I've been stuck on this seemingly-simple probability problem for about 30 mins... T^T

[shrewd grove]

Suppose you are taking a multiple-choice test with C choices for each question.

In Answering a question on this test, the probability that you know the answer is P.

If you don't know the answer, you choose one at random.

What is the probability that you knew the answer to a question, given that you answered it correctly?

[blissful bone]

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[timber briar]

@shrewd grove do you know baye's theorem?

[shrewd grove]

Nope

[timber briar]

well the probability they are asking for is calculated by
$$\frac{\text{answer is correct because you know it}}{\text{ answer is correct}}$$

[eager canyon]

scilent

[timber briar]

see if the answer is correct it might be for 2 reasons

either you knew the answer

or you got lucky

so probability of getting the answer correct is :

$p\times 1 + (1-p) \times 0.25$

[eager canyon]

scilent

[timber briar]

@shrewd grovedo you understand this?

[shrewd grove]

somewhat... Let me look more into this, give me a sec

[timber briar]

@shrewd grove ?

[shrewd grove]

sorry for the wait man I had to eat dinner... But anyways I don't understand where you got the 0.25 :((

[timber briar]

so you understand the p*1 part?

[shrewd grove]

I don't understand that as well, can you explain how you got to there one by one?

[timber briar]

see its given in the question p is the probability that you know the answer of the question, and if you know the answer to the question probablity of you getting it right is 1
therefore p*1

probablity of you not knowing the answer is therefore 1-p, and then you randomly choose an option, so the probablity of that option being correct is 1/4=0.25

wait its not 0.25

its 1/c

sorry

[shrewd grove]

ahhh I see(got a bit confused there lmao).

so probability of getting an answer right (if I know the answer) is 1 • p... Right?

[timber briar]

yeah sorry about that I didn't read the question properly

[timber briar]

yes

[shrewd grove]

And probability of getting an answer right from guessing is 1/c, so we have that too

[timber briar]

(1-p)/c

[shrewd grove]

OHHH NOW I GET IT

thank u sm!!

[timber briar]

nice

but

wait

question isn't over

@shrewd grove

[shrewd grove]

yeah?

oh wait It isn't?

[timber briar]

they are asking for probablity of you knowing the answer

if the answer is correct

[timber briar]

you have to use this

$\frac{p}{p+\frac{1-p}{c}}$

[eager canyon]

scilent

[shrewd grove]

Oh dang... so this is the final answer?

[timber briar]

yeah

[shrewd grove]

I'm starting to actually get it now,

but can you tell me why the probability of getting an answer right from guessing is (1-p)/c?

[shrewd grove]

Now I am aware that (1-p) is the probability of me not knowing the answer, correct?

[timber briar]

yes

and there are c options one of them is correct

[shrewd grove]

Oh wow that makes sense

[shrewd grove]

But why do we add (1-p)/c to p?

[timber briar]

thats the total probablity of getting a answer correct

[shrewd grove]

OH RIGHT I FORGOT ABOUT THIS

Thank you!! I understand how we got to the answer now 😭

[timber briar]

nice can you close the thread

[shrewd grove]

Oh okay

Sorry I'm a bit new to this server

[timber briar]

@shrewd grove 2.

[shrewd grove]

+close