A multiple-choice test has five possible answers for each questions. On any given question a student has probability .75 to know the right answer and probability .25 to have to guess. Since there are five possible answers, the probability he guesses correctly is .25. If a student answered the question correctly, what is the probability that he guessed?

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A multiple-choice test has five possible answers for each questions. On any given question a student has probability .75 to know the right answer and probability .25 to have to guess. Since there are five possible answers, the probability he guesses correctly is .25. If a student answered the question correctly, what is the probability that he guessed?

Probability
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give me one second to show my work
P(knows correct) = .75 P(guess) = .25 Is the probability he guesses correctly .25 or is it 1/5 = .2? I believe that i am looking for P(guess given that it is correct) ??
you are looking for a conditional probability

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and i believe you need baye's formula for this one
if you put \(A\) as the probability he gets the answer right, and \(B\) as the probability he guessed, you are looking for \[P(A|B)\] which you compute via \[P(A|B)=\frac{P(A\cap B)}{P(B)}\]
ok that was wrong, you are looking for \(P(B|A)\)
yes, P(guess given that he answers correctly)
using the sets i wrote above, you compute \[P(B|A)=\frac{P(A\cap B)}{P(A)}\]
so, can we say that P(guess AND correct) = P(guess) * P(correct)
yes, that is your numerator
oh wait no it isn't sorry
that is assuming the two are independent right
because they are not independent
so do i have to use the extended bayes formula
you compute \[P(A\cap B)=P(B)\times P(A|B)\] yes, it is baye's formula you need
both of those are easy to compute, \(P(B)=.25\) and \(P(A|B)=.2\) i think
this line from your question . Since there are five possible answers, the probability he guesses correctly is .25. is wrong i think
if there are 5 possible answers, and he guesses, the probability he guesses right should be \(.2\) not \(.25\)
I'm confused. Is P(B AND A) the same as P(A AND B)?
i think you should get for the numerator \(.25\times .2\) and for the denominator \[.25\times .2+.75\times 1\]
yes, they are the same \(P(A\cap B)=P(B\cap A)\)
you know how to compute \(P(A)\) the probability he guesses, and \(P(B|A)\) the probability he gets it right given that he guesses
where do i find the AND/intersection statement in the equation editor? i can't find it and it would help me explain my thinking.
i used \cap
thanks, ok I'm confused because I thought that changing the order of B and A in bayes formula changed the B and A in the intersection. \[P(B|A) = P(B \cap A)/P(A) \] and \[P(A|B) = P(A \cap B)/P(B)\]
let me get to a desktop, it is hard to write this on a lap top
ok that's better
both the formulas you wrote above are correct, and the numerators are both the same the point of baye's is that you can use the conditional probability you know (the easy one) to compute the conditional probability you want (the hard one) at the moment, we don't know \(P(A\cap B)\) so we have to use a conditional probability to find it
if \(A\) is you get the right answer, and \(B\) is you guess, we can know both \(P(B)=.25\) because it is given to us, and we can compute \(P(A|B)\) by thinking so we can compute \(P(A\cap B)=P(B)\times P(A|B)\)
\(P(A|B)\) means you guess, and you get the answer correct. since one out of the 5 answers is correct, if you guess you have a one in five (aka .2\) chance of getting it right, so \(P(A|B)=.2\) and therefore \[P(A\cap B)=P(B)P(A|B)=.25\times .2\]
we cannot compute \(P(B|A)\) because it is not obvious, it is what we are looking for so we have to use the one above
now we need \(P(A)\) the denominator.
btw after we do this the hard way i will show you a very quick snappy way to do it
ok, give me one second to compute the answer.
ok
then we will do it in our heads with almost no computation at all
P(B|A) = .05/.75 = .0666
that is not what i get
your denominator is wrong it should be \(P(A)=.75+.25\times .2=.8\)
oh yeah, I forgot to add the probability of guessing correctly
\(A\) is he gets it right, but he gets it right more than 75% of the time he gets it right if he knows it and also if he guesses correctly
ok now lets do it the snap way
ok
he answers 100 questions, and he gets 75 right because he knows them and of the remaining 25 he gets 5 right
therefore he gets 80 right total, and of those 80 he guessed on 5 your answer is therefore \(\frac{5}{80}\)
wasn't that easy?
interesting. I didn't think of that.
baye's formula becomes much less mysterious when you use numbers and think
thank you very much for all the help. I really appreciate it. Sometimes this probability can be quite confusing.
easier than saying \[P(B|A)=\frac{P(B)P(A|B)}{P(B)P(A|B)+P(B^c)P(A|B^c)}\]
yw
hint, on an exam if you get confused use numbers make them big
thanks for the tip.
yw
A little way up, you said that \[P(A|B) = .2\] I was thinking about that and tried doing a probability tree. |dw:1358740038530:dw| why isn't it .25*.2. (.25 for guessing and .2 for guessing it correctly.)? Does \[P(A|B) \] mean probability of B given that we know A has occurred? Now that be occurs what is the probability of A?
be should be 'B'
right after P(A|B) i messed up. I mean "Does P(A|B) mean probability of A given that we know B has occurred? Now that B occurs what is the probability of A?"

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