Assistant Interview Questions in New York, NY | Glassdoor

Assistant Interview Questions in New York, NY

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Jul 20, 2010

Feb 16, 2010
 You roll a die until the sum of the integers rolled is greater than 13. What number are you most likely to stop on?7 Answers14the probability is monotonically decreasing after the first stopping sum, i.e., Smith is correct. 14Can someone elaborate on this?Show More ResponsesEach roll has expected value of 3.5. So you're expected to get 14 after 4 rolls.We can express the probability of ending up on 14 recursively, in terms of the probabilities of ending up on the numbers from 8 to 13. If we end up on 14, the last roll we made must, of course, have been some number from 1 to 6, each with probability 1/6. We can then express p(14), the probability of hitting a 14, as p(8)*(1/6) + p(9)*(1/6) + ... + p(13)*(1/6). Similarly, we can express p(15) = p(9)*(1/6) + ... + p(13)*(1/6). Note that there is no p(14)*(1/6) term in p(15), because we cannot get to a dice value of 15 through a value of 14, since the game would be over in this case. Thus, p(15) < p(14), p(16) < p(15), ..., and we choose to stay with 14the question is actually much more tricky than it seems and the last post is very wrong. P(14)>P(8)*(1/6)+...+P(13)*1/6. however they got the answer right the key is to understand that each roll is independent and of course each outcome is equally likely. hence if I am at 13 I'm equally likely to end up at 14-19. if I'm at 12 I'm equally likely to end up at 14-18 since i can either get directly to these points or pass through 13 and get there. in similar fashion at 11 I'm equally likely to end up at 14-17. each each having probability 1/6+1/6*1/6+1/6(1/6+1/6*1/6). the same argument goes down to if I'm at 8 by recursiveness. this means that before the last role I have to be at 8-13 and at none of those positions am i less likely to end at 14 than anything else. in fact I'm mostly likely to end at 14 if I am at 8. since being at 8 has positive probability I'm mostly likely to end at 14.4 times, since expectation number = 3.5/each time

Feb 27, 2011
 you have five coins. One is double-headed. Pick one coin at random without looking and throw it 5 times. Suppose the outcome are five heads, what is the probability that the coin picked is the double-headed one?6 Answers8/9(1/2*1)/(1/2*1+1/32*4/5)use Bayesian's formula (1/5*1) / (1/5*1 + 4/5 * 1/32) = 8/9Show More ResponsesThis is basic Bayes theorem. Let X be the event that the double-headed coin is picked. Pr(X | Outcome) = Pr(Outcome | X) * Pr(X) / Pr(Outcome) Pr(Outcome | X) = 1 Pr(X) = 1/5 Pr(Outcome) = 1/5 + 4/5*1/32 = 36 / (5*32) Plug everything in, you've got: Pr(X | Outcome) = 32/36 = 8/9.sorry... its 8/9 (cant del my prev post)use Bayesian's formula (1/5*1) / (1/5*1 + 4/5 * 1/16) = 4/5

Nov 29, 2010

Feb 26, 2010
 How many digits are there in 2^50?5 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.

Oct 15, 2010
 Calculate 68% of 132 up to two decimal numbers without calculation4 Answersa = (132 * 70)= 9240 b = (132* 2) = 264 (a-b)/100 = (9240-264)/100 = 89.7668*132 = 100^2-32^2 = 10000-1024 = 8976The idea is: 1) (68+32)^2=100^2 2) (68+32)^2= 68^2 + 2*68*32+ 32^2 =68(68+2*32)+32^2=68(100+32)+32^2=68(132)+32^2. Letting 1) =2), yields 100^2-32^2=68(132). This works for multiplications of the formShow More ResponsesEven simpler, 132*68=(100+32)(100-32)=100^2-32^2

Jan 11, 2012
 Expected value of the products of a six-sided die and a eight-sided die.4 Answers15.75Write down the table with 48 entries (6 rows x 8 columns), sum them up and divide by 48: 756 / 48 = 15.75.EV(6) = 3.5 EV(8) = 4.5 3.5 * 4.5 = 15.75Show More ResponsesLast answer is right. Never ever try to tell them to write the whole table, its better to say u cant solve it.

Jan 11, 2012