Research Associate was asked...12 October 2012

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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. Less

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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

Research Analyst was asked...6 August 2014

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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. Less

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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. Less

Quantitative Researcher Summer Intern was asked...18 April 2011

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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. Less

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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 Less

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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. Less

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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 Less

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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! Less

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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. Less

Senior Research Executive was asked...19 October 2013

Quantitative Researcher was asked...18 November 2015

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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). Less

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Complete the square in the integral

Research Assistant was asked...16 May 2017

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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. Less

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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 Less

Quantitative Researcher was asked...21 December 2015

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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 Less

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

Quantitative Researcher was asked...23 July 2012

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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 Less

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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. Less

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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. Less

Quantitative Researcher was asked...26 December 2014

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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. Less

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sorry i was wrong

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