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give me one second to show my work

you are looking for a conditional probability

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

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

yes, P(guess given that he answers correctly)

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

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

yes, that is your numerator

oh wait no it isn't sorry

that is assuming the two are independent right

because they are not independent

so do i have to use the extended bayes formula

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

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

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

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

i used \cap

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

ok that's better

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

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

ok, give me one second to compute the answer.

ok

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

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

that is not what i get

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

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

ok now lets do it the snap way

ok

wasn't that easy?

interesting. I didn't think of that.

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

easier than saying
\[P(B|A)=\frac{P(B)P(A|B)}{P(B)P(A|B)+P(B^c)P(A|B^c)}\]

yw

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

thanks for the tip.

yw

be should be 'B'