research intern interview questions shared by candidates
What is a 1572?2 Answers
Statement of Investigator: An agreement between the PI and the FDA that he will conduct the study according to ICH GCP guildlines
How did I handle a stressful situation at work?1 Answer
Thinking logically, formulating a plan, and approaching the situation with a level-head
The professor quizzed me on several technical aspects of a journal paper I was a co-author of. Most of them were manageable. He asked me what exactly my contribution was in that paper and why I was not first author on it.1 Answer
I replied that the project was industry-funded and because I was a student at the time the industry I was working for did not share enough proprietary details with me like they did with my advisor who was the first author on the paper. As a result it was easier and much faster for my advisor to write the paper up and add me on as a second author despite the fact that I did all the experimental work that was featured in it. The rest of the interview was excellent and the Professor collected my references and then e-mailed them separately for recommendations. In ~4 weeks time I was hired.
How do you manage your time effectively?1 Answer
To do lists, calendars, reminders, prioritizing daily, communicate with team.
What research area are you interested in?1 Answer
Don't be so specific that you could only see yourself working for one prof. Read up on the faculty, pick out several you like and then widen your answer to potentially include any of them. Read up on these subjects beforehand so you can ask some intelligent questions about the research projects and see if it is what you really want. Be sure to talk with the profs at recruitment weekend!
If you had a machine that produced $100 dollars for life what would you be willing to pay for it today?74 Answers
No more than 100$ for 100$ for life
Depends entirely on how fast the machine produces the bills. Current price will be determined by estimated future yield which isn't defined here.
I would pay nothing for it. $100 dollars in a life? I can make that in one day...
I would not pay anything. Only the Federal Reserve can legally produce $100 dollar bills.
I would pay for everyone in my family to get a four year college degree and then their master's degree. I would cover the cost of books, supplies and dorm rooms or apartments near campus. I would buy a new car if they held a GPA over 2.00. I would buy them nice clothes and shoes so that they looked nice while attending class. I would also buy them a new coat to keep warm and an umbrella to keep the rain off of their heads. I would value education more than I can afford to do so now.
I already have the machine, why would I pay for it??
Nothing, I already have it.
absolutely nothing. $100 dollars for a life time is ridiculous
What is "$100 dollars"? That reads as "one hundred dollars dollars." What's a dollar dollar? Is it double value? Say $100 or one hundred dollars; do not say both, in order to avoid being redundant.
As much as I could lay my hands on. It could all be repaid instantly
I would pay nothing for a machine that produced dollars at a cost of $100 each. What would you offer me to haul the machine away for you?
Nothing. Fiat money is either rendered worthless by uncontrollable inflation, or $100 over a lifetime is not worth much.
Firstly, if it made $100 for life, it wouldn't be worth anything. Secondly, if it's making money, it's illegal.
10 to 20 years is what I would have to pay for a machine like that, life is only for murder with special circumstances.
Does such a machine exist? What? No? Then why are you asking this question ... MORON.
No. Not worth it.
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.
Nothing. According to the question, I already have it.
I am assuming the question is stated incorrectly. It should read: "$100 per year for life". Let's say I expect to live another 50 years. Then in nominal terms, the machine produces $50,000. My time-preference rate is 4 percent per year and let's suppose that the Federal Reserve does a good job of keeping inflation in their target range and averages 2 percent. Then the total discount rate is 6 percent per year. In present-value terms, the machine produces $100 this year, $94.34 next year [e.g. $100/((1+r)^n) ], $89.00 in two years, etc, for a sum of $1,676.19. This is the present value of the machine and I would pay this much or less to own the machine. Now suppose that the machine actually produces real $100 dollars (so increases the amount according to inflation). In that case the discount rate is 4 percent and I would pay a maximum of $2,248.22. Obviously I'm using a calculator (Excel) to make these calculations, but the idea is the same.
What are talking here? $100 per day? Week? Month? Only once in a lifetime? Oh, and does the IRS have to know about this?
Be specific. Does it produce $100 every minute of your life or just once in your life? Will it produce enough for me to make bail and hire a lawyer after I'm arrested for counterfeiting?
Press NPV button on calculator.
1) As many have stated, this is illegal. But putting that aside, 2) The way the question is presently phrased, technically "you" already have the machine so why would you pay anything? And also, there is a lack of clarity regarding the frequency of the machine's production of $100 - daily, weekly, monthly, etc? The output frequency changes the entire question (because if you only receive $100 once well why bother wasting x% of that $100 by paying for the machine itself?), 3) But assuming this machine provides $100 more than once, does it print a single $100 bill or 100 $1 bills (or other denominations)? Because based on that, what are the operating costs of this machine? How much does the paper cost? How much does the ink cost? How much does it cost to maintain this machine? The interviewer may or may not answer those questions depending on the interview style and format so I'd just go with "It's an illegal practice and I'd never do something so horrific and undermine the honest principles this country was founded upon!" :P
Somebody has been watching too much Twilight Zone.
Whoever formulated this question does not speak good English. The question has no real discernible meaning. They should be fired and a qualified person should be given the job.
I will pay with a blank cheque; the seller puts whatever he/she wants; I just have to print it & deposit to my bank account.
I assume I can print the $100 dollars unlimited number of times a day. I would tell the manufacturer that I can't pay for the machine now but I'm willing to pay them $1,000 per day for the rest of my life. That'll come up to just 10 pcs for them each day and I'll print another 100 pcs for myself each day. That'll be $10,000 for me everyday!
Please define what $100 dollars is. Is it $100 or 100 dollars? USD? When does it produce this undefined amount? For my life or its? What are its operating and disposal costs? Is that in $ dollars also? Where do you spend $ dollars? There is nothing to base an answer on.
Life is worth more than a hundred dollars!
If I already have it, to whom would I pay anything and why? Silliness aside, you would need to know at least: Payment frequency Your life expectancy Some discount yield (subjective or otherwise) Then it's a matter of PV annuity, or PV perpetuity (if you are allowed to pass on payments after death)
It produces $100 for each life it takes? Probably evil, would avoid.
I already have it so I don't need to pay for it. But if I didn't have it and had to buy its details as mentioned above need to be answered such as $100 a day month year number of times u can take it out. But you are basically putting your own money into a machine to give it back to you when you need it, sort of like an atm
Nothing, I already have it!
The wording makes no sense. Saving money by using a hybrid car, LED bulbs, efficient appliances and HVAC, or solar is easier to calculate and pays way more than, $100 a year.
With that 100$, i will buy another such machine. Now i have 2 machines, each of them generates 100$, which is 200$. With these 200$, i will buy another 2 such machines, now i have 4 machines. I will buy another 4 machines, with 400$, so on.. I will multiply so on, until i have enough money to solve everyone's problems of their lifetime !!!
I would use it to buy a truckload of English usage manuals and distribute them to people who insist on uninformed mash-ups like "$100 dollars". Are they anything like "ATM machines" or "square acres"?
I'd be really worried that if it put out $100 on the first day, I'd be dead the next day.
Honestly most people here haven't truly read the question. If I already had a machine that would make $100 bills for the rest of my life, why would I have to pay any money at all to get it?
The majority of ppl answering this question, haven't even comprehended the question properly. Seriously.
I would pay $6,500 for it. I plan to live to about 85 so I'd set up a payment plan and pay it off with the machine 1 hundred every year
i will not pay anything. $100 is the least I could get for free
$100. That's all it's worth if it pays that for life.
In the first place this is a trick question. It wasn't stated that the machine would produce 100 dollar bills. It stated 100 dollars for life. In the second place it would be counterfeiting so possession of this machine could get you a prison sentence, and no amount of money is worth loosing years of your life. I've noticed most of these questions are trick questions like how many people flew people don't fly plane's and birds do.
One dollar. Why? Because I know a value when I see one. My deal, my value. Secure that one and you have it made.
Anything less than 100$
I'd want at least a 500% profit so $20 at most.
I'd pay someone to stop coming up with these stupid questions. Makes me not want to work for anyone. No wonder so many jobs go unfilled. Who wants to work with or for people that start off our first meeting with stupid head games. Chances are that all the job involves is pushing buttons, referring to manuals, taking calls, reading emails and forwarding and escalating issues to other employees. If I wanted to solve riddles, I'd apply to the Riddle Factory where such questions would be relevant.
$100USD Dollars for life
Well, if I had the machine, as the question indicates, then I would not have to pay for it if I already owned the machine, which the beginning of the questions indicates.
I'd leave the interview with a company that produces such an incredibly poorly-worded question. "$100 dollars?" So "one hundred dollars dollars?" Ok. Also, it doesn't specify any sort of rate at which it produces money. It doesn't say if it produces $100 per day, nor does it say if it produces X amount of dollars per day until it reaches $100 in a lifetime. As it is written, this question makes no sense, and the writer of it should be ashamed.
I wouldn't pay anything for it today, or the day I got it. What are "dollars for life" anyway? And why do they cost $100?
I want to earn money from my work not from machine
Who is life, and why would I want to pay for his/her machine?
I would pay up to $50 for it. Make 60,70,80 or $90 then sell it for $50+
Why would I pay for something that I already have?
I would ask first the frequency of distribution, if that's daily/weekly/monthly etc.
I would say that I have a policy of never buying something without testing it first and then when I was stood in front of the machine I would say how much do you want for it? Money is no object! :)
I would first ask how easy is it to reproduce the machine. If it is less than $100 to reproduce the machine, I would just recreate as many machines as possible and earn the difference. Then I will do some math to calculate the optimized price to pay for a desired annual income. And maybe in the future, invest some of the income to reduce production cost, reuse of old machines, etc.
I think the question has probably been misquoted . Probably meant "How much would you pay for a machine giving you $100 per year for the rest of your life ". Its basically a PV of future cashflows ; simply if into perpetuity : = coupon / discount rate. The coupon being the $100 . The Discount rate is the rate of the prevailing interest rate - usually take the yield of 30 yr UST here (say 3% for arguments sake) and add in a bit more if you fell there is any risk in holding the machine. (like some fool with a loaded gun trying to steal it of you) ie 100/ 0.03 = $3,333 . This will be the maximum one should pay.
I'll pay for another machine which produce money in such an easy way for sure.
Nothing. It clearly is a scam
Learn to write a coherent question and then get back to me.
1) why would I pay for a machine I already own 2) $100 for life is just a one time payment of $100
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
expected earn is 25 cents. 1/2*1/2*1, prob of choosing to guess is 1/2, prob of guessing right is 1/2, and the pay is $1
I would start picking cards without making a decision to reduce the sample size. This is risky because I could just as easily reduce my chances of selecting red by taking more red cards to start, as I could increase my chances of selecting red by picking more black cards first. But I like my chances with 52 cards, that at some point, I will at least get back to 50% if I start off by picking red. Ultimately, I can keep picking cards until there is only 1 red left. But I obviously wouldn't want to find myself in that situation so I would do my best to avoid it, by making a decision earlier rather than later. Best case scenario, I pick more blacks out of the deck right off the bat. My strategy would be to first pick 3 cards without making a decision. If I start off by selecting more than 1 red, and thus the probability of guessing red correctly is below 50%, then I will look to make a decision once I get back to the 50% mark. (The risk here is that I never get back to 50%) However, if I pick more than 1 black card, then I will continue to pick cards without making a choice until I reach 51% - ultimately hoping that I get down to a much smaller sample size, and variance is reduced, while odds are in my favor that I choose correctly. The expected return, in my opinion, all depends on "when" you decide to guess. If you decide to guess when there is a 50% chance of selecting correctly, then your expected return is 50 cents (50% correct wins you $1 ; 50% incorrect wins $0 --- 0.5 + 0 = .5) If you decide to guess when there is a 51% chance of selecting red correctly, then the expected return adjusts to (0.51* $1) + (0.49 * $0) = 51 cents. So, in other words, your expected return would be a direct function of the percentage probability of selecting correctly. i.e. 50% = 50 cents, 51% = 51 cents, 75% equals 75 cents. Thoughts?
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.
scheme: guess when the first one is black, p(guess) x p(right) x 1=1/2 x 26/51=13/51
0.5, just turn the first card to see if it's red. I think it's more about trading psychology. If you don't know where the price is going, just get out of the market asap. Don't expect anything.
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
This should be similar to the brainteaser about "picking an optimal place in the queue; if you are the first person whose birthday is the same as anyone in front of you, you win a free ticket." So in this case we want to find n such that the probability P(first n cards are black)*P(n+1th card is red | first n cards are black) is maximized, and call the n+1th card?
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.
The answer above is not 100% correct, for second scenario, if you don't guess, and only look, the total probability of getting red is indeed the same. However, the fact that you look at the card means you know if the probability of getting red is x/(x+y)*(x-1)/(x+y-1) or y/(x+y)*x/(x+y-1). Therefore, this argument only holds if you don't get to look at the card, or have any knowledge of what card you passed
Doesn't matter what strategy you use. The probability is 1/2. It's a consequence of the Optional Stopping Theorem. The percent of cards that are left in the deck at each time is a martingale. Choosing when to stop and guess red is a stopping time. The expected value of a martingale at a stopping time is equal to the initial value, which is 1/2.
My strategy was to always pick that colour, which has been taken less time during the previous picks. Naturally, that colour has a higher probability, because there are more still in the deck. In the model, n = the number of cards which has already been chosen, k = the number of black cards out of n, and m = min(k, n-k) i.e. the number of black cards out of n, if less black cards have been taken and the number of red cards out n if red cards have been taken less times. After n takes, we can face n+1 different situations, i.e. k = 0,1,2, ..., n. To calculate the expected value of the whole game we are interested in the probability that we face the given situation which can be computed with combination and the probability of winning the next pick. Every situation has the probability (n over m)/2^n, since every outcome can happen in (n over m) ways, and the number of all of the possible outcomes is 2^n. Then in that given situation the probability of winning is (26-m)/(52-n), because there are 26-m cards of the chosen colour in the deck which has 52-n cards in it. So combining them [(n over m)/2^n]*[(26-m)/(52-n)]. Then we have to sum over k from 0 to n, and then sum again over n from 0 to 51. (After the 52. pick we don't have to choose therefore we only sum to 51) I hope it's not that messy without proper math signs. After all, this is a bit too much of computation, so I wrote it quickly in Python and got 37.2856419726 which is a significant improvement compared to a basic strategy when you always choose the same colour.
dynamic programming, let E(R,B) means the expected gain for R red and B blue remain, and the strategy will be guess whichever is more in the rest. E(0,B)=B for all Bs, E(R,0)=R for all Rs. E(R,B)=[max(R,B)+R*E(R-1,B)+B*E(R,B-1)]/(R+B). I don't know how to estimate E(26,26) quickly.
The question, to me, is not clear. Perhaps on purpose. If so, the best answers would involve asking for clarification.
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
the second ans should be 1/3
Actually the asnwer is 3/4. Lets go through the conditional probabilities. If the first child is named William and the second is NOT: the prob. of the second child being a boy is 1/2. If the second child is named William and the first is NOT: the prob. of the second child being a boy is 1. If both children are named William: the prob. of the second child being a boy is 1. Now, assuming equal probs. of the first, second, or both children being named William, the total probability of the second child being a boy is (1/3)(1/2)+(1/3)(1)+(1/3)(1)= 5/6
^^^^^^^^^^^^^ Correction... I meant to write the answer is 5/6
You're both incorrect. The first answer is 2/3 and the second answer is 4/5. Let B stand for boy (not named William), Bw be a boy named William, and G be a girl. 1. The sample space is: (B,B) (B,G) (G,B) It's easy to see that two of these cases have a boy as the second child so the probability is 2/3. 2. The sample space is: (Bw, Bw) (Bw, G) (G, Bw) (Bw, B) (B, Bw) 4 of the 5 cases have a boy as the second child, and therefore the probability is 4/5.
The first answer is 2/3, as mentioned. But the second question hasn't been answered correctly here. Given a person X, define e = P[X's name = William | X = boy]. Then, note that P[X's name = William] = P[X's name = William | X = boy] * P[X = boy] = e / 2 (the problem states that William is a boy's name). Letting C1 be child 1 and C2 be child 2, we are asked to find P[C2 = boy | Y], where for notational simplicity, Y denotes the event "C1's name = William or C2's name = William." By Bayes' rule, we have: P[C2 = boy | Y] = P[C2 = boy, Y] / P[Y] == Denominator: P[Y] = 1 - P[C1's name != William AND C2's name != William] = 1 - e/2 * e/2 = 1 - e^2/4. == == Numerator: P[C2 = boy, Y] = P[Y | C2 = boy] * P[C2 = boy] = 1/2 * P[Y | C2 = boy] = 1/2 * P[C1's name = William or C2's name = William | C2 = boy]. To compute P[C1's name = William or C2's name = William | C2 = boy], there are 3 cases. Case 1: C1's name is William, C2's name isn't William (given C2 is a boy). This is e/2 * (1- e). Case 2: C2's name is William, C1's name isn't William (given C2 is a boy). This is e * (1 - e/2). Case 3: both names are William (given C2 is a boy). This is e/2 * e. Summing these 3 cases gives e/2 - e^2/2 + e - e^2/2 + e^2 / 2 = 3e/2 - e^2/2, which is the numerator. == Dividing, we have (3e/2 - e^2/2) / (1 - e^2/4) = e * (3 - e) / (4 - e^2). As a sanity check, note that setting e = 1 implies that all boys are named William, and our probability is 2/3, as in the first question. Setting e = 0 implies that no boys are named William, in which case our probability is 0.
unless William's prior distribution is provided. The only information we got is that William is a boy name. Thus the event is equal to at least one of them is a boy = 2/3. Becareful when you guys calculate probability with outcomes UNEQUAL LIKELY.
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
1. P[C2 = boy | C1 = boy ] = 0.5 since these are independent events 2. P[C2 = boy | C1 = boy and is called William ] = 0.5 since these are independent events
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!
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.
i think you guys are doing way too much and this is a trick question, these are completely independent events, i could name my child william, i could name it lollipop - the chances of it being a boy are still .5, regardless of its brothers name as well . (im going with .5 for both questions) keep in mind this is a quick phone interview question so they wont give anything thats too calculation heavy, involving e^2 exponents fractions etc, because you're supposed to be able to do it in your head
Answer by indosaurabh is correct, i go to harvard
Given log X ~ N(0,1). Compute the expectation of X.13 Answers
This is a basic probability question.
exp(mu + (sigma^2)/2) = exp(0+1/2) = exp(1/2)
Let Y = log(X), then X = exp(Y) = r(Y), if we call the pdf of X f(X), then E[X] = integral(Xf(X)dX). By variable transformation, f(x) = g(r^-1(X))r^-1(X))', plug this into E[X] = integral(Xf(X)dX), we get integral( f(y)dy ), which equals to 1
Suppose the density function of Y is P(y) and the one for X is F(x), it obeys that P(y)*dy = F(x)*dx; then the expectation of X is E(x) = Integral( x*F(x)*dx ) = Integral( Exp(y) * P(y) * dy ); if you plug the gaussian function and standard deviation in, you will find E(x) = Integral( Exp(1/2) * P(y-1/2)*d(y-1/2) ) = Exp(1/2) So, mojo's ans is correct.
I m not that sure, as I got E(x) = 4 I substituted log X = y e^y = X ;and e^2y = t and plz do not forget to change the integration limits
Do they care if you explain the theory or not? I just looked at it, it's standard normal, therefore x=50%
Sorry misread the problem. ignore.
X has a log-normal distribution, so yes the mean is exp(mu+sigma^2/2)=exp(1/2)
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).
Complete the square in the integral
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