"John Knight" <johnknight at usa.com> wrote in message
news:5YFZ8.7306$Fq6.534235 at news2.west.cox.net...
> "mat" <mats_trash at hotmail.com> wrote in message
> news:43525ce3.0207180723.642fd8f9 at posting.google.com...> > >
> > > If you're asked a question which has four multiple choice answers, and
> you
> > > haven't got a clue what the answer is, what is the probability of
> getting a
> > > correct answer? Since you have once chance in four of getting the
right
> > > answer, your probability is 0.25. If you guess on two questions, your
> > > probability is .5, and three it's .75, and four, it's 1.0.
> >
> > So you are certain to get a correct answer if you guess on four
> > questions? and you claim to understand probability?!
> >
> > P(one questions correct given 4 choices and if guessing) = 0.25
> > P(two questions correct) = 0.25 x 0.25 = 0.0625
> > P(three questions correct) = 0.25 x 0.25 x 0.25 = 0.015625
> > .
> > .
> > .
> > P(n questions correct) = 0.25^n
> >
> > its essentially the same as asking the probability of getting a one
> > (or any specified number) on sequential throws of a dice. Overall you
> > expect to get 1/6 (the relative frequency), but getting four ones in a
> > row is equal to 1/(6^4)
> >
>> Nice try, "mat", but you're way off. Thanks for giving it a try, though.
> P1 = P2 = P3 = P4 = 0.25
> In other words, the probability of getting one question right by guessing
at
> a 4 part multiple choice question is 0.25
> But to figure the probability of getting four of the same questions right,
> not the *same* answer on each question, but the *correct* answer, you must
> add them up, which is P1 + P2 + P3 + P4 = 1.0
Nope, that's not true.
Now, you seem to be stating that the probability of answering 4 multiple
choice questions correctly is higher than that of answering 1 correctly.
This is not true. Actually, that is logically impossible, since the
probability of the superset (getting first one right) must be >= to the
probability of the subset (getting them all right).
And if you meant that the probability of getting one correct out of the
four, then that can be shown below.
P(answer a question correctly) = 0.25
P(at least one of two correct) = P(both correct) + 2 * P(one correct and
other incorrect) = 0.25*0.25 + 2 * 0.25 * 0.75 = 0.4375
It's not the 0.5 like you thought.
P(at least one of three correct) = 1 - P(all 3 wrong) = 1 - (0.75*0.75*0.75)
= .578125
P(at least one of 4 correct) = 1 - P(all 4 wrong) = 1 - (0.75)^4 = 0.6835
(etc)
Clearly not equal to 1.
