anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
give me one second to show my work
anonymous
  • anonymous
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) ??
anonymous
  • anonymous
you are looking for a conditional probability

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anonymous
  • anonymous
and i believe you need baye's formula for this one
anonymous
  • anonymous
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 \[P(A|B)=\frac{P(A\cap B)}{P(B)}\]
anonymous
  • anonymous
ok that was wrong, you are looking for \(P(B|A)\)
anonymous
  • anonymous
yes, P(guess given that he answers correctly)
anonymous
  • anonymous
using the sets i wrote above, you compute \[P(B|A)=\frac{P(A\cap B)}{P(A)}\]
anonymous
  • anonymous
so, can we say that P(guess AND correct) = P(guess) * P(correct)
anonymous
  • anonymous
yes, that is your numerator
anonymous
  • anonymous
oh wait no it isn't sorry
anonymous
  • anonymous
that is assuming the two are independent right
anonymous
  • anonymous
because they are not independent
anonymous
  • anonymous
so do i have to use the extended bayes formula
anonymous
  • anonymous
you compute \[P(A\cap B)=P(B)\times P(A|B)\] yes, it is baye's formula you need
anonymous
  • anonymous
both of those are easy to compute, \(P(B)=.25\) and \(P(A|B)=.2\) i think
anonymous
  • anonymous
this line from your question . Since there are five possible answers, the probability he guesses correctly is .25. is wrong i think
anonymous
  • anonymous
if there are 5 possible answers, and he guesses, the probability he guesses right should be \(.2\) not \(.25\)
anonymous
  • anonymous
I'm confused. Is P(B AND A) the same as P(A AND B)?
anonymous
  • anonymous
i think you should get for the numerator \(.25\times .2\) and for the denominator \[.25\times .2+.75\times 1\]
anonymous
  • anonymous
yes, they are the same \(P(A\cap B)=P(B\cap A)\)
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
i used \cap
anonymous
  • anonymous
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 \[P(A|B) = P(A \cap B)/P(B)\]
anonymous
  • anonymous
let me get to a desktop, it is hard to write this on a lap top
anonymous
  • anonymous
ok that's better
anonymous
  • anonymous
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
anonymous
  • anonymous
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 \(P(A\cap B)=P(B)\times P(A|B)\)
anonymous
  • anonymous
\(P(A|B)\) 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 \[P(A\cap B)=P(B)P(A|B)=.25\times .2\]
anonymous
  • anonymous
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
anonymous
  • anonymous
now we need \(P(A)\) the denominator.
anonymous
  • anonymous
btw after we do this the hard way i will show you a very quick snappy way to do it
anonymous
  • anonymous
ok, give me one second to compute the answer.
anonymous
  • anonymous
ok
anonymous
  • anonymous
then we will do it in our heads with almost no computation at all
anonymous
  • anonymous
P(B|A) = .05/.75 = .0666
anonymous
  • anonymous
that is not what i get
anonymous
  • anonymous
your denominator is wrong it should be \(P(A)=.75+.25\times .2=.8\)
anonymous
  • anonymous
oh yeah, I forgot to add the probability of guessing correctly
anonymous
  • anonymous
\(A\) 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
  • anonymous
ok now lets do it the snap way
anonymous
  • anonymous
ok
anonymous
  • anonymous
he answers 100 questions, and he gets 75 right because he knows them and of the remaining 25 he gets 5 right
anonymous
  • anonymous
therefore he gets 80 right total, and of those 80 he guessed on 5 your answer is therefore \(\frac{5}{80}\)
anonymous
  • anonymous
wasn't that easy?
anonymous
  • anonymous
interesting. I didn't think of that.
anonymous
  • anonymous
baye's formula becomes much less mysterious when you use numbers and think
anonymous
  • anonymous
thank you very much for all the help. I really appreciate it. Sometimes this probability can be quite confusing.
anonymous
  • anonymous
easier than saying \[P(B|A)=\frac{P(B)P(A|B)}{P(B)P(A|B)+P(B^c)P(A|B^c)}\]
anonymous
  • anonymous
yw
anonymous
  • anonymous
hint, on an exam if you get confused use numbers make them big
anonymous
  • anonymous
thanks for the tip.
anonymous
  • anonymous
yw
anonymous
  • anonymous
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?
anonymous
  • anonymous
be should be 'B'
anonymous
  • anonymous
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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