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Mar 20, 2014
 I was asked about if a machine failed and a frustrated production worker was on that machine when it failed, how do you deal with the production worker while you working to fix that machine as they tend to get more frustrated the longer the machine is down? 1 Answer I said that I would give the production worker the knowledge of knowing what was wrong with the machine, and an approximate time as to how long it may take. I would then try to conversate and diffuse the situation with humor.

Mar 11, 2011

Oct 4, 2011
 Say I take a rubber band and randomly cut it into three pieces. What's the probability that one of the pieces has length greater than 1/2 of the original circumference of the rubber band. 9 Answers 3/4 Suppose you have two cuts on the rubber band placed randomly. The probability of having one segment greater than half the circumference is the probability that the third cut will be inside the combined range of 90* to either side of the cuts. Since the average distance between the first two cuts is also 90*, the combined range is 270*, or 3/4 of the circle. You need 3 cuts to end up with 3 pieces. The first cut doesn't matter. The second cut can also be anywhere and the largest piece will still be at least half the circumference. What matters is the third cut, which should lie in the same half as the second cut. So the probability is actually 1/2. Show More Responses The correct answer is 3/4, as this problem is equivalent to the famous 3-points-on-semicircle problem. Why? If one of the pieces has length greater than 1/2 the circumference, then the three points of cutting must lie in the same semicircle. On other hand, if the three points of cutting lie on the same semicircle, then the longest piece must be at least 1/2 of the circumference. For reference to the 3-points-on-same-semicircle problem, see e.g., http://godplaysdice.blogspot.com/2007/10/probabilities-on-circle.html 1/4 1 -3/4 suppose I have two points whose minor arc distance is t <= 1/2. Then the range of semicircles covering both points gives an arc length of (1/2+1/2)-t = 1-t. say we fix the first point, tracing the second point around gives minor arc lengths from 0 to 1/2 and then 1/2 to 0. Therefore the answer is 2*integral (1-t) from 0 to 1/2, which is 2(1/2-1/8) = 3/4 It's 3/4. Cut it into 1 piece make a line. Cut it close. Pretend the length is 100. If you cut the first at x=1, as long as it isn't between x=50-51, it will have a length greater than 50% so there's 99% chance. You can imagine that if the cut was infinitely close to the end it would be about 100%. Now cut at x=2 you can't do between x=50-52. For x=3 it's 50-53 etc. So when you get to right to infinitely close to 50 it is pretty much between x=50-100 so there is a 50% chance you hit your spot. (obviously 50-50 is 100%, but since this length is continuous there's little chance it lands on that point). Obviously since this is symmetrical you can see this pattern going from 50% to 100% at the other end. Since each point on the continuous line has the same probability of happening the answer is clearly 75%. This problem is also equivalent to the probability that, if you have a line segment from 0 to 1 and you make 2 random cuts on that line segment, what is the probability that the three resulting pieces do NOT make a triangle?

Jan 8, 2012
 I have 10 cards face-down numbered 1 through 10. We play a game in which you choose a card and I give you the corresponding dollar amount. a) What is the fair price of this game? b) Now, after picking a card you can either take the dollar value on the card or \$3.50. Also, cards worth less than 5 are now valued at \$0. What is the maximum price you are now willing to pay for the game? 9 Answers Fair value is 5.5 2nd game: FV = (3.5 x 5 + 6 + 7 + 8 + 9 + 10)/10 = 4.75 You decide on taking the dollar value on the card or \$3.50 *before* seeing the number on the card. And you get \$5, if you open that card, not \$0. So: 2nd game: (1/2)*3.5 + (1.2)*(1/10*[5+6+7+8+9+10]) = 4 3.5(.4) + (4.5)*.6 = \$4.10 3.5 because this is what 40% will select, 4.5 as this is the avg of 5-10*1/10. There is not a 50% chance of selecting 3.5 @Your are WRong @ EASY. There is not a 10% chance of receiving \$3.5 as numbers under 5 are worthless this will /should change your perspective. Show More Responses FJM is correct. "Cards worth less than 5 are now valued at \$0" - that does not include card number 5. how does 4.10 make sense if 4.5 is the ev of taking the card..ii think you are incorrect. can you explain your reasoning. using 1/10 to take your average and then multiplying by .6 seems like you are double counting Total amount=3.5*4+(5+6+~+10) Expected value=total/10=5.9 @FJM no idea where your 4.5 coming from OK agreed with xinzhuo. Should be 7.5 instead of 4.5 in FJM's calculation, which gives 5.9. Agreed. I double counted. 3.5*.4 + 7.5*.6 3.5 was given 7.5 =(5+6+7+8+9+10)/6 "You decide on taking the dollar value on the card or \$3.50 *before* seeing the number on the card." 2nd game: Obviously, the game price should be no less than 3.5, otherwise there is an opportunity for arbitrage. Now that the price is higher than 3.5, it makes no sense for a rational player to choose just take 3.5 dollar and exist the game. The player should always choose to bet. And the expectation for betting is (5+6+...+10)/10 = 4.5.

Oct 13, 2010

Apr 14, 2010
 How many ways can you arange five people at a round table, such that there they are in increasing age order (clockwise or anti-clockwise)? 5 Answers Trick question, you can't. Oldest will always meet youngest Was asked this question (phrased as "what is the probability that 5 people with different birthdays will sit in increasing age order around a round table (clockwise or counter-clockwise), and the recruiter said that you should ignore the jump from the oldest to youngest. In other words, it's not a trick question -- you actually have to solve the problem. The answer is 1/12. The question is answer how many ways, nothing about probability. There are 10 ways to seat the people, 5 clockwise and 5 counterclockwise Show More Responses Assuming there are five seats at the table then there are ten different ways to seat the five people in increasing age order. There are (5-1)! ways of seating 5 people around the table. Of these, there are 2 permutations wherein there will be ordered sitting (one clockwise & the other anti-clockwise). Hence, the probability is 2/((5-1)!) i.e., 1/12

Mar 12, 2012
 If i roll two dice and multiply the two outcomes, what is the probability of a perfect square? 6 Answers 2/9 I think 1/6 - just the probablity that the numbers on the die are the same Total Outcomes = 36 Perfect square= 4 or 9 Odds of getting 4 are 3/36 Odds of getting 9 are 4/36 Probability of landing perfect square= 7/36 = 0.19 Show More Responses 2/9 is correct. 6 combinations that are the same on both plus 1,4 and 4,1 -> 8/36 = 2/9 ic is talking about the sum, which is not what's asked 1,4,9,16,25,36 1-1 rolls 1 (1 combo) 1-4 (2 combos) and 2-2 (1 combo) 3-3, 4-4, 5-5, 6-6 roll a combo each So the answer is 8/36 which reduces to 2/9 2/9

Aug 17, 2011
 We play a game. You flip two coins, if you get both heads I give you \$1. If you get both tails you flip again. If you get one of each, I pay you nothing. What's the expected value of this game? 5 Answers Dismiss TT since it is not part of the sample space (ie there is no outcome from it) You are twice as likely to get one of each (H-T or T-H) versus getting both heads answer: (2/3)(\$1) + (1/3)(0) = 66.7 cents E=0.25x1+0.5x0+0.25xE 0.75xE=0.25 E=0.33 interviewer told me .667 is correct Show More Responses HT and TH add up to 1/2 of the probability with zero payout. HH is 1/4 probability with \$1 payout. TT results in a "reset" so we can disregard. After we normalize the probabilities, we get: E[payout]= (1/3)*\$1 + (2/3)*\$0 = \$0.33 You can also see it a geometric series. By writing E=0.25x1+0.5x0+0.25xE, Kerzane could as well have written E?0.25x1 + 0.5x0 + 0.25x(0.25x1 + 0.5x0 + 0.25x(0.25x1 + 0.5x0 + 0.25x(....))) This is quite simply the geometric sum: (n-> infinity) E=0.25 + 0.25^2 + 0.25^3....0.25^n which is equal to 0.25/(1-0.25) = 0.25/0.75 = 1/3 or 0.333

Sep 28, 2010
 What is 29^2? 4 Answers 841 Mental math. You could do it mentally. 29 is close to 30. So the upper bound is 30*30 = 900. You would have to subtract 30 and 29. 900- 59 = 841. or, take 29 x 30 to get 870. Take away one 29, get 841. Same idea, but more elegant. You've multipled 29 thirty times, you only need 29 of them, so take one away. Show More Responses Easiest way I can think of is to mentally calculate (30-1)(30-1) and then mentally foil the answers. You end up getting 900 - 30 - 30 + 1 = 841 which is pretty easy to do in your head.

Sep 28, 2010
 What is the expected value to you of the following game: You and another person have 3 coins each. You both flip all of your coins and if your three coins show the same number of tails as the other person, you pay them \$2, otherwise they pay you \$1. 4 Answers 1/16 Here's the explanation to the above (Since none was given): There are four different number of Tails that can be shown when you flip 3 coins: 0,1,2,3. The Probability of 0 Tails, P(0T) = 1/16, P(3T) = 1/16, P(1T) = 3/16, P(2T)=3/16. Since your flips and your opponent's flips are independent, we know that the Probability that I roll 0 tails and you roll 0 tails = P(0T)*P(0T). Thus, square each of the above probabilities and sum them: you get 1/64+1/64+9/64+9/64=20/64. Thus P(same number of Tails) = 20/64 = 5/16. Now it is asking you for expected value, so plug the above probability into an expected value equation: (5/16)(-2) + (11/16)(1) = 1/16 ie, 5/16 of the time you roll the same number of tails and lose two dollars, 11/16 of the time you don't and you win a dollar, thus your expected value of each play of the game is 1/16 of a dollar. (In the long run if you play many times you should expect to win that much per play) The above has the correct answer, but the probabilities are 1/8, 1/8, 3/8, 3/8. Then square them....and so on. Show More Responses Probabilities are: p(0)=1/8, p(1)=3/8, p(2)=3/8, p(3)=1/8.
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