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A multiplechoice 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?
 one year ago
 one year ago
A multiplechoice 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?
 one year ago
 one year ago

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tomoBest ResponseYou've already chosen the best response.0
give me one second to show my work
 one year ago

tomoBest ResponseYou've already chosen the best response.0
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) ??
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
you are looking for a conditional probability
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
and i believe you need baye's formula for this one
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
if you put \(A\) as the probability he gets the answer right, and \(B\) as the probability he guessed, you are looking for \[P(AB)\] which you compute via \[P(AB)=\frac{P(A\cap B)}{P(B)}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
ok that was wrong, you are looking for \(P(BA)\)
 one year ago

tomoBest ResponseYou've already chosen the best response.0
yes, P(guess given that he answers correctly)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
using the sets i wrote above, you compute \[P(BA)=\frac{P(A\cap B)}{P(A)}\]
 one year ago

tomoBest ResponseYou've already chosen the best response.0
so, can we say that P(guess AND correct) = P(guess) * P(correct)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
yes, that is your numerator
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
oh wait no it isn't sorry
 one year ago

tomoBest ResponseYou've already chosen the best response.0
that is assuming the two are independent right
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
because they are not independent
 one year ago

tomoBest ResponseYou've already chosen the best response.0
so do i have to use the extended bayes formula
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
you compute \[P(A\cap B)=P(B)\times P(AB)\] yes, it is baye's formula you need
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
both of those are easy to compute, \(P(B)=.25\) and \(P(AB)=.2\) i think
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
this line from your question . Since there are five possible answers, the probability he guesses correctly is .25. is wrong i think
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
if there are 5 possible answers, and he guesses, the probability he guesses right should be \(.2\) not \(.25\)
 one year ago

tomoBest ResponseYou've already chosen the best response.0
I'm confused. Is P(B AND A) the same as P(A AND B)?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i think you should get for the numerator \(.25\times .2\) and for the denominator \[.25\times .2+.75\times 1\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
yes, they are the same \(P(A\cap B)=P(B\cap A)\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
you know how to compute \(P(A)\) the probability he guesses, and \(P(BA)\) the probability he gets it right given that he guesses
 one year ago

tomoBest ResponseYou've already chosen the best response.0
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.
 one year ago

tomoBest ResponseYou've already chosen the best response.0
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. \[P(BA) = P(B \cap A)/P(A) \] and \[P(AB) = P(A \cap B)/P(B)\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
let me get to a desktop, it is hard to write this on a lap top
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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 \(P(AB)\) by thinking so we can compute \(P(A\cap B)=P(B)\times P(AB)\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
\(P(AB)\) 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 \(P(AB)=.2\) and therefore \[P(A\cap B)=P(B)P(AB)=.25\times .2\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
we cannot compute \(P(BA)\) because it is not obvious, it is what we are looking for so we have to use the one above
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
now we need \(P(A)\) the denominator.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
btw after we do this the hard way i will show you a very quick snappy way to do it
 one year ago

tomoBest ResponseYou've already chosen the best response.0
ok, give me one second to compute the answer.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
then we will do it in our heads with almost no computation at all
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
that is not what i get
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
your denominator is wrong it should be \(P(A)=.75+.25\times .2=.8\)
 one year ago

tomoBest ResponseYou've already chosen the best response.0
oh yeah, I forgot to add the probability of guessing correctly
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
\(A\) 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
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
ok now lets do it the snap way
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
he answers 100 questions, and he gets 75 right because he knows them and of the remaining 25 he gets 5 right
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
therefore he gets 80 right total, and of those 80 he guessed on 5 your answer is therefore \(\frac{5}{80}\)
 one year ago

tomoBest ResponseYou've already chosen the best response.0
interesting. I didn't think of that.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
baye's formula becomes much less mysterious when you use numbers and think
 one year ago

tomoBest ResponseYou've already chosen the best response.0
thank you very much for all the help. I really appreciate it. Sometimes this probability can be quite confusing.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
easier than saying \[P(BA)=\frac{P(B)P(AB)}{P(B)P(AB)+P(B^c)P(AB^c)}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
hint, on an exam if you get confused use numbers make them big
 one year ago

tomoBest ResponseYou've already chosen the best response.0
A little way up, you said that \[P(AB) = .2\] 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 \[P(AB) \] mean probability of B given that we know A has occurred? Now that be occurs what is the probability of A?
 one year ago

tomoBest ResponseYou've already chosen the best response.0
right after P(AB) i messed up. I mean "Does P(AB) mean probability of A given that we know B has occurred? Now that B occurs what is the probability of A?"
 one year ago
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