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

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

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

Quantitative Researcher was asked...22 May 2013

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http://wolfr.am/1i1XT4P

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http://www.johndcook.com/blog/2010/06/17/covariance-and-law-of-cosines/

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0.98 & 0.46

Quantitative Research was asked...17 March 2014

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Z sigma_x^2/ (sigma_x^2 + sigma_y^2). This is because X and Z are jointly normal and their covariance is equal to the variance of x. Therefore, the correlation coefficient is equal to sigma_x/sigma_z, and as E(X|Z)= rho. (sigma_x/sigma_z). Z, replacing the fact that the variance of the sum is the sum of the variance for independent (normal) R.V.s will give us the answer! Less

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The answer is 0.5 EX - 0.5* EY + 0.5z (EX EY is not equal to zero)

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[ z * sigma_y^ (-2) ] / [ sigma_x^ (-2) + sigma_y^ (-2) ] From signal processing point of view, x is the signal, y is the noise, and z is the observation. We know X has a prior distribution X ~ N(0, sigma_x^ 2 ), noise Y has distribution Y ~ N(0, sigma_y^ 2 ) and the value Z = z, the questions is what is the MMSE estimate of X given Z, i.e., E(X|Z)? Using Bayesian theorem, or Gauss Markov Theorem, one can show that : E(X|Z) = [ z * sigma_y^ (-2) + 0 * sigma_x^ (-2) ] / [ sigma_x^ (-2) + sigma_y^ (-2) ] Comments: 1. This kind of problems are very common so please keep it in mind in Gaussian case the best estimate of X is a weighted linear combination of maximum likelihood estimate (z in this problem ) and the prior mean (0 in this problem). And the weights are the the inverse of variance. 2. In multi dimension cases where x, y, z are vectors, similar rules also apply. Check Gauss Markov Theorem 3. In tuition here is the larger variance of noise y, the less trust we will assign on ML estimate, which is sigma_y^ (-2) . Correspondingly, the more trust we put on the prior of X. Less

Quantitative Research Analyst was asked...9 November 2015

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between 0 and 0.96

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bc + - sqrt(1-b^2) * sqrt (1-c^2) if b=0.6 and=0.8, then a can be between 0 and 0.96 Less

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... bc^2 - 0.96 *bc <= 0 so 0 <= bc <= 0.96 to keep the determinant of the correlation matrix greater than or equal to zero. Less

Quantitative Researcher was asked...9 August 2013

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The number of ways of picking k 3point darts to hit is 500Ck, and then you need 1500-3k 1 point darts to hit. The number of ways fulfilling the required conditions is then (500 Choose k)*(500 Choose 1500-3k). You can clearly see that k needs to be bigger than 333, and making a plot of the function shows that k=398 gives the maximum probability. Less

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TK answer is right, but I think they probably want you to approximate with Gaussians. Ignoring a lot of constants, and the standard deviations since they don't matter, the distribution for x 1pt hits and y 3pt hits is: p(x,y) ~ exp(-(x-250)^2-(y-250)^2) log likelihood: l(x,y) ~ -(x-250)^2-(y-250)^2 use x+3y=1500 l(y) ~ -(1250 -3y)^2 - (y-250)^2 l'(y) = 6(1250-3y) -2(y-250) Find maximum: 6*1250 - 18y = 2y -500 y = 8000/20 = 400 This way you get 400 3pt hits and 300 1pt hits, which is pretty close to the real answer 398/306, and doesn't require evaluating massive binomial coefficients over the phone. Less

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average score per hit: (1+3)/2=2, then expected hits: 1500/2=750, each gets 750/2=375 hits Less