# Quantitative researcher Interview Questions

# 3K

Quantitative Researcher interview questions shared by candidates### How to measure 9 minutes using only a 4 minute and 7 minute hourglass

14 Answers↳

Start both timers together. When the 4 minute timer is done, flip it. 7 minute timer will have 3 minutes left. When the 7 minute timer is done, the 4 minute timer will have 1 minute left. Now you can count to 9 minutes by simply leaving the 4 minute to expire (1 min), flip it and let it expire (4 min), flip it again and let it expire (4 min). 1 + 4 + 4 = 9 Less

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The key is understanding that you will have to use the two hourglasses together. Since this problem could be asked in many ways using different values for the hourglasses and the total amount of time, it's more important to understand how you use the tools rather than memorize a specific example. The question is used to determine those who can apply their knowledge to solve problems vs. those who memorize answers "from the book". Start both timers. After four minutes, the four-minute timer will have expired and the seven-minute timer will have three minutes remaining. Flip the four minute timer over. After seven minutes, the seven-minute timer will have expired and the four-minute timer will still have one minute left. Flip the seven-minute timer over. After eight minutes, the four-minute timer will have expired for the second time. The seven-minute timer will have accumulated one minute after it's last flip. Flip over the seven-minute timer and when it expires nine minutes will have elapsed. For extra measure, you can always throw in something like, "assuming the timers can be flipped over nearly instantly..." Less

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1st timer 2nd timer time count 4 7...................start both timers 3 6................. 1min 2 5..................2mins 1 4..................3mins 0(flip) 3..................4mins completed 4 3..................4mins(assuming flip takes no time ideally) 3 2..................5mins 2 1..................6mins 1 0(flip)..........7mins 1 7..................7mins(again ideal flip) 0 6..................8mins(flip 2nd timer to count 1min) 0(as it is) 7..................9mins... Less

### There are 3 coins. One coin has heads on both sides, one coin has tails on both sides, the third one has head on one side and tail on the other side. Now I pick up one coin and toss. I get head. What is the chance that the coin I picked has heads on both sides?

14 Answers↳

2/3

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Because you have 1/3 chance to get double head coin and you will surely get head, 1/3 chance to get single head coin and then 1/2 chance to get head. So the probability of choosing double head coin and get head is 1/3, while choosing single head coin and get head is 1/6. Then, given you get head after tossing, then chance that you chose double head coin is (1/3)/(1/3+1/6) = 2/3 Less

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2 heads on double headed coin, 1 head on the other, P(head is coming from double headed) = 2/3 Less

### 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 Answers↳

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

### 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 Answers↳

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

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

12 Answers↳

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

### 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 Answers↳

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

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f(9)=1366 answer 1000 is wrong

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What are the last 4 digits of 2015^(2013^2014) ? 9375

### 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 Answers↳

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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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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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

### 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 Answers↳

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

### You have 10 mice and 1000 bottles of wine. You also have 24 hours before a party, and one of the bottles has been tainted with a slow acting poison, which takes 24 hours to kill a mouse. In the 24 hours you have remaining, how many bottles can you guarantee safe for human consumption (assume humans and mice react identically)? Assume the lethal dosage is insignificant relative to the size of the bottle.

8 Answers↳

999. It is like a binary problem. First mouse tests the first #1-500 (mixed). Second tests #1-250 and #501-750. Third one tests #1-125, #251-375, #501-625, #751-875, and so on. 10 mice with 2 status each (death/alive) could encode number of bottles up to 2^10=1024. So 10 mice is enough to find out the single bottle that tainted. Less

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The answer is 1023. You need to think in bit-wise way. 1023 can be represented in binary as (1111111111). Your goal should be: representing each wine label (i-th number) to each binary representation. 1000th wine will be represented with 1111101000 meaning (1,2,3,4,5,7th) mices will be used to check the toxicity of this wine. In binary way, you can assign label to up to 1023 wines. So by analysing the rats that die after 24 hrs, you can actually identify which wine is toxic or not. hope this helps. Less

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Brian was quite close, but to represent 1024 wines, you actually need 11 mice. So the maximum # of wines that one can guarantee is up to 1023. Less

### If X, Y and Z are three random variables such that X and Y have a correlation of 0.9, and Y and Z have correlation of 0.8, what are the minimum and maximum correlation that X and Z can have?

8 Answers↳

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