# 5 k

Questions d'entretien pour Strategist partagées par les candidats

## Principales questions d'entretien

Trier: Pertinence|Populaires|Date
On a demandé à Associate Account Strategist...2 décembre 2014

### If you wanted to bring your dog to work but one of your team members is allergic to dogs what would you do?

11 réponses

I would leave my dog at home

Give him benedryl

Non-drowsy of course

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### If you look at a clock and the time is 3:15, what is the angle between the hour and the minute hands?

9 réponses

7.5 deg. The hour hand moves 30 deg for a full minute hand rotation 60 min (time). So when the minute hand moves 15 min, the hour hand makes 1/4th of 30 deg = 7.5 deg. Moins

7.5. But ask them, do they mean a digital or analog clock...

1 clock wise 12 hours is 360 degrees. 15 mins is equivalent for 29 degrees. Why? 360/15 is 29 Moins

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### If Google decided to charge g-mail users, how would you recommend implementing this? Would it be sustainable and what would be the advantages and disadvantages?

9 réponses

Since it's a Google Interview question, we've got to give an answer, even if a paying Gmail doesn't make sense since the business model is on advertising. I would go for a freemium model with an enhanced functionalities+ awesome interface Gmail client, working offline, unlimited space (1To for instance). Target would be 5% customer base of 1billion =50millions paying customers Fee : $2 /month Revenues =$1.2b /year Moins

C Free

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### Suppose you toss a fair coin 400 times. What is the probability that you get at least 220 heads? Round your answer to the nearest percent.

7 réponses

This is a central limit theorem question. The trick is to view each toss as a random variable that returns 1 if a head is tossed and 0 if a tail is tossed. Then each such random variable has expected value 1/2 and variance 1/4. So your Z-variable (for using the central limit theorem) will be: (220-200)/(sqrt(400*(1/4))) = 20/10 = 2 So we've reduced the question to asking what's the probability that Z takes a value bigger than 2. Recall that on the standard normal, the probability that z takes values between -2 and 2 is about 95%, so the probability that it takes values less than 2 is about 97.5% (it's actually more like 97.7% but just estimating). So the probability that we are bigger than 2 is a little less than 2.5%, which after rounding to the nearest percent gives us 2%. Moins

a typical "binomial distribution with normal approximation" type of question. let r.v. X be # of heads in n=400 bernoulli trials then X~ binomial (400, 0.5) the conditions of using the normal approximation are 1. n being large so there are enough discrete values to approximate ==&gt; 400 is large enough 2. p being "in the middle" (not near 0 or 1) so the binomial is nearly symmetric as is the normal ==&gt; p=0.5 is indeed in-the-middle and a (conservative) rule of thumb of using the normal approximation is min(np, n(1-p)) &gt; 5, which is well satisfied in this case. then P(X ge 220) = 1 - P (X &lt; 220) and have a normal r.v. Y~ normal (np, np(1-p)) (note: sigma_sq=np(1-p)=100 so sigma = 10) P(X&lt;220) approximately = P(Y &lt;220) = P( Z&lt; (220-200)/10) = P(Z &lt; 2) (standard normal distribution) = (1-0.9544)/2 = 0.0228 reference: Statistical Inference by George Casella, page 105, example of normal approximation Moins

What? With 400 tosses you are summing 400 binomial distributions. That's more than enough to use central limit theorem. I agree that you can do it directly from the binomial distribution (which you clearly did), but there is no way someone expects you to sum up those kinds of numbers in your head. So I'm afraid I'll have to disagree with your disagreement. :-P Moins

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### Consider a random walk on a graph in the shape of the capital letter Y, with nodes at A and B at the top, C at the bottom, and O in the center. With probability 1/3, there is a transition from O to A, B, or C that takes 1 time unit to complete. Find the expected time to get to C, starting at A.

7 réponses

All branches go to O with prob 1. So Ec_a = 1 + 1/3 + 2/3Ec_c since the time from b to a is the same by symmetry. Solving we have Ec_a = 4 Moins

I thought this was a geometric series question and came up with the answer of 2, which was clearly wrong. The interviewer tried to help me out, but he and I were using different notation, which we only discovered after 10 minutes of very confusing discussion, in which we both probably thought the other was either insane or stupid. He kept prodding me to "use the symmetry of the situation" (i.e., A and B are basically the same thing), but I wasn't able to see how the symmetry translated into a mathematical result. We moved on without discovering the answer. Moins

if no probability is associated with A to O (A to O is a certain thing to happen) then it would be 1+ 1*(1/3) = 4/3. correct me if im wrong Moins

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### There are 100 prisoners in a cell and you are the warden and you have one bullet. How can you prevent all of them from escaping.

6 réponses

Do not open the cell

Write a note saying that they tried to kill you before committing a suicide.

Shoot one of them and only 99 will escape

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### Q3: You are playing this game with flipping a coin. You can play it as many times as you want until there is a head. At the nth time, when you get a head, they will pay the 2^n dollars. How much are you willing to play this game? What's the fair value?

5 réponses

Ok, not sure why the last sentence gets messed up, trying once more here. The condition is not very enlightening, but it is clear if the two functions are plotted in 3D.: _ Exp[a]+Exp[b] only if a &gt; b-Log[-1+Exp[b]] Moins

raf's answer to the probability question is wrong. P(X+Y+Z 1) = 5/6

The answer to Q1 above here is wrong: there is no unique answer. To see this just consider the easy examples: _ a=b=0 implies Exp[a] + Exp[b] &gt; Exp[a+b] _ a=b=1 implies Exp[a] + Exp[b] Exp[a]+Exp[b] only if a &gt; b-Log[-1+Exp[b]] . Not very enlightening, but it is clear if the two functions are plotted in 3D. Moins

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### Went through your resume how you have the approximate of square root of 37 ? find the location of the number in an array. desrcibe the codes.

5 réponses

We know that the root is greater than 6. Let \sqrt(37)=6+e. Then (6+e)^2=37, so, e^2+12e=1. we know e is small so e^2 is much smaller. Ignoring it, e=1/12\approx 0.083. So the answer is 6.083. Moins

How did you get from 1/36 to 1/72

\sqrt(37) = 6 + 1/(12+1/(12 + 1/12+...))

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### 1) Manufacture and option from plain vanillas that makes money in case of low volatility and does not lose too much in the opposite scenario. 2) We are in a junction. p is the probability that at least one car will pass through it within the next 20 minutes. Give me the probability that no car will pass within the next 5 minutes.

5 réponses

Let q be the probability of seeing no car in any given 5 minutes. Then the prob of not seeing a car in 20 mins is q^4 (or 1 - p). So q = (1-p)^0.25 Moins

For question 2, one needs to use the Poisson distribution. The number of cars passing through within 20 mins distributes according to a Poisson distribution with parameter lambda. 1-p is the probability that no car will pass through within the next 20 minutes. Therefore 1-p=e^(-lambda). The number of cars passing through within 5 mins distributes according to a Poisson distribution with parameter lambda/4. Therefore, the probability that no car will pass within the next 5 minutes is e^(-lambda/4)=(1-p)^(1/4) Moins

1) Butterfly: Call(S0 - K) -2*Call(S0) + Call(S0 + K), K = Sigma*Sqrt(T), T: option expiration. Moins

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### If you flip two coins, one is a head,what is the probability that the other is a head

5 réponses

For crying out loud, it's a half.

1/3 Before conditioning on info, possible states are: HH HT TH TT Since one is a head, only three of the states above are possible. Moins

The problem was probably (at least intended to be): I flip two coins and tell you "one of them is heads, what is the probability the other is also heads". The info you are getting is one of two is heads. As the person above said the initial outcomes are TT, TH, HT, HH. The info you have only eliminates the first possibility, the others are equally probable. So 1/3. The way it was asked above could be interpreted as you flipping one coin and looking at it and seeing heads and then flipping another. In that case they are totally independent since you know the specific coin being discusses. That bit of info changes things. Really depends on how the question was specifically asked. Moins

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