Associate Account Strategist was asked...December 3, 2014

Associate Account Strategist was asked...October 16, 2013

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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 Less

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The first thing Gmail users will want to know is WHY...all of a sudden...Gmail is not FREE?? And Google should expect to cause a panic in the minds of their customers. Because, if "I have to pay for Gmail...that must mean that soon, everything on the internet will follow suit." In this regard, I hold that charging money to access and utilize Gmail is not a sustainable business concept. In fact, the feeling of betrayal within the hearts and minds of Google's customers is THE most significant disadvantage. However, if it MUST be so that Google charges users to access and utilize new and existing Gmail accounts; well, quite candidly, the greatest advantage would be a significantly positive impact on profitability. Considering, of course, that some new "breakthrough" is implemented into Gmail (first as an Upgrade option for paid service, then standardized across the board for all users). The said "breakthrough" created by Google, in this example, would most certainly demand a marketing campaign centered around coaching the global public on HOW and WHY Google is changing the way we all think about email. Wow! -- The "WOW" factor created in the hearts and minds of Google and Gmail customers will need to be embedded, repeated, delivered in as unique a way as you wish your customers to identify Google with innovation--changing the world...not changing my bank account!! Thus, the decision to charge Gmail users should take into account the emotions of all Google customers, and all Internet users. And the families that will be affected by this net deficit to multiple family members, simultaneously, all paying for email service; and the take-home pay of those family members that will be stretched further than it already is, to include paying for ONE (or 2, 3, 4...) MORE fixed cost every month. Less

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C Free

Strategist was asked...January 17, 2012

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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%. Less

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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 ==> 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 ==> 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)) > 5, which is well satisfied in this case. then P(X ge 220) = 1 - P (X < 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<220) approximately = P(Y <220) = P( Z< (220-200)/10) = P(Z < 2) (standard normal distribution) = (1-0.9544)/2 = 0.0228 reference: Statistical Inference by George Casella, page 105, example of normal approximation Less

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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 Less

Strategist, Credit Flow Desk was asked...March 14, 2014

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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 Less

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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. Less

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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 Less

Account Strategist was asked...November 1, 2011

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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. Less

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

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Zero

Strategist was asked...November 3, 2012

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Do not open the cell

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

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Shoot one of them and only 99 will escape

Strategist was asked...January 20, 2015

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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 > b-Log[-1+Exp[b]] Less

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

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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] > Exp[a+b] _ a=b=1 implies Exp[a] + Exp[b] Exp[a]+Exp[b] only if a > b-Log[-1+Exp[b]] . Not very enlightening, but it is clear if the two functions are plotted in 3D. Less

Strategist was asked...September 24, 2015

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

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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. Less

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

Strategist was asked...December 6, 2013

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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 Less

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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) Less

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

Deployment Strategist was asked...February 6, 2015

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For crying out loud, it's a half.

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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. Less

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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. Less