The number of days a particular stock increases in a given five-day period is uniformly distributed between zero and five inclusive. In a given five-day trading week, what is the probability that the stock will increase exactly three days? I guessed the right answer. But it doesnt make any sense.
Can you post the question?
I’d say this is the straight binomial probability distribution formula, n!/(k!*(n-k)!)*(probability of success^#of successful trials)*probability of failure ^(1-#of successful trials) [5!/(3!*2!)]*(0.5^3)*(0.5^2)=10*0.5^5=0.3125
I did what you did map. The answer is .1667 I think the question is just incorrectly worded all around.
The model may make no sense, but it clearly states that the probability of number of days of increase is uniform over {0, 1, 2, 3, 4, 5}, so P(N = 3) = 1 / 6.
actuaryalfred Wrote: ------------------------------------------------------- > The model may make no sense, but it clearly states > that the probability of number of days of increase > is uniform over {0, 1, 2, 3, 4, 5}, so P(N = 3) = > 1 / 6. agree
Let’s say the problem didn’t state that the # of days a stock increases is “uniformly distributed”, would map1 be correct?