# Pass CFA by chance

Do you think CFA will give a question like this? 240 questions. 70% passing rate. 25% to get each question correct. What’s the probability that you’ll pass it by chance? Is it? 240 C 168 *.25^168 .75^72

If they did ask that they’d have to admit that there is a tangible pass rate…

Wouldn’t it be .25^168? Which I had to use excel to get, b/c my calculator just says 0.00 Excel says: 0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000071436711955142200000000000%

pepp Wrote: ------------------------------------------------------- > Do you think CFA will give a question like this? > 240 questions. 70% passing rate. 25% to get each > question correct. What’s the probability that > you’ll pass it by chance? > > > Is it? > 240 C 168 *.25^168 .75^72 That would be the prob of exactly 168 correct.

FIAnalyst Wrote: ------------------------------------------------------- > Wouldn’t it be .25^168? > > Which I had to use excel to get, b/c my calculator > just says 0.00 > > Excel says: > 0.000000000000000000000000000000000000000000000000 > 00000000000000000000000000000000000000000000000000 > 071436711955142200000000000% that would be the prob of getting the first 168 correct

joey, looks like we are all failing. hahaha

Use DeMoivre-Laplace CLT - Mean = 0.25*240 = 60 Variance = 0.25*(1-0.25)*240 = 45 s.d. = sqrt(45) = 6.71 X = Score on exam by guessing then P(X > 0.7*240) = P(X > 168) and using CLT and continuity correction P(Y >= 168.5) where Y is normal mu = 60 sd = sqrt(45). = P(Z > (168.5 - 60)/sqrt(45)) = P(Z > 16.09) = unbelievably small.

double post

JoeyDVivre Wrote: ------------------------------------------------------- > that would be the prob of getting the first 168 > correct As soon as I posted it, I knew something was off, but I can’t figure it out.

JoeyDVivre Wrote: ------------------------------------------------------- > Use DeMoivre-Laplace CLT - > > Mean = 0.25*240 = 60 > Variance = 0.25*(1-0.25)*240 = 45 > s.d. = sqrt(45) = 6.71 > > X = Score on exam by guessing then P(X > 0.7*240) > = P(X > 168) > > and using CLT and continuity correction P(Y >= > 168.5) where Y is normal mu = 60 sd = sqrt(45). > > = P(Z > (168.5 - 60)/sqrt(45)) = P(Z > 16.09) = > unbelievably small. All of this makes sense to me other than the variance calc. Suppose I should spend more time studying and less time posting

I think variance formula is same for binomial variance.

LOL it’s funny to see the exact same post on L1, usually a few days before the exam. Needless to say that Joey always comes to the rescue.

You need to use the Bernoulli formula. You are looking for 65 successes out of 100 trials with the probability of success on each trial being 25%.