# Summer intern Interview Questions in Australia

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Summer Intern interview questions shared by candidates### Assume you are the people who help balance between student campus and the society. The student campus set a budget of 50%, but the society set a budget of 75% after conference. And both of them don't want to decrease their budget. What will you do ?

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Got a similar question. Interviewed on the 26th of January. Will hear back in a couple of weeks too! Really scared. Less

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Got the offer!

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I also haven't heard back

### Flip a coin until either HHT or HTT appears. Is one more likely to appear first? If so, which one and with what probability?

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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

### 3) Poker. 26 red, 26 black. Take one every time, you can choose to guess whether it’s red. You have only one chance. If you are right, you get 1 dollar. What’s the strategy? And what’s the expected earn?

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There is symmetry between red and black. Each time you pull a card it is equally likely to be red or black (assuming you haven't looked at the previous cards you pulled). Thus no matter when you guess you odds are 50% and the expected return should be 50 cents. Less

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The problem should be random draw card and dont put it back. Every draw you have one chance to guess. So the strategy is after first draw you random guess it's red. If correct you get one dollar, next draw you know there is less red than black. So you guess black on next draw. Else if first guess you are wrong, you guess red on next round. It's all about conditioning on the information you know from the previous drawings Less

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The problem statement is not very clear. What I understand is: you take one card at a time, you can choose to guess, or you can look at it. If you guess, then if it's red, you gain $1. And whatever the result, after the guess, game over. The answer is then $0.5, and under whatever strategy you use. Suppose there is x red y black, if you guess, your chance of winning is x/(x+y). If you don't, and look at the card, and flip the next one, your chance of winning is x/(x+y)*(x-1)/(x+y-1) + y/(x+y)*x/(x+y-1) = x/(x+y), which is the same. A rigorous proof should obviously done by induction and start from x,y=0,1. Less

### Nel test scritto erano presenti domande di geometria, logica generale e fisica

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Ok grazie,io ho fatto il colloquio orale tecnico giovedì scorso e non ho ancora ricevuto l'esito. Less

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Ok grazie,io ho fatto il colloquio orale tecnico giovedì scorso e non ho ancora ricevuto l'esito. Less

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Qualcuno che ha fatto il colloquio per summer job e ha avuto esito dell'andamento del colloquio? Dopo 2 settimane non ancora ricevuto notizie Less

### A bad king has 1000 bottles of wine. A neighboring king plots to kill the bad king, and sends a servant to poison the wine. the servant is able to poison only one of them before he is caught. The guards don't know which bottle was poisoned, but they do know that the poison is so potent that even if it was diluted 1,000,000 times, it would still be fatal. Furthermore, the effects of the poison take one month to surface. The king decides he will get some of his prisoners in his vast dungeons to drink the wine. At a minimum how many servants does he need to kill to find out the poisioned bottle of wine and will still be able to drink the rest of the wine in 5 weeks time?

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Simply saying "binary search tree" does not solve this problem in its entirety because of the five week time limit. If you just naively used a binary search, you'd take four weeks for the first iteration alone (the problem says the poison's effects take a month to surface, so that definitely wouldn't work. 10 prisoners lined up. Number the bottles from 0 - 999. Now, divide the bottles into blocks of powers of 2. Prisoner 9 drinks from 0 - 511. Prisoner 8 drinks from 0 to 255 and 512 to 999. Prisoner 7 drinks from 0 to 127, 256 to 383, 512 to 639, 768 to 896 Prisoner 6 drinks from 0 to 63, 128 to 191, 256 to 319, 384 to 447, 512 to 575, etc. Prisoner 1 drinks from every second bottle A month later, line up the Prisoners and treat the dead ones as 1s and the living ones as 0s - you'll have the answer in binary and find the dreaded bottle! Less

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10 servants

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The way the question is phrased, doesn't the king only need to sacrifice 1? If 1000 servants each sip a little of one of the wines, then after 5 weeks, only 1 will die? Less

### You are playing a game where the player gets to draw the number 1-100 out of hat, replace and redraw as many times as they want, with their final number being how many dollars they win from the game. Each "redraw" costs an extra $1. How much would you charge someone to play this game?

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This is all very interesting and I'm sure has some application...but to trading? I don't think so. I own a seat on the futures exchange and was one of the largest derivatives traders on the floor. Math skills and reasoning are important but not to this level. I would associate day trading/scalping more to race car driving i.e. getting a feel for what's going on during the day, the speed at which the market is moving and the tempo for the up and down moves. If I were the interviewer at one of these firms, I throw a baseball at your head and see if you were quick enough and smart enough to duck. Then if you picked it up and threw it at my head I'd know that you had the balls to trade. I know guys who can answer these questions, work at major banks, have a team of quants working for them and call me up to borrow money from me because they're not making money. At the end of the day, if you want to be a trader then...be a trader. If you want to be a mathematician then be a mathematician. It's cool to multiply a string of numbers in your head, I can do this also, but never in my trading career did I make money because in an instant I could multiply 87*34 or answer Mensa questions which...realistically the above answer is: it depends on the market as the market will dictate the price. You may want to charge $87 to play that game but you'd have to be an idiot to play it. In trading terms this means that when AAPL is trading at $700 everyone would love to buy it at $400. Now that it's trading at $400 everyone is afraid that it's going to $0. Hope this helps. No offense to the math guys on this page, just want to set the trading record straight. Less

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All of the above answers are way off. For a correct answer, see Yuval Filmus' answer at StackExchange: http://math.stackexchange.com/questions/27524/fair-value-of-a-hat-drawing-game Less

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Let x be the expected value and n be the smallest number you'll stop the game at. Set up equation with x and n, get x in terms of n, take derivative to find n that maximizes x, plug in the ceiling (because n must be integer) and find maximum of x. ceiling ends up being 87, x is 87.357, so charge $87.36 or more Less

### a family have 2 kids. if you have seen 1 girl , what is the proberbl the other kids is a boy

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answer is 1/2 - the question is very subtle...

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chainsaw is right - the question is subtle. Outcome space is (b1,b2), (b1,g2), (g1,b2) and (g1,g2). Now - if you HAVE SEEN the girl (specific girl let's say girl 1) that removes outcomes (b1,b2)&(b1,g2) leaving us with p=1/2 of choice between outcomes (g1,b2) and (g1,g2). If you KNOW (was told) that one is a girl, but do not know which one, than only (b1,b2) is removed from outcome space and the chance is 2/3. Less

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isnt it 1/2? BG (boy and girl) and GB (girl and boy) are the same. order is not important. or am i missing something? Less

### how many times the hour and minute hands of the clock form right angle during one day?

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Let me show you a mathematical approach. Common sense dictates that the minute hand moves at a faster rate of 5.5 degrees a minute (because the hour hand moves 0.5 degrees a min and the minute hand moves 6 degrees a minute). We start at 12 midnight. The hands are together. For subsequent 90 degree angles to occur, the minute hand must "overtake" the hour hand by 90 degrees, then 270 degrees, then 360 + 90 degrees, then 360 + 270 degrees, then 360 + 360 +90 degrees.. and so on. This can be re-expressed as: (1)90, 3(90), 5(90), 7(90), 9(90), 11(90)... n(90). The number of minutes this takes to happen can be expressed as (1)90/5.5, 3(90)/5.5, 5(90)/5.5, 7(90)/5.5, 9(90)/5.5, 11(90)/5.5... n(90)/5.5. In one day, there are 24 hr * 60 mins = 1440mins To find the maximum value of n, n(90)/5.5 = 1440 n = 88 but as seen from above, n must be an odd number (by pattern recognition and logic) hence n must be the next smallest odd number (87) counting 1,3,5,7,9,11......87, we see that the number of terms = (87-1)/2 +1 = 44. In other words, the minute hand "overtakes" the hour hand on 44 occasions in 24 hours in order to give a 90 degree angle. Therefore the answer to your question is 44. Less

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Relative speed is 5.5 degree/min. Time is 24*60 mins. Total distance is 5.5*24*60 degrees. How many full circles it is? 5.5*24*60/360 = 5.5*4 = 22. Each full circle contains 2 right angles (90 and 270). So answer is 22*2 = 44. Less

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4 times?

### 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.

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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. Less

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For reference to the 3-points-on-same-semicircle problem, see e.g., http://godplaysdice.blogspot.com/2007/10/probabilities-on-circle.html Less

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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? Less

### At a party everyone shakes hands, 66 hand shakes occur, how many people are at the party?

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12

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You can suppose there are n people in the room and think of them in a row. The first one has to shake hands with (n-1) people (because he doesn't have to shake hands with himself). The second one has already shaken hands with the first one, so he has (n-2) shakes remaining... and so on. So you have to sum: (n-1)+(n-2)+(n-3)+...+1= (n/2)*(n-1) Then you have to solve (n/2)*(n-1)=66 and you get n=12. Less

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n(n-1)/2=66 so n=12