I can't find how to calculate the probability of winning an exam knowing it has 40 questions and 4 choices in it, so I must do 20 good answers. Any help on that matter ?
I've tried (1/4)^20 but I don't feel like 0.0000000000009095 is the good answer.
Am I supposed to use the random event / N of possible events formula ? Was it wrong to 1/5 ^40 knowing this an independant event, the odds of answering right doesn't change the second odds of answering right
Update : I've done this
((3/4)^20 ) / (4/4)^40 = 0.003171211938934
I feel like this one is the good answer
+130516 We want to make sure that any 20 of 40 questions are correct....so the number of possible different sets of correct questions = C(40, 20)
We have a 25% =.25 probability of answering each question correctly and a 75% = .75 probability of an incorrect answer.....so......the probability of getting any 20 questions correct is given by :
C(40,20) * (.25)^20 * (.75)^20 = 0.0003976 = about .03976%
Mmmm......doesn't seem that guessing is going to help much !!!!
C(40,20) * (.25)^20 * (.75)^20 = 0.0003976 = about .03976%
Okay I get the logic behind this, but one thing i'm still not sure of is : Does this sum up every possibility or only the possibility of doing 20/40 at the exam, another way to ask it would be : Is it only 20/40 or at least 20/40 ?
How I am supposed to calculate the sum of every possibilty from getting 20/40, then 21/40, then 22/40, ... 40/40 ?
+130516 This only covers the probability of getting exactly 20 questions correct
To compute the probability of getting 20/40, 21/40, 22/40. 23/40, etc., correct......we would have :
C(40,20) (.25)^20 (.75)^20
+ C(40.21) (.25)^21 (.75)^19 = ???
+ C(40.22) (.25)^22 (.75)^18 = ???
+ C(40,23) (.25)^23 (.75)^17 = ???
.
.
etc......
There may be faster methods, but I'm not familiar with them....
