Passing by randomness

What are my chances?

120 questions

need atleast 84 right, (don’t mind getting 85, or 90 or 100 right)

1/3rd chance of getting right in each question.

you’re a battler mate

(1/3)^84 = zero basically.

But, you can do a little better because there are questions you know their answers for sure and some you can easily eliminate 1 of the choices in them, so you can say your chance is (1/2)^84 which is still basically zero. Even if your chance of getting the right answer is 0.99, still your chance of passing is less than 50%!!!

That’s weird!

Try it this way: Assume there are only 5 questions and the probbaility of getting each question correct is 0.50. What’s the probbaility of passing all 5? Amazingly, it’s only 0.03… (0.50)^5 = 0.03. This can’t be right, can it?

Dreary, time to go back to level 1. haha. how you doing? i am struggling here. didn’t realize it was may and as usual i have so much left to study, this time around planning on avoiding AF.

1/3^84 = the probablity of getting 84 questions right in a row.

it however is true that by pure guessing the chance of passing this exam is 0% but not because of what you state.

if you have 5 questions, and chance of getting right is .5 on each one of them. then to get all 5 right, you are right. the prob = .5^5 = .03% chance. another way to think about this would be to create the sample you could have for 5 questions.

TTTTT, TFFFF,TFTTT,TFFTT… (the number of possiblities is 32). and you have only 1 outcome that you want, so you are effectively with 1/32 = .03.

Haha Nice!!

Dreary your math is off. Your calc would assume it’s an 84 question test and you’d have to get 100% to pass. In which case 1 nerd in Asia would get the CFA every year while everyone else in the universe fails. Too early to do it, but it’s a variation off of 120 c 84. Still ridiculously low though.

True, but the chances of getting 5 out of 6 questions correct is: 6*(1/2)^6= 9.4% much better.

I think we need the combination function to do it right…I can’t look it up now, but but it is something like 5 Choose 6, choose any combination of 5 questions out of 6, to do it right.

hahahah, well it’s not zero, but very close, you are a fighter! I hope you don’t mind some rounding errors.

By just guessing the mean number of answers correct is 40, with a standard deviation of 5.16497, which is sqrt(120*(1/3)*(2/3))

So, (40-84)/5.16497= -8.520563362 SDs

-8.520563362 = .00000000000000000793895 (17 zeros)

***Edit Note: I just ran it in Excel, the easy way: (1-BINOMDIST(84,120,(1/3),TRUE)) =

.00000000000000510703 (14 zeros)

Not sure what std deviation has to do with it! You want to figure out how many combinations of 84 questions (in each combination) that there are in 120 questions. Knowing that tells you in how many ways you can pass…from there then you find the probability of passing each combination, multiply and you’re done. No probability guys out there to chime in?

you people are overthinking this. spend the time studying! laugh

Dreary, just take a second to check it out, it’s a binomial probability, I first calculated it with approximation statistics. It’s very simple to do. I then showed the probability in my edit.

you might be right, I have not thought about it in detail, but it is an interesting problem by itself. Back to studying.

well to give the correct answer:

here is what i came up with (ofcourse after reading up on binomial distribution)

n = 120, k = 84, p=1/3

so the probability that you get *atleast* 84 right is the probability of getting *exactly* 84 right + probablity of getting exactly 85 right + probability of getting 86 right … all to way to probability of getting exactly 120 right.

so ill give you answer on how to solve for getting exactly 84 right, and you can sum up the rest.

p(84 right) = (120! / 84!36! ) * (1/3)^84 * (2/3)^36

Then jaeestrada’a number is correct, because I get about 10^-14 as answer! So your chance of passing by guessing, pepp, is about one in 100 trillion!!!

LOL you guys are wayy into this…Dreary fell for the gamblers fallacy…he failed to realize that every question is independent of each other.