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
 Have a truck starting that can carry a maximum of 1000 apples, and starts with 2000 apples at a depot. What's the maximum number of apples that can be carried across a 1000 mile desert assuming 1 apple falls off the truck for every mile travelled?4 Answers500Not sure if this is optimal, but: Load up 1000 apples, drive 1/2 of the way there. Repeat. [The number 1/2 comes from the observation that this gives you your maximum carrying capacity at the 1/2 way point. You pay 2 apples per mile when you have to take two separate trips to move a load. So you want to take as few trips as possible.] You now have 1000 apples at the 1/2 way point. Now drive the remainder of the way: You lose 1000/2 apples en route, but keep 500. So you get that many to your destination. What if the numbers were different? Say that you were trying to move 3000 apples instead. Then, following the strategy above, you would load 1000 apples and drive 1/3 of the way there. Take two more such trips. So now you have 2000 apples at the 1/3 point. Now load up 1000 apples and drive 1/2 of the total distance. Repeat. You now have 1000 apples at the 1/2 + 1/3 = 5/6 point. Now load up those 1000 and drive the remaining distance. You keep 5/6 of the 1000 apples, or roughly 833. Again, I'm not sure if that's optimal. It's just one strategy. If someone thinks of a better thing: Please post it!I don't think above is correct. You drop an apple when you make your way back. I got 1000/3Show More ResponsesI haven't gone completely through the math, but in the case where there are a total of two stops along the way.... Let the first stop be A and the second be B (given two points we'll assume the trip goes from left to ri. The picture should then look like this. start-------A---------------B--------------------------finish The number of applies that are left at stop A are 1000-A and 1000-B respectively, assuming that each trip is made separately. Let's now assume that we're going to go to A, pick up all of A's apples, go to B, and finally then go to the rest of the way. Let's pick up 1000-A apples at point A, and then lose B-A of these applies to yield 1000-B apples still left in the truck once we arrive at B. Then pick up the 1000-B apples at B. ....At this point its clear that we have 2000-2B apples in the truck. This value is less than or equal to 1000. The optimal value is to have all of the trucks space filled, or have 2000-2B= 1000. This means that B=500 From here, a total of 500 apples are delivered. Of the 2000 apples, A+B are used. Given that B=500, and you want the minimal value used with the 500 delivered. Since 1000 are wasted total, given the loss of one apple per mile, A=500 as well. Generalizing this strategy. you can make 2 starting trips of 1000 apples and drop them both off at the 500 mile marker. However, as Blah above me concluded, the maximum number of apples you can deliver is 500. This is based on the condition that the truck is loaded at the start point twice, each time up to its maximum capacity of 1000 apples.