# Fragen in Vorstellungsgesprächen in Yung Shue Wan, Hongkong

Vorstellungsgespräche bei HSBC Holdings in Yung Shue Wan

www.hsbc.com / Hauptsitz: London, UK

234 Vorstellungsgespräche in Yung Shue Wan (von 3.156)

Vorstellungsgespräche bei Bloomberg L.P. in Yung Shue Wan

www.bloomberg.com / Hauptsitz: New York, NY

182 Vorstellungsgespräche in Yung Shue Wan (von 5.452)

Vorstellungsgespräche bei AlphaSights in Yung Shue Wan

www.alphasights.com / Hauptsitz: London

167 Vorstellungsgespräche in Yung Shue Wan (von 2.123)

## Fragen im Vorstellungsgespräch für in Yung Shue Wan

There are 3 coins. One coin has heads on both sides, one coin has tails on both sides, the third one has head on one side and tail on the other side. Now I pick up one coin and toss. I get head. What is the chance that the coin I picked has heads on both sides? 15 Antworten2/3 i got this question too, but i am still not sure how it is calculated why not 1/2? Mehr Antworten anzeigen Because you have 1/3 chance to get double head coin and you will surely get head, 1/3 chance to get single head coin and then 1/2 chance to get head. So the probability of choosing double head coin and get head is 1/3, while choosing single head coin and get head is 1/6. Then, given you get head after tossing, then chance that you chose double head coin is (1/3)/(1/3+1/6) = 2/3 That can't be right. We know that the chosen coin landed heads up, so we can disregard the third (Tails/Tails) coin. Of the remaining 2 coins, there's a 1/2 chance that the chosen one is the double-headed coin. The answer should be 1/2, no? 2/3 is correct... review http://en.wikipedia.org/wiki/Bayesian_inference Let HH, TT, HT be the events defined by picking the head-head, tail-tail, and head-tail coins, respectively. Given that we pick one of the coins randomly, we obtain P(HH)=P(TT)=P(HT)=1/3. Let H and T be the head and tail events respectively. The probability of event H given HH is clearly 1. That is, P(H|HH)=1. Furthermore, P(H|HT)=1/2 and P(H|TT)=0. The probability of a heads is: P(H)=P(H int HT)+ P(H int HH)+P(H int TT)= P(H|HT)P(HT)+ P(H|HH)P(HH)+P(H|TT)P(TT)=(1/2)(1/3)+ (1)(1/3)+(0)(1/3)= (1/3)(3/2). Note that P(H int HH)=P(H|HH)P(HH)= 1/3. Therefore, the probability of event HH given H is: P(HH|H)=P(H int HH)/P(HH)= (1/3)/ ((1/3)(3/2))=2/3. 2 heads on double headed coin, 1 head on the other, P(head is coming from double headed) = 2/3 conditional probability? probability the double headed given a head = probability (head unoin probability double headed coin)/ probability head is thrown. = 1/3 / 1/2 = 2/3 most of you are missing a fact that can't be ignored. Read the question - is says if you draw heads, meaning only 2 coins can have heads, what are the odds that you drew the 2 headed coin is 50%! C1 := 2 tail coin C2 := 1 head 1 tail coin C3 := 2 head coin P(C3|H)=P(H|C3).P(C3)/P(H) = 1.(1/3)/(1/2) = 2/3 P(C2|H)= P(H|C2).P(C2)/P(H) = (1/2).(1/3)/(1/2) = 1/3 This is kind of like the Monty Hall Problem so check that out too. Easiest way to see that it is definitely not 1/2 is to think of the same problem with more possible out comes than just heads or tails: Imagine there are 101 boxes each with 100 balls in The first box has all blue balls, the next all red... etc with different colours in each box until the 100th which is all yellow say. The 101st has 1 ball of each colour. Suppose you randomly pick one box and pick one ball out at random. The ball is blue. Which is more likely? That you happened to pick the 1 multi coloured box out of the 101 boxes and picked the one ball in that box that was blue. Or That you picked one of the 100 boxes that were all one colour and that colour happened to be blue. Hopefully you can see that these are not equally likely events. 2/3 : this is an application of Bayes theorem Mehr Antworten anzeigen So your coins are H/H H/T T/T You see an H so you can't be holding the T/T coin. You are either holding the H/H coin or the H/T coin Now lets mark the H/H coin as H1/H2 and the H/T coin as H3/T If you're looking at the H1 then the other side is H2 If you're looking at H2 then the other side is H1 If you're looking at H3 then the other side is T So 2 out of the three times the other side would be heads. Fun problem pretty easy though. Another way to look at is between the H/H and H/T coin there's 3 heads and 1 tails. You're already looking at a head so there's 2 more heads and 1 tail left. so 2/3 of the time the other side will be heads. |

There are 25 horses. Each time you can race 5 horses together. Now you need to pick the top three horses among them. How many races do you need to conduct? 13 Antworten7 times. after the 5th race the top five will be established; so, after the 6th the top three could be established. Answer is 6. doesn't that assume that 1st in one race is faster than 2nd/3rd/4th/5th in all the other races? can we be sure of that? 1st round: 25 horses and 5 heats => drop bottom 2 in each race 2nd round: 15 horses and 3 heats => drop bottom 2 again 3rd round: 9 horses and 2 heats => drop bottom 2 from large heat and 1 from the other 4th round: 6 horses and 2 heats; first heat has 5 and drop bottom 2 then race again with the 6th horse included 12 heats in total but there's probably a more efficient way Mehr Antworten anzeigen I think it is 7. Kannp, the top five are not established after 5 because the second in one race might be better than the first in another race. Have five races of five each and keep the top 3 in each races. Then, take each of the winners and race them against eachother. The two bottom and the 4 who lost to them are discarded. The two who lost to third place is discarded. And, the one who got 3rd in the race with the horse who gets second in the 6th race is discarded. Now, there are six horses left. Race all but the horse who won twice and keep the top two, combined with the horse who sat out. Now you are done in 7 races. you're over thinking this...... run 5 heats with 5 horses in each one. Using a stopwatch, you take the 3 fastest........ but if you don't have a stop watch - anonymous is wrong... marty is right - anonymous you can't take the winner of each race, because you need the top 3. if a 2nd place from one race is faster than the other 4 winners, you don't have it right. 1st time thru 5 horses by 5 heats drop bottom 2 = 15 left 2nd time thru 5 horses by 3 heats drop bottom 2 = 9 left 3rd time thru one race with 5, other with 4. keep top 3 of each race = 6 left now run any 5 horses, keep top 3 and for last one, run remaining 4,drop last. 12 heats total 7 1) 5 heats take the first 3 =15 horses left 2) run 3rd against each other, take the first, same thing for 2nd and for the 1st take the best 3 = 3 heats 5 left 3) final heat 9 total He gave a very good explanation. I'll decline to explain why prm is incredibly wrong. 7 is correct in assumption that each round ends up with a clear 1-st, 2-nd and 3-rd places. If horses can come simultaneously - than we need 11 rounds (When we have 9 horses left - we set up a round for first 5 and set up a round for the last 4 and a horse that took 3-rd place in the previous round. This leaves us 5 horses to race in the final 11th round). I got stuck with his explanation but once I understood it, it seems very brilliant. Let me put it in this way: 5 groups, each of 5 horses -> race (heat) take the 1st from each group and race (semi-final) discard the 4th and 5th one and all other horses which lost to them in the heat race keep the 3rd one but discard all horses lost to it in the heat race keep the 2nd one but discard the 3,4,5-th horses who raced against it in the heat race keep the 1st one but discard the 4,5-th horses who raced against it in the heat race Now there are 6 horses, namely the 1st from semi-final and the two just following him in heat the 2nd from semi-final and the one just following him in heat the 3rd from semi-final the 1st from semi-final is champion. ignore it and race all other 5 horses in final so 7 in total definitely Russ considers more cases but I am not interested in making it too complicated 7 Hey guys, revicing the things here. Can you guys be more explicit on the 7 figure ? Even with the explainations it’s not clear for me. You say run the 5 winners of the 1st round and discard the last 2, but what if the 5 winners of the first round are actually the 5 firsts ? Forget it I read with my eyes closed, I was lookin for top 5 and it’s top 3 |

Toss a fair coin between two people. If it end with HHT, A wins the game. If it ends with HTT, B wins the game. Whats the probability of A wining the game. 9 Antworten2/3 if order isn't important, than you have for possible outcomes. HHH, TTT, HTT, TTH. You can see immediately there's 1 chance out of 4 for A to win. If order is important, there are 8 possibilities. 1 chance out of 8 for A to win. 1/4 srsly? http://www.glassdoor.com/Interview/Flip-a-coin-until-either-HHT-or-HTT-appears-Is-one-more-likely-to-appear-first-If-so-which-one-and-with-what-probabili-QTN_46824.htm Mehr Antworten anzeigen Once the second person flips their first H, they cannot ever go back to the state where they must roll an H to move on. So after P1 and P2 have both flipped their first H, P2 needs 2 flips to go his way, P1 needs 3, thus has half of the probabilty of A to win. Prob(HHT) * 0.5 = Prob(HTT), Prob(HHT) + Prob(HTT) = 1, ===> Prob(HHT)=2/3 Pa=2/3, Pb=1/3 Pa = 3/8 Let p be the probability player 1 wins. Nothing matters until you roll a H, so let's fast forward to there. If another H is rolled, then player 1 wins automatically so that's (1/2). If a T is rolled, then if T rolls again you lose but if H rolls now then you reset to the past situation. So p = 1/2 + 1/4p so p = 2/3. Clearly A has the upper hand, because A retart from 2 if 3 fails, but B restarts from 1 if 3 fails |

We play a game. Let’s guess the sum of two dices. Who make a guess that is closer to the outcome is the winner. What’s your strategy? 5 AntwortenI choose 7 because if you pick say 6, the closest digit with lower value than 7 (meaning greater range of winning numbers), I get from [7,12], but you only get from [2,6]. And if I choose 7 and you choose say 8, the closest digit with higher value than 8, I get from [2,7], but you only get from [8,12]. Therefore, if I pick 7 I always have a greater chance of being closer to the correct number. the EV of rolling a dice is 3.5, so if you roll a dice twice the number with a greater chance is 7 I have you pick first and bet one step closer to 7 . eg you say 4 I say 5, unless you say 7 then I say 7. Mehr Antworten anzeigen Ben's answer was what I came up. But Pierreou's one is an even better strategy. Pierreou's answer assumes that your opponent does not have the concept of expectation. It is so easy to win if your opponent is ignorant......So I suggest you take the initiative to bet a guess on 7 then the probability of winning is >=6/11 no matter what your opponent's guess is |

You have answered five mental math questions, what’s your confidence that three answers are correct? How about four answers are correct? 5 Antwortenmy confidence is that five mathmatical equations are correct. I'm totally confident that the answer to all 5 questions are correct. What does it matter? Lets move on to the next question. Mehr Antworten anzeigen It matters if you have already answered the interviewed about how confident you are to get the correct answer to each question. For example, if you said that you are 99% confident that you get each of five answers correctly, then the probability that you get 3 answers correctly is 10*(99%)^3(1%)^2 ~ 1%. Now you know it matters! good point by tmbtw. If the question was, you are 90% confident you get a question right, what are the odds you got all 5 right, then you could say (0.90)^5 was the right answer. That's not easy to do in your head, but it comes out to about 59%. The odds of missing all five is more easy, it's (.10)^, which is .00001, or about .001% |

It is noon. The hour hand and minute hand overlap. When is the next time these two hands overlap again? 4 Antworten13:05:300/11 You need to be very accurate into fractions of seconds 1:05:5/11 13:05:5/11 is correct. there is a simple way to think of it. during 12 hours, 11 times of overlapping. so each duration of overlapping is 12/11=1 and 1/11 hour. So the first time of overlapping after noon is 13:05:5/11 Mehr Antworten anzeigen Here is the longer way. The minute hand of a clock travels at a rate of dM/dt=(360 degrees)/(60 minutes)=6 degrees/minute. Note that every hour of a clock spans (360/12 degrees)=30 degrees. The hour hand of a clock travels at a rate of dH/dt=(30 degrees)/(60 minutes)=(1/2) degrees/minute. Now, in your head, imagine the minute and hour hand at 12:00. The minute hand will travel faster than the hour hand and wrap around the clock once. When the minute hand is at 0 minutes or 0 degrees again, the hour hand is at 5 minutes or 1 o'clock or 30 degrees. Now we determine the point at which the minute hand catches-up to the hour hand. To do this, we set the two "degree" functions equal to each other. For 0<=t<=60, M(t)=6t and H(t)=30+(1/2)t. Setting M(t)=H(t), we get that t=60/11 minutes= 5 + 5/11 minutes. Now, we find out how many seconds, approximately, are in 5/11 minutes. There are 60 seconds in a minute, and hence 5/11 minutes= (60*5)/11 seconds= 300/11 seconds= 27 + 3/11 seconds. Therefore, the time at which the minute and hour hand meet is 1:05:27. |

If you have four coins and I have four. We both throw the four and if your four sides equal to mine, I will give you 2 dollar and otherwise you give me 1. Will you do it? 4 AntwortenThe answer is no, as the expected value of the game is (1/256 + 1/16 + 9/64 + 1/16 + 1/256) * 2 - (1 - (1/256 + 1/16 + 9/64 + 1/16 + 1/256)) = -1/4 0 0.0625 0.0625 0.9375 -0.8125 -0.05078125 1 0.25 0.25 0.75 -0.25 -0.0625 2 0.375 0.375 0.625 0.125 0.046875 3 0.25 0.25 0.75 -0.25 -0.0625 4 0.0625 0.0625 0.9375 -0.8125 -0.05078125 Roll P(Roll) P(WIN|Roll) P(LOSE|Roll) Expected Gain Given Roll Expected Gain Expected Return: -0.1796875 In this game of you four tosses, there are five possible combination {HHHH, HHHT, HHTT, HTTT, TTTT}. The probability of my four tosses matching any particular combination is 1 out of 5 (20%). So you shouldn't engage in any bet unless there's atleast a 5-for-1 payout. In the question there's only a 2-for-1 payout so you shouldn't engage in the bet. Mehr Antworten anzeigen output probability 0000 (1) --> 1/16 0001 (4) --> 1/4 0011 (6) --> 3/8 0111 (4) --> 1/4 1111 (1) --> 1/16 \sum \sqr{prob} = 2 / 16 / 16 + 2 / 16 + 9 / 64 = 1/128+1/8+9/64 = 0.27 (+2 dollars) < 1/3 so No. |

hh ht ht tt, one h, probability of the other is h? 3 Antworten1/2, not to use bayes rule ?? Dont understand the question... Assuming your question is 'flipped two coins, one is a head what is the probability the other is a head' then it is 1/2 by independence. If the question is 'flipped a coin twice, one of the outcomes was a head what is the probability the other outcome was a head' then the required probability is P(HH | H) = 1/3 |

Would you feel comfortable working for a person with less experience than you? 2 Antwortenits nothing about experience.... all u get to know at the end is his or hers talent its nothing about experience.... all u get to know at the end is his or hers talent |

The problems are not hard. 3 AntwortenBut you may need well preparation. o(nlogn) solution based on binary search below. An o(n) solution based on hashtable is also possible but consume o(n) memory. public static void main (String[] args) throws java.lang.Exception { Pair p = findPairBySum(15, new int[] {1,2,3,4,5,6,7,8,9,10}); System.out.println(p.x); System.out.println(p.y); } public static Pair findPairBySum(int sum, int[] sortedArr) { for (int i=0; i= 0) return new Pair(first, expected); } return null; } public static int getIndex(int num, int[] sortedArr, int startIndex, int endIndex) { if (startIndex > endIndex) return -1; int pos = startIndex + (endIndex - startIndex) / 2; int valAtMiddle = sortedArr[pos]; if (valAtMiddle == num) return pos; if (num < valAtMiddle) return getIndex(num, sortedArr, startIndex, pos-1); return getIndex(num, sortedArr, pos+1, endIndex); } public static class Pair { public int x; public int y; public Pair(int x, int y) { this.x = x; this.y = y; } } Note that the array is sorted already. O(n) time with O(1) space is feasible. |

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