Procurement Internship was asked...March 10, 2014

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Call 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 Less

Internship was asked...July 8, 2019

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I would really love to work at nick so that I can prove that I can handle tough competition like this. Also to inspire girls in my area Less

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I would really love to work at nick so that I can prove that I can handle tough competition like this. Also to inspire girls in my area Less

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So as get rid of stage fright and to improve my talent

Assistant Trader Summer Internship was asked...February 26, 2010

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HHT is more likely to appear first than HTT. The probability of HHT appearing first is 2/3 and thus the probability of HTT appearing first is 1/3. Indeed, both sequences need H first. Once H appeared, probability of HHT is 1/2 (b/c all you need is one H), and probability of HTT is 1/4 (b/c you need TT). Thus HHT is twice is likely to appear first. So, if the probability that HTT appears first is x, then the probability that HHT appears first is 2x. Since these are disjoint and together exhaust the whole probability space, x+2x=1. Therefore x=1/3. Less

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Let A be the event that HTT comes before HHT. P{A} = P{A|H}P{H} + P{A|T}P{T} = .5P{A|H} + .5P{A|T} P{A|T} = P{A} therefore, P{A|H} = P{A|T} P{A|H} = P{A|HH}P{H} + P{A|HT}P{T} = (0)(.5) + P{A|HT}(.5) Therefore, 2P{A|H} = P{A|HT} P{A|HT} = P{A|HTT}P{T} + P{A|HTH}P{H} = (1)(.5) + P{A|H}(.5) 2P{A|H} = .5 + P{A|H}(.5) P{A|H} = 1/3 and P{A|H} = P{A}, therefore, P{A} = 1/3 So, HHT is more likely to appear first and it appears first 2/3 of the time. Less

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Above link is the best solution I have seen for this problem http://dicedcoins.wordpress.com/2012/07/19/flip-hhh-before-htt/ Less

Quality Engineer Internship was asked...August 11, 2022

Internship was asked...December 14, 2016

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divide the te ams

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looking the problem size and his effect and try to solve athore wise request for helping the admin Less

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Im currently working in danger place who had the biggest enemies to humanity ( taliban , al qaeda ....) . When i started working i was explained about my employer idea and i convience all of them ,we are not enemies we are just trying to help humanity Less

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1/3

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1/3 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. Less

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Marchov chain with 2 and -1 absorbing OR geometric series 1/4^n from 1 to inf=1/3 Less

Quant Internship was asked...May 6, 2011

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Put one red marble in one drawer and all the others in the other drawer.

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In 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 = .7475 Less

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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. Less

Software Engineer Internship was asked...February 25, 2013

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First, you tackle with a small 3x3 grid and see if the sum of the numbers match the expected sum of numbers from 1 through 9 because each number has to be distinct in a sudoku. Then you repeat this for 9 mini-sudokus. Finally, you check if for 9 rows and 9 columns if there are any missing numbers or duplicate numbers. Less

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But there could be a duplicate number in the small 3X3 grid. This would make it invalid sudoku but you would not catch this if you just look at the expected sum Less

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A duplicate number in the small 3x3 grid would not give the desired sum, because the desired sum in a 3x3 Sudoku grid is a constant i.e 55 Less

IBD Strats Internship was asked...April 21, 2012

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how do you get 2/3?

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2/3

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Use 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)). Less