Internships Interview Questions | Glassdoor

# Internships Interview Questions

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## Top Interview Questions

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Jun 24, 2012

### Procurement Internship at Mars was asked...

Mar 10, 2014
 Your supplier ran out of resin and you have a delivery to a customer today. What do you do? How would you improve this package? Name three cost savings that could be made to this package? 1 AnswerCall supplier ask why they ran out of resin and get name of vendor the supplier purchases resin from . Place a resin order with suppliers vendor @ their cost or @ an even lower cost and ask what action they are to take to prevent this on future resin orders. Due to inventory discrepancy @ supplier negotiate the below cost supplier should agree with. 1) purchase from suppliers vendor @ supplier cost 2) supplier responsible for freight bill 3)Supplier pay cost for same day delivery fee to expedite

Feb 26, 2010

Dec 14, 2016

### Quant Internship at Jane Street was asked...

May 6, 2011
 You have 100 marbles, 50 are blue, 50 are red. You want to distribute them between two drawers, in such a way that none is left outside and no drawer is left empty. After distributing them you are gonna select a drawer randomly and from that drawer you are gonna remove one marble randomly. How do you distribute the marbles in such a way that the probability of getting a red marble is maximized?6 AnswersPut one red marble in one drawer and all the others in the other drawer.It doesn't matter how you distribute the marbles. The fact that they are split into two drawers doesn't affect the sample space. The probability of choosing either draw is 1/2 (i.e. the same).The first answer is correct. Say you put one red marble in drawer 1, and drawer 2 has 24 red and 25 blue. The chances of choosing drawer 1 and a red marble are .5 or 49/98 since all the marbles are red in drawer 1. The chances of choosing drawer 2 and a red marble are .5*(24/49) so 12/49 or 24/98. The total chances of choosing a red marble in this case are 73/98, much higher than 1/2.Show More ResponsesIn drawer #1: 1 RED 0 BLUE In drawer #2: 49 RED 50 BLUE 50% chance of selecting drawer #1 - where there is a 100% chance of selecting RED = .5 * 1.00 = .5 50% chance of selecting drawer #2 - where there is a 49/99 or .4949% chance of selecting RED = .5 * .4949 = .2474 .5 + .2474 = .7475Stephen is correct. The answer is indeed 1/2*(1/1) + 1/2*(49/99) = 1/2*(148/99) = 74/99 = 74.75%. "P" has the right approach but has assumed the wrong number of marbles, as there should be 100 marbles in total, not 50 :)Here's a really intuitive way to see derive the solution without any actual math. You can obviously achieve 50% odds by putting all the red marbles in one drawer and all the blue marbles in the other. If you were to take a blue marble from the blue drawer and put it into the red drawer, then you slightly decrease the odds overall. Conversely, if you were to take a red marble from the red drawer and put it into the blue drawer, then you slightly *increase* the odds overall, since if you pick the red drawer you still have a 100% chance (at that point) of getting a red marble, and if you pick the blue drawer you now have a nonzero chance of getting a red marble. Repeating this, we arrive at the optimal solution posted by others: one drawer should contain one marble while the other drawer should contain all other marbles.

### IBD Strats Internship at Goldman Sachs was asked...

Apr 20, 2012
 You have 4 cards, 2 black and 2 red. You play a game where during each round you draw a card. If it's black, you lose a point. If it's red, you gain a point. You can chose to stop at any time. What's the expected value of this game?7 AnswersUse a results table and start from the ending state (0,0) = no cards left, calculate it's expected value, and work backwards from that state knowing that you can always choose to continue the game until you break even (reach (0,0)).2/3how do you get 2/3?Show More ResponsesI got 3/4 ~Draw a tree. If you get 1 point, you prefer to stop because that's 1/2 expected value, and otherwise you have (1/2*1/3) to get 2, which is 1/3 expected value. Than find that in the other direction you have to play (0.5*(2/3)*0.5) just to get 1.It's 1. The person who suggested the tree is correct. You add 1/2 + 1/3 + (0.5*(2/3)*0.5) you get 1. In this game you will never lose.Could you elaborate on getting the answer 1? I get 2/3 as the final answer: With probability 1/6 you get 2 ( draw two red cards and stop). With probability 1/2x2/3x1/2 you get 1-1+1 (red-black-red cards and stop). With probability 1/2x2/3x1/2 you get -1+1+1 (black-red-red cards and stop). So the expected value is 2x1/6+1x1/6+1x1/6=2/3. All the other situations, i.e., brbr, bbrr, rbbr give you zero dollars.

Feb 26, 2010
 How many digits are there in 2^50?6 Answerslog2(2^50) / log2(10) = 50 / 3.3 ~= 152^50 / 10 = 16 * 2^46 *1.616 digits: 2^50 = (2^10)^5 = 1024^5Show More Responsesevery 3 powers of 2 the number of digits increases by 1. 50/3 = 16.67, so there are 16 digits.With these brain teasers, is it allowed to use scratch paper? Is there a favored approach between trying to work quickly and mentally as opposed to thoroughly and methodically? I can see the pro's and con's of both, and I'm just curious whether I should practice one way or another.First answer was close, but realize its going to be 50*log(3) ~ 15.05. Since this will be 1.xxx * 10^15, thats 16 digits. (1*10^1) is 2 digits

### Summer Internship - Quantitative Investment Strategies at Goldman Sachs was asked...

Apr 10, 2012
 Suppose that you have a fair coin. You start with \$0. You win 1\$ each time you get a head and loose \$1 each time you get tails. Calculate the probability of getting \$2 without getting below \$0 at any time.5 AnswersDraw a tree to have an idea of how to compute the probability. The flips are all independent with each other.1/31/3 Because if he requires 2 heads = \$2, then it means he require HH out of {HH,HT ,TH,TT} We cannot consider TT because it would turn to zero. Therefore, he has 1 chance out of 3. Hence 1/3.Show More ResponsesEach flip has a .5 possibility of being head. (.5)(.5)=.25 or 1/4Marchov chain with 2 and -1 absorbing OR geometric series 1/4^n from 1 to inf=1/3