think of X,Y as lying in the 2D plane, one axis corresponding to X values and one axis corresponding to Y values). Thus it can only be nonzero in the region where X and Y are both nonzero, i.e. the rectangle [−5,5]×[10,10]. If it were nonzero anywhere else non-trivially (i.e. for a region R outside this rectangle with some 'thickness'), then if we integrated the joint distribution over the plane to find the marginal probabilities, we would obtain that the marginal probabilities of either X or Y are nonzero over the x or y projections of R, violating the assumptions of the question.

Suppose the continuous random variable X has a pdf that is nonzero in the region [−5,5] and the continuous random variable Y has a pdf that is only nonzero in the region [−10,10]. Then the joint distribution of (X,Y) has a pdf that is nonzero in a subset of which region? A. The interval [−5,5]∪[−10,10]=[−10,10] B. The ellipse x2/25+y2/100=1 C. The rectangle [−5,5]×[−10,10]

.705; The probability is P(A)=P(A|Study)P(Study)+P(A|Doesn't study)P(Doesn't study)=.9∗.7+.25∗(1−.7)=.705.

Suppose the probability of getting an "A" on a test conditional on studying is 0.90 and the probability conditional on not-studying is 0.25. The probability of studying is 0.70. What is the unconditional probability of obtaining an A?

The probability of getting an "A" is P(A)=P(A|Study)P(Study)+P(A|Doesn't study)P(Doesn't study). However, we don't have P(Study) (or equivalently, P(Doesn't study)=1−P(Study), and different values of P(Study) lead to different final P(A). Thus P(A) as stated is not possible to determine.

Suppose the probability of getting an "A" on a test conditional on studying is 0.90 and the probability conditional on not-studying is 0.25. The unconditional probability of getting an "A" is: A) .873 B) .575 C) .45 D) Not possible to determine

False. The probability density function must integrate to 1 over its entire domain, but this does not mean that the probability density must be less than 1 at all points in the domain. Take for instance a uniform distribution with range [0,0.5]; the probability density at every point is 2.

True or false: The conditions for a probability density function to be valid include that the density at each point is less than 1 and that the function must integrate to 1.

B; API stands for Application-Programming Interface. APIs help users directly harvest data from certain sites (such as Facebook, Twitter, or Google Maps) with relative ease, often in conjunction with a programming language like Python. As mentioned, many major companies provide their own API tools by they may come at a cost.

What is an API? A. Official guidelines for web scraping B. A programming interface, typically constructed by the developers of an application, that among other things helps users obtain certain structured data from the application more easily C. A popular website containing free datasets D. A programming language, similar to Python and R

A, C; Since Ω3={HT,TH}, and Ω4={HH,TT}, we know that (i) they are mutually exclusive since they have empty intersection (i.e. Ω3∩Ω4=∅), and they are (ii) exhaustive since their union comprises the whole sample space (i.e. Ω=Ω3∪Ω4 )

Which of the following expresses the relationship between the two sets Ω3={outcomes which contain at least one heads and one tail} and Ω4={outcomes where both coin throws are identical}? (Select all that apply) A) They are mutually exclusive B) They are identical C) They are exhaustive D) They have non-empty intersection

Ω1={HH,HT,TH}

Which of the following sets Ω1 represents all the outcomes which contain at least one head? Ω1={HH,HT} Ω1={HH,HT,TH} Ω1={HT,TH}

B, C; Note the distinction between the value of the joint PDF and the probability of the joint PDF. Just as in the case of the PDF for a single random variable, the joint probability density, fXY(x,y), at any particular point is non-negative, and the joint PDF must integrate to 1 over the x−y plane.

Joint probability density functions (joint PDF) for continuous random variables exhibit which of the following properties? (Select all that apply) A. The value of the joint PDF at any particular point is zero B. The value of the joint PDF at any particular point is non-negative C. The joint PDF integrates to 1 over the entire domain D. The joint PDF integrates over the entire domain to the number of variables. For example, for 2 variables x and y, the joint PDF integrates to 2

x

Let z be a continuous random variable with a uniform probability distribution between 0 and 1. What is the value of its CDF, FX(x) for any given value x such that 0≤x≤1?

Ω={HH,HT,TH,TT}; Each outcome looks like XY, where X is the result of the first coin, and Y is the result of the second coin. Each of X,Y can be H or T, independently of each other.

Say we flip two coins. Which of the following sets represents the entire sample space Ω of outcomes of this experiment? Ω={HH,HT,TH,TT} Ω={HH,TT} Ω={H,T} Ω={HT,TH}

B; The spikes approximate a discrete uniform probability distribution where the only possible outcomes are the integers 1 through 6, hence option 1 is incorrect. Option 2 is correct: A fair dice has an equal probability,16, of showing any of its 6 faces when rolled, hence the probability distribution for this experiment is a discrete uniform distribution defined only at integers 1 through 6. Finally, the number of realizations of a particular outcome in a series of dice rolls follows the binomial distribution, not the uniform distribution, and is therefore not approximated by the distribution shown above.

Consider a probability density function which has a narrow "spike" centered at each of the fixed points 1,2,...,6, and is 0 otherwise. The area of each spike is 16, so the probabilities indeed integrate to 1. Such a distribution approximates which of these discrete random variables? A. A random real number from 1 to 6 B. A single roll of a fair dice C. In a series of 6 dice rolls, how many dice rolls result in a "1"

A, B, C; A simple sample space is defined as a sample space where each of the possible outcomes is equally likely. Since there are four outcomes exhausting the sample space, they each have probability 1/4. Furthermore, if p denotes the probability of getting a heads in each coin toss, then P(HH)=p2=1/4, so p=1/2 and the coin is unbiased.

If someone told you the setting for a coin toss thrown twice was a simple sample space, what do you immediately know about it? (Select all that apply) A. All outcomes are equally likely B. The coin is unbiased C. Each outcome {HH,HT,TH,TT} has probability 1/4. D. Each outcome {HH,HT,TH,TT} has probability 1/2.

A, B; For each x, fX(x) captures distinct probabilities and these have to be less than 1 and add to 1 over the domain of X (the sets of values that X takes). However, the set where they are nonzero doesn't have to be finite. See the geometric distribution for an example.

The probability mass function fX of a discrete random variable X has the properties: (Select all that apply) A. 0≤fX(x)≤1 B. ∑xfX(x)=1 C. The set x where fX(x) is nonzero has to be finite.

A, C; The sample space is of size 2n, because at each of the n steps there are two possibilities for the coin's outcome. However, this can also be counted in another way. (n0) counts the number of outcomes in which there are 0 heads; (n1) in which there are 1 heads; and so on until (nn) counts the number of outcomes in which all n coins result in heads. Because each of these cases is disjoint, and every outcome of n coin flips must have some number of heads from 0 to n, summing all these terms covers the whole sample space. Thus 2n=(n0)+....

Consider the sample space of n tosses of a coin. What is the size of this space? (Select all that apply) A. 2^n B. n! C. (n \choose 0)+(n \choose 1)...(n \choose n) D. (n \choose 1)+(n \choose 2)...(n \choose n)

B; The number of ways to choose n items, including k of the required ones, is (N−k (choose) n−k), since we only have options in the N−k non-required items, and we want to choose n−k of them.

Consider x=k in the above distribution. (kk)=1, so the hypergeometric distribution reduces to (N−k (choose) n−k)/(N (choose) n). Which of these events has this probability? A. When randomly choosing n items from N possible items without replacement, the probability that none of k forbidden items are chosen. B. When randomly choosing n items from N possible items without replacement, the probability that all of k required items are indeed chosen C. When randomly choosing n items from N possible items without replacement, where k of them are recommended to be chosen, the probability that at least one of the recommended items is chosen

A, D; Deidentifying data involves more than just removing names. Any information that can identify an individual needs to be removed (such as the residential address). Gender, race/ethnicity, and state would not be considered identifiable information in this dataset. ICPSR has published guidelines for handling identifiable information and for documenting datasets.

Data needs to be de-identified or anonymized before datasets can be made public. Suppose you have a dataset on all registered voters in the United States that only contains the information below. Which variables should be removed before publishing the data? (Select all that apply) A. Full Name B. Gender C. Race/Ethnicity D. Residential address E. State

B; We are looking for unique dice combinations that add up to 8. The second answer exhausts all the possibilities and is therefore correct. Note that a dice combination like {2,6} which means "2 on the first dice, 6 on the second", is different from {6,2}, which means "6 on the first dice, 2 on the second". The third answer thus undercounts. Meanwhile, the first answer involves throwing a "7" on a dice roll, which is impossible.

Now consider the random variable X:Ω→N that tracks the sum of the outcomes of two dice rolls. Which of these captures X−1(8), i.e. the set of pure outcomes that add to 8? A. {{1,7},{2,6},{3,5},{4,4},{5,3},{6,2},{7,1}} B. {{2,6},{3,5},{4,4},{5,3},{6,2}} C. {{2,6},{3,5},{4,4}}

False

True or false: The Belmont Report was designed for social science applications.

A, C, D; Respect for persons, beneficence, and justice are the three core principles of the Belmont principles. Informed consent is part of respect for persons, but doesn't cover everything under the respect for persons principle.

Which of the following are core principles of the Belmont principles? (Select all that apply) A. Respect for Persons B. Informed consent C. Beneficence D. Justice

A, B, C; Weight, inflation rate, and time can take on any real value within their respective possible ranges and are hence continuous. While the number of people that will vote for the democratic presidential candidate must be integer, which means it is a discrete quantity. Note a technicality, which is that sometimes when we report the value a continuous random value takes, we have to report it on a discrete scale. But if this scale is highly precise, the variable is approximately continuous, which suffices to a consider as a continuos random variable.

Which of the following are examples of continuous random variables? (Check all that apply) A. The weight of a fish from the ocean B. Next year's inflation rate C. When a bird will next pass by your window D. How many people will vote for the Democratic presidential candidate

A, B, D; The first option involves the number of people, a discrete quantity. The second involves a binary choice, and is hence also discrete. The third involves the exact weight of an object, and this quantity is continuous as it can take any value on the real line. Finally, the last option involves the weight to the nearest whole number so it is indeed discrete.

Which of the following are examples of discrete random variables? (Select all that apply) A. Number of customers at a grocery store on a certain day in the future B. Whether someone will vote for a Democratic presidential candidate in the next election C. The exact weight of a box of books someone hands you D. The weight reading on a scale with precision to the nearest kilogram, of a small box of books that someone hands you

B; You can think of the inclusion-exclusion principle here as correcting for overcounting. To find the probability of at least one head in the throws, we first independently find the probabilities that the first throw is a head, and the second is a head, and add them. However, these events are not disjoint, so we subtract their intersection (the probabilities that both the first and second throw are heads).

Which of the following demonstrates the law of inclusion-exclusion applied to find the probability that in two coin flips, at least one heads occurs? A. P(at least one head)=P(first throw is a head)+P(second throw is a head) B. P(at least one head)=P(first throw is a head)+P(second throw is a head)−P(both throws are heads) C. P(at least one head)=P(first throw is a head)+P(second throw is a head)+P(both throws are heads)

B, C; Ω is the set of all outcomes, so any subset of outcomes such as Ω1 will satisfy Ω1⊂Ω. The third equality captures the definition of Ω1 in set theory notation.

Which of the following expresses the relationship between Ω1 and Ω above? (Select all that apply) A) Ω ⊂ Ω1 B) Ω1 ⊂ Ω C) Ω1 = Ω∩{H appears in the outcomes}

B; We need a statement of the form Pr(X∪Y)=Pr(X)+Pr(Y), where X=A and Y=B∖A.

Which of the following is true? A. P(A∪B)=P(A∖B)+P(B∖A) B. P(A∪B)=P(A)+P(B∖A) C. P(B∖A)=P(A)+P(A∪B)

A, D; Any set intersected with the empty set is the empty set. B∩Ac is the set of elements in B but not in A; intersecting this with A gives the empty set, not A. Instead, taking this set's union with A results in A∪B.

Which of the following set properties hold generally? (Select all that apply) A) A∩∅=∅ B) A∩∅=A C) A∩B=A∩(B∩Ac) D) A∪B=A∪(B∩Ac)

Ω2={HT,TH,TT}

Which of the following sets Ω2 represents all the outcomes which contain at least one tail? Ω2={HT,TH,TT} Ω2={TH,TT} Ω2={HT,TT}

D; There are n choices for each sample, and k samples, for n^k total enumerations.

Which of these enumerates sampling k items from n items with replacement? A) n!/k! B) n(n−1)...(n−k+1) C) n!/(n−k)! D) nk

B, C; There are n choices for the first sample, n−1 for the second (can't choose the same as the first sample), n−2 for the third, and so on until n−k+1 choices for the kth sample. Taking the product of the choices gives the enumerations above.

Which of these enumerates sampling k items from n items without replacement? (Select all that apply) A. n^k B. n(n−1)...(n−k+1) C. n!/(n−k)! D. n!/k!

A, D

Which of these pairs of events are independent events? Hint: check your answer using the multiplicative criterion. (Select all that apply) A. A dice roll is both at most 4 and odd B. A dice roll is both at most 5 and odd C. In a sequence of 5 coin flips: the first coin landing on heads and there being at least 3 heads D. In a sequence of 5 coin flips: the first coin landing heads and the total number of heads is even

A, B; If we go through the elements of Ω sequentially to find the outcomes that meet the condition, we see that only {HT,TH} meet it. However, Ω1∩Ω2 also captures the same outcomes, which makes same sense conceptually: Ω1 captures all the outcomes that include a head, and Ω2 captures all outcomes that include a tail, so their intersection captures the outcomes that include both.

With possible reference to the previous sets, which of the following sets Ω3 represents all the outcomes which contain at least one head and at least one tail? (Select all that apply) A) Ω3={HT,TH} B) Ω3=Ω1∩Ω2 C) Ω3=Ω1∪Ω2