# Questions d'entretiens - Research project coordinator

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Questions d'entretien pour Research Project Coordinator partagées par les candidats## Principales questions d'entretien

### If you were to get rid of one state in the US, which would it be and why?

65 réponses↳

Based on a highly sophisticated algorithm I concocted while you were asking the question, I would find the states who take in the most federal dollars and pay out the least federal tax in return. I would take the number of right wing radio stations in those states complaining about that self-same federal gummymint, and divide that by the literacy rate. Then I would choose Texas anyway. Moins

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The state of confussion!

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The state of dependency because everyone needs to find a way to play it forward

### If you had a machine that produced $100 dollars for life what would you be willing to pay for it today?

65 réponses↳

I already have the machine, why would I pay for it??

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I would not pay anything. Only the Federal Reserve can legally produce $100 dollar bills. Moins

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Why is this considered a top 10 odd interview question? It's a basic accounting question that applies to any applicant at a financial institution. Let's assume the proper phrasing of the question is "If you had a machine that produced a free $100 dollars per year for life, what would you be willing to pay for it today?" Given that Aksia is a financial firm, they're basically asking what is the present value of a perpetuity with a $100 annual payment. PV=pmt/r where: PV=PResent value PMT= payment per period r= discount rate Given current US fed reserve discount rate is 0.75%, the Present value of such a device would be $13,333.33 Answer varies obviously if discount rate changes or if proper phrasing was meant to be $100 for a different time period. Moins

### 3) Poker. 26 red, 26 black. Take one every time, you can choose to guess whether it’s red. You have only one chance. If you are right, you get 1 dollar. What’s the strategy? And what’s the expected earn?

13 réponses↳

There is symmetry between red and black. Each time you pull a card it is equally likely to be red or black (assuming you haven't looked at the previous cards you pulled). Thus no matter when you guess you odds are 50% and the expected return should be 50 cents. Moins

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The problem should be random draw card and dont put it back. Every draw you have one chance to guess. So the strategy is after first draw you random guess it's red. If correct you get one dollar, next draw you know there is less red than black. So you guess black on next draw. Else if first guess you are wrong, you guess red on next round. It's all about conditioning on the information you know from the previous drawings Moins

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The problem statement is not very clear. What I understand is: you take one card at a time, you can choose to guess, or you can look at it. If you guess, then if it's red, you gain $1. And whatever the result, after the guess, game over. The answer is then $0.5, and under whatever strategy you use. Suppose there is x red y black, if you guess, your chance of winning is x/(x+y). If you don't, and look at the card, and flip the next one, your chance of winning is x/(x+y)*(x-1)/(x+y-1) + y/(x+y)*x/(x+y-1) = x/(x+y), which is the same. A rigorous proof should obviously done by induction and start from x,y=0,1. Moins

### The first question he gave me was not hard. 1. You call to someone's house and asked if they have two children. The answer happens to be yes. Then you ask if one of their children is a boy. The answer happens to be yes again. What's the probability that the second child is a boy? 2. (Much harder) You call to someone's house and asked if they have two children. The answer happens to be yes. Then you ask if one of their children's name a William. The answer happens to be yes again.(We assume William is a boy's name, and that it's possible that both children are Williams) What's the probability that the second child is a boy?

12 réponses↳

1. BB, BG, GB, GG 1/4 each, which later reduced to only BB, BG, GB with 1/3 probability each. So the probability of BB is 1/3 2. Let w is the probability of the name William. Probability to have at least one William in the family for BB is 2w-w^2, For BG - w, GB - w, GG - 0. So the probability of BB with at least one William is (2w-w^2)/(2w+2w-w^2) ~ 1/2 Moins

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The answer by Anonymous poster on Sep 28, 2014 gets closest to the answer. However, I think the calculation P[Y] = 1 - P[C1's name != William AND C2's name != William] should result in 1 - (1- e /2) ( 1- e / 2) = e - (e ^ 2 ) / 4, as opposed to poster's answer 1 - (e^2) / 4, which I think overstates the probability of Y. For e.g. let's assume that e (Probability [X is William | X is boy]) is 0.5, meaning half of all boys are named William. e - (e ^ 2) / 4 results in probability of P(Y) = 7/16; Y = C1 is William or C2 is William 1 - (e ^ 2) / 4 results in probability of P(Y) = 15/16, which is way too high; because there is more than one case possible in which we both C1 and C2 are not Williams, for e.g. if both are girls or both are boys but not named William etc) So in that case the final answer becomes: (3e/2 - (e^2)/2) * 0.5 / (e - (e ^ 2) / 4) = 3e - e^2 / 4e - e^2 = (3 - e) / (4 - e) One reason why I thought this might be incorrect was that setting e = 0, does not result in P(C2 = Boy | Y) as 0 like Anyoymous's poster does. However I think e = 0 is violates the question's assumptions. If e = 0, it means no boy is named William but question also says that William is a Boy's name. So that means there can be no person in the world named William, but then how did question come up with a person named William! Moins

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I think second child refers the other child (the one not on the phone) In this case answer to first is 1/3 and second is (1-p)/(2-p) where p is total probability of the name William. For sanity check if all boys are named William the answers coincide. Moins

### Given log X ~ N(0,1). Compute the expectation of X.

12 réponses↳

exp(mu + (sigma^2)/2) = exp(0+1/2) = exp(1/2)

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Expanding on the correct answers above: E[X] = E[exp(logX)], and logX is normally distributed. So: E[X} is the moment-generating-function (mgf) of a standard normal distribution, evaluated at 1. The mgf of a normal distribution with mean mu, SD sigma is exp(mu*t + (1/2) * sigma^2 * t^2), now set mu = 0, sigma = 1, t = 1 to get exp(1/2). Moins

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Complete the square in the integral

### Tell me about yourself

11 réponses↳

I will do my assigned work with full dignity

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My name is Stephanie. I have 30 years of experience working with seniors and special needs people. I’m caring and maternal and look after others needs first. I am punctual and reliable. Moins

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My name sandeep Bains I live north side edmonton I am health care aid and I have 2 years India hospital work experience Moins

### What are the last 4 digits of 2015^(2013^2014]? The three distinct entries of a 2 x 2 symmetric matrix are drawn from the uniform distribution [-60, 60]. What is the expected determinant of the matrix? How many times a day do the hour and minute hands of an analog clock form a right angle? If WHITE=000, RED-101, BLUE-110, and PURPLE-100, then what three-digit string corresponds to YELLOW? There are 4 green and 50 red apples in a basket. They are removed one-by-one, without replacement, until all 4 green ones are extracted. What is the expected number of apples that will be left in the basket? Given two assets that have expected excess returns of 7 and 4, and given their expected covariance matrix: {1,1}{1,2} What is the maximum expected Sharpe ratio that you can achieve by combining the two assets into a portfolio? Consider a polynomial f(x) and its derivative f (x) that are related according to: f(X) - f‘(X) = X"3 + 3*X"2 + 3*X + 1 What is f(9)? A pedestrian starts walking from town A to town B. At the same time, another pedestrian starts walking from town B to town A. They pass each other at noon and continue on their paths. One of them arrives at 4 PM, the other at 9 PM. How many hours had each walked before passing each other? Seven people are in an argument, but potentially some or all of them are liars. They give the following statements: Bob: "No one lies." Jennifer: "No one tells the truth." Conrad.: "Jennifer is not a liar." Tom: "Conrad and Sherry always lie at the same time." Sherry: "Danny never lies." Danny: "Sherry is a liar." Adam: "Danny sometimes lies.” How many of them are lying? A city is composed of three parallel east-west streets and four parallel north-south streets: Note there are 12 intersections and 17 street segments. A policeman needs to visit every street segment, but he wants to take the shortest path. The policeman can start at any intersection, and he can only traverse streets, going from one intersection to another. How many street segments are there in the shortest path that visits each street segment at least once? Three riflemen A, B, and C take turns shooting at a target. The first rifleman to hit the target gets 2002 dollars. A shoots first, B second, and C third, after which the cycle repeats again with A, until one of the riflemen hits the target. Each hits the target with probability 0.5. What is rifleman A's expected winnings in dollars? Nine boys and seven girls are seated randomly around a circular table with 16 seats. Find the expected number of girl-boy neighbors. For example, in the seating below there are four such pairs. GBBBBB G B G B GGGBBG

11 réponses↳

f(9)=1366 answer 1000 is wrong

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Yellow = 001 it`s my calculation but i`ve also found the same answer in one chinese forum Moins

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E = sum(k=1, 50) (53-k)(52-k)(51-k)x4xk/(54x53x52x51) = 10

### Game: I throw 1 die 4 times, trying to reach at least one 6, you throw 2 dice 24 times and try to reach at least one double 6 (6,6). Who has greater chance of winning

10 réponses↳

To estimate, compare (5/6)^4 and (35/36)^24 this is 5/6 and (35/36)^6 this is 30/36 and (35/36)^6 notice that 30/36 is missing six 1/6th from 1 (36/36) and taking powers of (35/36)^6 will reduce the number by nearly 1/36th each time, but less than than, so that (35/36)^6 is greater than 1-6/36=30/36. Therefore the probability of not getting any double six is greater than probability of not getting any 6, and you should choose to roll one die. To understand the reasoning, think about taking powers of 0.90, 0.90^1 = 1-1x0.10 0.90^2 > 1-2x0.10 0.90^3 > 1-3x0.10 and so on Moins

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It all comes to which is greater: 1-(5/6)^4 or 1-(35/36)^24. They will expect you to calculate this (which is greater, not actual numbers) without a calculator. Moins

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First comment is correct, second comment is wrong since it asks for at least one 6 or at least one (6,6). This also includes the outcomes 2 or more sixes or double (6,6). Hence the easiest way of calculating this is by calculating the complementary probabilities P(no six) and P(no double six), respectively, to get P(at least one six) = 1 - P(no six) and P(at least one double 6) = 1 - P(no double six), which gives the result in the first comment. Moins

### There is a solar system with three planets orbiting around the sun. One of them has a translation period of 60 years, another one of 84 years and another one of 140 years. Today, the three planets are aligned with the sun. When is the next time the three planets will be aligned with the sun?

10 réponses↳

Should be 210 years

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These planets can be aligned on either the same side of the sun or opposite sides. So the answer is a number x that is the least common multiplier of 30, 42 and 70, which is 210. Moins

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sorry i was wrong