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anonymous
 4 years 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?
anonymous
 4 years 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?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0give me one second to show my work

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0P(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) ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you are looking for a conditional probability

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and i believe you need baye's formula for this one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if 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 \[P(AB)=\frac{P(A\cap B)}{P(B)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok that was wrong, you are looking for \(P(BA)\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes, P(guess given that he answers correctly)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0using the sets i wrote above, you compute \[P(BA)=\frac{P(A\cap B)}{P(A)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so, can we say that P(guess AND correct) = P(guess) * P(correct)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes, that is your numerator

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh wait no it isn't sorry

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is assuming the two are independent right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because they are not independent

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so do i have to use the extended bayes formula

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you compute \[P(A\cap B)=P(B)\times P(AB)\] yes, it is baye's formula you need

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0both of those are easy to compute, \(P(B)=.25\) and \(P(AB)=.2\) i think

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this line from your question . Since there are five possible answers, the probability he guesses correctly is .25. is wrong i think

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if there are 5 possible answers, and he guesses, the probability he guesses right should be \(.2\) not \(.25\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm confused. Is P(B AND A) the same as P(A AND B)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think you should get for the numerator \(.25\times .2\) and for the denominator \[.25\times .2+.75\times 1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes, they are the same \(P(A\cap B)=P(B\cap A)\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you know how to compute \(P(A)\) the probability he guesses, and \(P(BA)\) the probability he gets it right given that he guesses

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0where do i find the AND/intersection statement in the equation editor? i can't find it and it would help me explain my thinking.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks, 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 \[P(AB) = P(A \cap B)/P(B)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let me get to a desktop, it is hard to write this on a lap top

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0both 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if \(A\) 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 \(P(A\cap B)=P(B)\times P(AB)\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(P(AB)\) 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 \[P(A\cap B)=P(B)P(AB)=.25\times .2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we cannot compute \(P(BA)\) because it is not obvious, it is what we are looking for so we have to use the one above

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now we need \(P(A)\) the denominator.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0btw after we do this the hard way i will show you a very quick snappy way to do it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, give me one second to compute the answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then we will do it in our heads with almost no computation at all

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0P(BA) = .05/.75 = .0666

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is not what i get

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0your denominator is wrong it should be \(P(A)=.75+.25\times .2=.8\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh yeah, I forgot to add the probability of guessing correctly

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(A\) 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok now lets do it the snap way

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0he answers 100 questions, and he gets 75 right because he knows them and of the remaining 25 he gets 5 right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0therefore he gets 80 right total, and of those 80 he guessed on 5 your answer is therefore \(\frac{5}{80}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0interesting. I didn't think of that.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0baye's formula becomes much less mysterious when you use numbers and think

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0easier than saying \[P(BA)=\frac{P(B)P(AB)}{P(B)P(AB)+P(B^c)P(AB^c)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hint, on an exam if you get confused use numbers make them big

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0right 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?"
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