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# Trader development program Interview Questions

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Dec 28, 2011
 There is a 91% chance of seeing a shooting star in the next hour, what is the probability of seeing a shooting star in the next half hour?12 Answers70%How could this be the answer? surely it is 40.5%; half of the 91% probability?sorry I meant 45.5%Show More Responsesprobability shooting star in half hour = x probability of NOT seeing shooting star in 1 hour = (1-x)^2 probability of seeing shooting star in 1 hour = 1 - (1-x)^2 so we solve 1 - (1-x)^2 = 0.91 for xAccording to the question, suppose we have a random variable X that is the time when we see a shooting star, and let F(x) be the CDF of X. Hence, F(now+1h)-F(now)=0.91. And this is unfortunately the only thing we know about F(x). In other words, any valid F(x) satisfying such requirement could be the CDF of X. Therefore, we have no idea of the value of F(now+0.5h)-F(now) because F(now+0.5) could be any value between F(now) and F(now)+0,91. Answer: unknown result depending on the distribution.The second answer is incorrect. Take an extreme case as an example, if the probability of seeing a shooting star in 1 hour is 1, then the root of the function would be 1, which means that the probability of seeing a shooting star in the first 30 minutes is 1. An interesting question followed by that would be: what about in the first 15 (second 15) mins? According to this method, we can again conclude that the probability of having a shooting star in the first 15mins is still 1. And this can go on for ever, and result in the conclusion that at any time interval, P(shooting_Star) would be 1 which is obviously mistaken.let p= probability to see the star in half hour, (1-p)=won't see, (1-p)^2=can't see a star in an hour=1-0.91, solve we have p=0.7The answer should be 0.7 but most of the explanations here make no sense. It is reasonable to assume a exponential distribution here and solve the probability.Probability is 30%. The arrival of star follows a Poisson process. Seeing K arrivals in time T is P = (lambda*T)^K * exp(-lambda*T)/K!. Now we know there has been no arrival in 1 hour, T = 1, K = 0. P = exp(-lambda) = 1-0.91 = 0.09. Now T = 0.5 (half an hour), P = exp(-lambda*1/2) = sqrt(0.09) = 0.3. The probability of seeing a star is 1-0.3 = 0.7, or 70%.I would argue for the 91/2% = 45.5% chance of seeing the star. The wording of the problem implies this is not a poisson process. Rather, I would interpret as the weather channel telling residents in an area that there is a 91% chance of seeing the shooting star between 7pm and 8pm (this is an unusual occurrence). Hence, with 9% chance you can't see a star in the next half hour. Then, in the 91% of the time where there is a shooting star in the next hour, one could make the reasonable assumption that the time it comes is uniformly distributed in the hour. So, 45.5%.Like, think about it in the real world. It is kind of unreasonable to model seeing shooting stars as a Poisson process with parameter .91 stars per hour.P(no star in an hr) = 1 - 0.91 = 0.09 P(no star in half an hour)^2 = 0.09 P(no star in half an hour) = 0.3 P(star in half an hour) =1 - 0.3 = 0.7 = 70%

Dec 15, 2011