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Summer Intern interview questions shared by candidates

## Top Interview Questions

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Investment Banking Summer Analyst was asked...July 11, 2012

### What are the methods of valuing a company?

It's DCF(Discounted casfhflow) &amp; Relative method, where FCFE(Free cash flow to Equity) is vital for equity shareholders &amp; FCFF(Free cash flow to Firm) is vital for the company. Less

### Tell me about a time when you were working in a team and your opinion was challenged.

I asked the team member to explain their opinion. No one is perfect, so keeping an open mind when listening to a team member’s response is key. There were times a team member could convince me they were right and other times when I could explain why my opinion may be the better option. Less

### 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 ?

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

Got the offer!

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?

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

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

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?

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

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

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

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

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

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?

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

10 servants

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?

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

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

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)&amp;(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

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

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