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?

- anonymous

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- anonymous

give me one second to show my work

- 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

you are looking for a conditional probability

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## More answers

- anonymous

and i believe you need baye's formula for this one

- 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

ok that was wrong, you are looking for \(P(B|A)\)

- anonymous

yes, P(guess given that he answers correctly)

- anonymous

using the sets i wrote above, you compute
\[P(B|A)=\frac{P(A\cap B)}{P(A)}\]

- anonymous

so, can we say that P(guess AND correct) = P(guess) * P(correct)

- anonymous

yes, that is your numerator

- anonymous

oh wait no it isn't sorry

- anonymous

that is assuming the two are independent right

- anonymous

because they are not independent

- anonymous

so do i have to use the extended bayes formula

- anonymous

you compute
\[P(A\cap B)=P(B)\times P(A|B)\] yes, it is baye's formula you need

- anonymous

both of those are easy to compute, \(P(B)=.25\) and \(P(A|B)=.2\) i think

- anonymous

this line from your question
. Since there are five possible answers, the probability he guesses correctly is .25.
is wrong i think

- anonymous

if there are 5 possible answers, and he guesses, the probability he guesses right should be \(.2\) not \(.25\)

- anonymous

I'm confused. Is P(B AND A) the same as P(A AND B)?

- anonymous

i think you should get for the numerator \(.25\times .2\) and for the denominator \[.25\times .2+.75\times 1\]

- anonymous

yes, they are the same \(P(A\cap B)=P(B\cap A)\)

- 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

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

i used \cap

- 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

let me get to a desktop, it is hard to write this on a lap top

- anonymous

ok that's better

- 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

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

\(P(A|B)\) 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

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

now we need \(P(A)\) the denominator.

- anonymous

btw after we do this the hard way i will show you a very quick snappy way to do it

- anonymous

ok, give me one second to compute the answer.

- anonymous

ok

- anonymous

then we will do it in our heads with almost no computation at all

- anonymous

P(B|A) = .05/.75 = .0666

- anonymous

that is not what i get

- anonymous

your denominator is wrong
it should be \(P(A)=.75+.25\times .2=.8\)

- anonymous

oh yeah, I forgot to add the probability of guessing correctly

- 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

ok now lets do it the snap way

- anonymous

ok

- anonymous

he answers 100 questions, and he gets 75 right because he knows them and of the remaining 25 he gets 5 right

- anonymous

therefore he gets 80 right total, and of those 80 he guessed on 5
your answer is therefore \(\frac{5}{80}\)

- anonymous

wasn't that easy?

- anonymous

interesting. I didn't think of that.

- anonymous

baye's formula becomes much less mysterious when you use numbers and think

- anonymous

thank you very much for all the help. I really appreciate it. Sometimes this probability can be quite confusing.

- 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

yw

- anonymous

hint, on an exam if you get confused use numbers
make them big

- anonymous

thanks for the tip.

- anonymous

yw

- 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

be should be 'B'

- 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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