 # binomial problem???

Hi lads, just a binomial problem for a bit of craic. What’s the probability of passing (>70%) the exam of 240 questions if you had to guess every single answer? Probability of a correct answer = 0.25. Any ideas???

the probability of passing would be precisely that —> 0.25??? CP

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.

Wouldn’t this work JDV? [240! / (72! * 168!)] * 0.25^240 * 0.75^72

Make that: [240! / (72! * 168!)] * 0.25^168 * 0.75^72

.25^168 = 0 so if you guess, it is telling you, you are doomed!.. guess intelligently, you have a little more of a chance. guess after eliminating two out of the 4 possibilities, you are upwards of 50%!!!

Dreary Wrote: ------------------------------------------------------- > Make that: > > [240! / (72! * 168!)] * 0.25^168 * 0.75^72 Nope - that’s the probability of exactly 168 correct (it’s also a little difficult to calculate that probability without fancy software since 240! is a monster number and 0.25^168 is a minisicule number)

cpk123 Wrote: ------------------------------------------------------- > .25^168 = 0 > > so if you guess, it is telling you, you are > doomed!.. > > guess intelligently, you have a little more of a > chance. > > guess after eliminating two out of the 4 > possibilities, you are upwards of 50%!!! So suppose that the probability of guessing correctly is 0.5. Use DeMoivre-Laplace to estimate the probability of passing the exam.

Demoivre Laplace: Mean = 0.5*240 = 120 Variance = 0.5^2*240 = 60 s.d. = sqrt(60) = 7.75 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 = 120 sd = 7.75. = P(Z > (168.5 - 120)/sqrt(60)) = P(Z > 6.25) = is still unbelievably small. I see your point, JDV. You cannot guess all, is the ONLY output of this exercise. You need to do something and take educated guesses on others, maybe 20-40 questions, after elimination of the recognised bad choices.

And what’s the probability of passing all three stages, each on the first try? Assuming the pass rates are 40% for each stage? 0.4^3 = 6.4% Holy cow!

But Pr(passing L2 | Passed L1 on First try) should be taken into account…it is not 0.40.

Pr (L2 | L1 ) – aren’t L2 and L1 independent?

cpk, no they are not. If somone passed L1 after 10 trials (God forbid), do you think his chance of passing L2 is same as a wiz kid who went through L1 in no time?

I think so. Given the nature of the exam, all that matters is how you handle the sections, and what you learn in each section, and how you learn each section. The whiz kid who passed at one shot, has more chances of thinking he knows everything, and thus tend to prepare much less for the beast called L2. That is not captured in any probability matrix. I would think passing each exam is independent, and it depends clearly on how people learn and prepare.

If you had to bet, cpk, which one would you place your money on? We tell you, look this guy passed L1 from first time. Over there, there’s this guy who’s been taking L1 every 6 months for the last 10 years, and thank God he just passed. Now you are going to wager \$100,000 on the winner. You may reason about the psychology part, but I am not sure you would place your bet on the 20-trial guy! Would you?

You give me the \$100,000, I’ll put it into my account, and not wager. That’s another story altogether. Or is it? If the problem is that you pick your guy, and you either receive \$100k if he wins, or pay \$100k if he loses. In that case, even picking the wiz kid is a bad idea, because your expected value is -\$20k, i.e., 0.40 *\$100k + 0.60 (-\$100k) = -\$20k !

Pr (L2 | L1 ) The bar implies “given that” so it is NOT INDEPENDENT…these are DEPENDENT EVENTS. L2: Recession will Occur L1: Interest Rates Rise “Given” Internet Rates Rise, Recession Will oCCur

Lucai Wrote: ------------------------------------------------------- > Hi lads, > just a binomial problem for a bit of > craic. > > What’s the probability of passing (>70%) the exam > of 240 questions if you had to guess every single > answer? > > Probability of a correct answer = 0.25. > > Any ideas??? first take .7*240 = 168 72 questions that you can miss. Now take the favorable .25^168*.75*72 Now we still are done here. there 240!/168!*72! ways to arrange the numbers above. So really its 240!/(168!*72) * .25^168*.75*72 way too ridiculous to calculate

O wait this is for only 70%. hrmmm