# Analyst intern Interview Questions

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Analyst Intern interview questions shared by candidates### An analytical question about: Say you have ten stacks of coins with ten coins in each stack, each coin weighing 1g. However, one of the stacks has 2g coins instead of 1g coins. If you have a weighing scale and can only make one measurement, how do you determine which stack of coins weighs 2g each coin?

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It's simple. Take 1 coin from the first stack, 2 coins from the second, 3 from the thirds - you see where this is going - finally 10 from tenth stack. put them on weighing scale. If in result you get 56 - it was first stack, 57 - second.... 65 - tenth stack. Less

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1-If all the coins are made from the same material and same diameter, then the heavier coin stack should be slightly taller. 2- Weight all the stacks at once, and remove each stack until you see a dip in the tota weight . Less

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I didn't quite understand what you meant. So there are 10 coins, 9 weigh 1 gram, and one weighs 2 grams? The second sentence seems contradictory. If you can only make one measurement, then you would divide the stacks into two. Place each stack on either scale, and whichever one is heavier contains the 2g coin. Less

### I'm giving you a choice between two games. In the first game, you roll two six-sided dice. For every number by which their sum exceeds 10, I will give you $1 and $0 if their sum is equal to or less than 10. In the second game, you still have two dice but once is already set to 5 so you can only roll one die. The same rules apply: you get a dollar for each number over 10 and get nothing if the sum is equal to or less than 10. Which game would you prefer to play, and how much would you be willing to pay me to play the preferred game over the other game?

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Expected value of the first game is (2/36)($1) + (1/36)($2) = 1/9 of $1. The expected value of the second game is (1/6)($1) = 1/6 of a dollar. Therefore, the second game is preferred and you would pay (1/6 of $1) - (1/9 of $1) = 1/18 of $1 to play the second game instead of the first game. Less

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I believe the interview candidate's answer is incorrect. The expected value of game #1 is (3/36)($1) + (33/36)($0) = 1/12 of $1. The expected value of game #2 is (1/6)($1) = 1/6 of $1. The second game is preferred, and you'd pay 1/12 of $1 to play the second over the first. Less

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The interview candidate's answer is correct. The wording states that in the first game, you receive $1 for every number by which the sum of the dice exceeds 10. This means that rolling a sum of 12 gives you $2 and rolling a sum of 11 gives you $1, so the expected value of the first game is indeed 1/9 of $1. Less

### A robot encounters a bunch of coins spread out on the ground. It examines each coin and, if the coin is heads, turns it to tails with 100% probability. If the coin is tails, it turns it to heads with 50% probability. He examines each coin once per cycle. If you have the robot cycle through the coins a very large number of times, what will eventually happen to the proportion of heads and the proportion of tails?

3 Answers↳

Each coin is independent; let p(H) denote the probability of the coin being heads and p(T) denote the probability of the coin being tails in the long run. To get a head, note that we must be at a tail, and transition with 0.5 probability. So, p(H) = 0.5 * p(T). To get a tail, note that we can (1) be at a tail and transition to tail with 0.5 probability, or (2) be at a head and transition to tail with 1 probability. So, p(T) = 0.5 * p(T) + p(H). Lastly, p(H) + p(T) = 1. Note that the first two equations give the same thing: p(T) = 2 p(H). Substituting into p(H) + p(T) = 1, we get p(H) = 1/3, p(T) = 2/3. So, over time, we'll end up with 1/3's of the coins as heads and 2/3's of the coins as tails. Less

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It will eventually reach and remain at 1/3 heads and 2/3 tails.

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The answer above is good. Another way to phrase it is that at round n, the prob for a single coin to be H is p_H (n) = 1/2 p_T (n-1). Parenthesis indicate functional dependence. So we can say p_H(n) = 1/2 (1-p_H(n-1)). If we assume a stable solution exists at infinity, then it must be independent of n. So guess p_H(n) = const. Substitute and solve for const to get 1/3. Less

### Most of them were basic questions. One was " Write down Java classes and methods that you required for a Transaction" "How do you convey a Technical banking transaction to a street man"

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how do you convey a technical banking transaction to a street man? Can someone answer this?! Less

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so what are the classes and methods required for a transaction?

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I think every one knew it.

### What is put option?

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An insurance that you pay to hedge against price changes in the future. buy put to hedge against price going down. Less

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Where you “put” options on high or low before a date

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A put option is a right but not obligation to sell the underlying asset at the strike price at maturity. One put option contains 100 shares of the underlying asset. A put option can be an effective tool for managing risk and uncertainty. Using a protective put one can hedge the potential downside risk of the portfolio, with only giving up a small portion of the upside potential, which is the premium paid for the option(s). This is known as the breakeven point. Less

### if you had two classes, a circle class and an oval class, which class would be the super class and which one would be the subclass

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Oval would be the super class and circle would be the sub class which extends oval. As every circle is an oval whereas every oval is not a circle Less

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I dunno they're both pretty class. but circles would be my fave, like when ur on a roundabout and you keep going round with a big spliff in ur hand. it's sooo good Less

### The number of street lights in your hometown?

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Hey can you help me with assumptions for these?

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Do justify your calculations including the assumptions.

### What are 3 numbers the CFO would look at on a daily basis?

2 Answers↳

Anybody have an answer for this?

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ROI, percent change, Futures