Question

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?

Answer 1

give me one second to show my work

Answer 2

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) ??

Answer 3

you are looking for a conditional probability

Answer 4

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

Answer 5

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
\[P(A|B)=\frac{P(A\cap B)}{P(B)}\]

Answer 6

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

Answer 7

yes, P(guess given that he answers correctly)

Answer 8

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

Answer 9

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

Answer 10

yes, that is your numerator

Answer 11

oh wait no it isn't sorry

Answer 12

that is assuming the two are independent right

Answer 13

because they are not independent

Answer 14

so do i have to use the extended bayes formula

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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.

Answer 24

i used \cap

Answer 25

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 \[P(A|B) = P(A \cap B)/P(B)\]

Answer 26

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

Answer 27

ok that's better

Answer 28

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

Answer 29

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 \(P(A\cap B)=P(B)\times P(A|B)\)

Answer 30

\(P(A|B)\) 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
\[P(A\cap B)=P(B)P(A|B)=.25\times .2\]

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

ok, give me one second to compute the answer.

Answer 35

ok

Answer 36

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

Answer 37

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

Answer 38

that is not what i get

Answer 39

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

Answer 40

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

Answer 41

\(A\) 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

Answer 42

ok now lets do it the snap way

Answer 43

ok

Answer 44

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

Answer 45

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

Answer 46

wasn't that easy?

Answer 47

interesting. I didn't think of that.

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

yw

Answer 52

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

Answer 53

thanks for the tip.

Answer 54

yw

Answer 55

A little way up, you said that \[P(A|B) = .2\] I was thinking about that and tried doing a probability tree.


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?