Is it going to matter after all, debotosh?
(If not, then Baye's Theorem should settle this without much difficulty - I think)
Coz when an answer is correct, it's taken that both options are ticked correctly....what are ur views?
in a test an examinee either guesses or copies or he knows the answer to a multiple choice question, with more than one answer, and having 4 alternatives. the probabilty that he makes guess is 1/3 and that he copies the answer is 1/6. the probability that his answer is correct , given that he copied it is 1/8 .find the probability that he knew the answer to a question, given that he correctly answered it !
Is it going to matter after all, debotosh?
(If not, then Baye's Theorem should settle this without much difficulty - I think)
Coz when an answer is correct, it's taken that both options are ticked correctly....what are ur views?
Well I don't know what i was thinking last time when I wrote this stuff above....[78]
What's only going to change due to the fact that it's multiple-option-correct is the probability that he gets the answer correct, given he guesses it....
Rest can be done as usual considering it to be single option correct, and then the std way of using Baye's theorem.....
Use the classical definition for getting the prob that he correctly guesses the ans viz....
Number of cases possible for the answer to be correct is only 1, since the right combination has to be unique....
Lets now see how many casses are possible for this ......
The boy does not know how many options are correct......
So the reqd no. of possible cases are :-
1) He selects 1 option...
2) He selects 2.....like this,
Thus reqd prob=14C1+4C2+...4C4=115
I hope after this organic can complete it.....if u need the full soln, nudge me!