A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 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
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?

This Question is Closed

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

tomo
 one year 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) ??

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1you are looking for a conditional probability

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1and i believe you need baye's formula for this one

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1if 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 \[P(AB)=\frac{P(A\cap B)}{P(B)}\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1ok that was wrong, you are looking for \(P(BA)\)

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1using the sets i wrote above, you compute \[P(BA)=\frac{P(A\cap B)}{P(A)}\]

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1yes, that is your numerator

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1oh wait no it isn't sorry

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1because they are not independent

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1you compute \[P(A\cap B)=P(B)\times P(AB)\] yes, it is baye's formula you need

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1both of those are easy to compute, \(P(B)=.25\) and \(P(AB)=.2\) i think

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1this line from your question . Since there are five possible answers, the probability he guesses correctly is .25. is wrong i think

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1if there are 5 possible answers, and he guesses, the probability he guesses right should be \(.2\) not \(.25\)

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1i think you should get for the numerator \(.25\times .2\) and for the denominator \[.25\times .2+.75\times 1\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1yes, they are the same \(P(A\cap B)=P(B\cap A)\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1you know how to compute \(P(A)\) the probability he guesses, and \(P(BA)\) the probability he gets it right given that he guesses

tomo
 one year 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.

tomo
 one year 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 \[P(AB) = P(A \cap B)/P(B)\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1let me get to a desktop, it is hard to write this on a lap top

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1both 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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1if \(A\) 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 \(P(A\cap B)=P(B)\times P(AB)\)

satellite73
 one year ago
Best 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 \[P(A\cap B)=P(B)P(AB)=.25\times .2\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1we cannot compute \(P(BA)\) because it is not obvious, it is what we are looking for so we have to use the one above

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1now we need \(P(A)\) the denominator.

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1btw after we do this the hard way i will show you a very quick snappy way to do it

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1then we will do it in our heads with almost no computation at all

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1that is not what i get

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1your denominator is wrong it should be \(P(A)=.75+.25\times .2=.8\)

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

satellite73
 one year ago
Best 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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1ok now lets do it the snap way

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1he answers 100 questions, and he gets 75 right because he knows them and of the remaining 25 he gets 5 right

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1therefore he gets 80 right total, and of those 80 he guessed on 5 your answer is therefore \(\frac{5}{80}\)

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1baye's formula becomes much less mysterious when you use numbers and think

tomo
 one year 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.

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1easier than saying \[P(BA)=\frac{P(B)P(AB)}{P(B)P(AB)+P(B^c)P(AB^c)}\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1hint, on an exam if you get confused use numbers make them big

tomo
 one year 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?

tomo
 one year 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?"
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.