If there are x questions on a multiple-choice exam with 4 possible answers (A,B,C,D), what is the chance that 3 C's will occur in a row at least once in the exam? Write a generalized formula with x questions.
Only the geniuses get this one
Noah Barnes
Jayden Bailey
Ok i have the answer for x < 3 gimme a while to type out the equation
Aaron Hughes
Should be 0%...
Camden Miller
0.25 × 0.25 × 0.25
> should be 0%
mongoloid. Cant tell me this isnt homework.
Cameron Green
(x-2)/64 for x>3
Ez
Noah Foster
Have A B C D the same probability?
Lucas Turner
2 (3/4 * 1/4 * 1/4 * 1/4)
t. 213 IQ regular Rick and Morty watcher
Nathaniel Perez
Ok so obviously if x = 3 your chance is just 1/(4^x) = 1/(4^3) = 1/64
Makes sense, your only possible permutation is
CCC
If x = 4, questions 2 and 3 have to be C, and at least one of the other two have to be C as well. To calculate the chance of at least one, that's 1 minus the chance of 0.
So that would be [ 1/(4^2) ] * [ 1 - (3/4)^2 ] = [ 1/16 ] * [ 5/16 ] = 7/256.
Im pretty confident of that, but just to be sure, i'll list the permutations.
ACCC
BCCC
CCCA
CCCB
CCCC
CCCD
DCCC
Ok, that's 7 so i'm on the right track.
How can we generalize this for x though. Hm.
The chance that any preselected 3 in a row questions are all C is 1/64, as determined by when x = 3.
There's x - 2 ways you can pick exactly which 3 questions, say for example x = 5, there's 3 ways; the first few questions, the middle few questions, and the last few questions. This much seems pretty firm.
But for each question that isn't part of the 3 we selected, you can have 4 possibilities; A through D.
It's gotta be more complicated than [ 4 ] * [ x - 2 ] * [ 1 / 64 ], otherwise x = 4 would have yielded 8/64 rather than 7.
I think it's gotta be that CCCC is only 1 permutation, but with the formula above is counted twice; the first time being if we select the first 3 questions to be our 3-in-a-row C streak with the 4th irrelevant question that just happens to also be C, and the second time being the last 3 questions being our 3-in-a-row streak and the first queston being the irrelevant question being C.
Ok, so generalized we have to subtract redundant permutations.
Let's investigate x = 5 for more clues.
Levi Young
>Write a generalized formula with x questions.
Most people with IQ above 100 can read. Interesting.
Xavier Bell
Incoherent question - it doesn't account for the agency of the person setting up the test. Not a pure chance scenario.