Did you guys notice a string of consecutive Bs or As as answers in PM session? Wheever such things happen, I always panic and start checking each and every answer again.
Yes
@Nuppal: That was honest and quite quick.Do you keep refreshing AF every 2 seconds or what?
Yeah, I am permabored at my current job.
Well you should’ve expected that since there are only 3 options to choose from A, B, or C
with only 3 answers, the probability of having the same answers in a row is pretty high. not sure on how to calculate this, I think it is permutation,
cfagoal2 Wrote: ------------------------------------------------------- > with only 3 answers, the probability of having the > same answers in a row is pretty high. > > not sure on how to calculate this, I think it is > permutation, I dare you to do it.
Are you implying the data series for CFA answers is justified by weak form efficient market hypothesis?
3 nPr 1
I had a 4 in a row and a 5 in a row
Well I looked it up, the way to calculate k success in a row in N trials is a Bernoulli distribution. I’m not sure how to do it, maybe one of the actuaries on this forum will know
Ethics and Alt Inv sections if I am not wrong?
There was a 6 b in a row that was almost positive was correct, that was in the am
Yeah, who cares, i never let a previous answer effect what I know is a correct answer.
I had some trends up in that piece. Lots of ‘a’ and ‘b’ streaks of 3-5.
I analyzed some trend analysis but instead decided to sacrifice diligence for speed and just answered all ‘b’
The easiest way to solve it is to find the probability that the longest series is four in a row. Let’s just assume we are dealing with the problem of four "A"s in a row, and we’ll extend the solution for the B’s and C’s. Probability of an A: 1/3 Probability of not A: 2/3 Given four questions, possibility of four A’s: (1/3)^4 = .0123 Given four questions, possibility of not four A’s: 1-(1/3)^4 = .9877 Number of sets of four consecutive answers in 120 consecutive questions: 30, 29, 29, 29 (this technique will treat the 120 answers as chunks of 4 answer segments, split in 4 different ways. {{1 2 3 4}, {5 6 7 8}, etc., {117 118 119 120}} (there are 30), the next is {{2 3 4 5} {6 7 8 9}, … , {114 115 116 117}} (there are 29) … etc. Probability of none of the thirty chunks having four A’s: (.9877)^30 = .6889 Probability of none of the twenty-nine chunks having four A’s: (.9877)^29 = .6975 Probability of none of the (30 + 29 x3 ) 117 blocks having four As: (.6889)(.6975)^3 = (.9877)^117 = .2338 Probability of no string of 4 consecutive A’s, B’s or C’s: (.2338)^3 = .0128 The probability of a string of 4 or more consecutive answers of the same in 120 consecutive questions is 98.7% Using the same approach, probability of a string of 5 or more consecutive answers of the same in 120 consecutive questions is 76.2%, for 6 it is 37.7%, 7 is 14.5%, and 8 is 5%.
For the AM portion, did anyone of you have bunch of c’s (4 or 5 in a row) in the begginning of the second to last column…I think it was in the derivatives section…I forgot what my form number is, but can you guys confirm?
lisun Wrote: ------------------------------------------------------- > For the AM portion, did anyone of you have bunch > of c’s (4 or 5 in a row) in the begginning of the > second to last column…I think it was in the > derivatives section…I forgot what my form number > is, but can you guys confirm? I think I had 5 in a row but went back and changed one of them.
hawgdriver Wrote: ------------------------------------------------------- > The easiest way to solve it is to find the > probability that the longest series is four in a > row. Let’s just assume we are dealing with the > problem of four "A"s in a row, and we’ll extend > the solution for the B’s and C’s. > > Probability of an A: 1/3 > Probability of not A: 2/3 > > Given four questions, possibility of four A’s: > (1/3)^4 = .0123 > Given four questions, possibility of not four A’s: > 1-(1/3)^4 = .9877 > Number of sets of four consecutive answers in 120 > consecutive questions: 30, 29, 29, 29 > > (this technique will treat the 120 answers as > chunks of 4 answer segments, split in 4 different > ways. {{1 2 3 4}, {5 6 7 8}, etc., {117 118 119 > 120}} (there are 30), the next is {{2 3 4 5} {6 7 > 8 9}, … , {114 115 116 117}} (there are 29) … > etc. > > Probability of none of the thirty chunks having > four A’s: > > (.9877)^30 = .6889 > > Probability of none of the twenty-nine chunks > having four A’s: > > (.9877)^29 = .6975 > > Probability of none of the (30 + 29 x3 ) 117 > blocks having four As: > > (.6889)(.6975)^3 = (.9877)^117 = .2338 > > Probability of no string of 4 consecutive A’s, B’s > or C’s: (.2338)^3 = .0128 > > The probability of a string of 4 or more > consecutive answers of the same in 120 consecutive > questions is 98.7% > > Using the same approach, probability of a string > of 5 or more consecutive answers of the same in > 120 consecutive questions is 76.2%, for 6 it is > 37.7%, 7 is 14.5%, and 8 is 5%. you lost me at the part about chuknks, how did you get 30, 29, 29, 29