Probability of answer being correct if multiple student marked it correct I got a situation where I want to calculate the probability of a answer being correct if multiple student marked it as correct.
So, here is the situation. I have a multiple choice question with 3 options, only one of them is correct. We are given that the student is 80% of time correct.
p(option is correct ans|student selected that option) = 0.8
Then what will be the probability of the option to be correct answer if 2 student marked it as correct. Certainly it will be more than 0.8. But I am not able to find how to calculate.
i.e. p(option is correct ans| student s1 and s2 selected that option) = ?
Is the data sufficient for calculation of this probability.
Earlier I was modeling this as finding probability of cancer if two tests are positive, but then in that case we were given the prior probability of cancer. Here we don't have prior probability of answer being correct. Can I assume that it is 1/3. Also then in denominator there will be a joint probabilty of two student selecting the option, which I think is not independent. It is independent conditionally given the answer is correct, but not in general.
Please correct me if my above model of this problem is wrong and how can i calculate this.
Can this be generalised when we have N options and X out of Y student selected the option.
 A: Isn't using conditional probability easy?:
Suppose the options are A,B,C and correct is B.
They both can select $AA,AB,AC,...$and so on with $3^2=9$ choices in total.
So they'll select B with probability $0.8$ and rest two with let $a$ and $c$ respectively for A and C.
Now $a+c+0.8=1\iff a+c=0.2$ and $a,c\ge0$
$$\newcommand{\t}[1]{\text{#1}}
\zeta=\t P(\t{option X is correct answer | 2 student marked X as correct})\\=\frac{\t P(\t{option X is correct answer $\cap$ 2 student marked X as correct})}{\t P(\t{2 student marked X as correct})}\\=\frac{0.8*0.8}{0.8*0.8+a^2+c^2}$$
This would range from:
$$\frac{0.64}{0.66}\Bigg|_{a=c=0.1}\t{ to }\qquad\frac{0.64}{0.68}\Bigg|_{\substack{a=0,c=0.2\\\t{ or }c=0,a=0.2}}$$
And both of them are approximately:
$$94.11\%\le\zeta\le96.96\%$$
Quite larger than 80% :D.
A: Let $A_i$ be the random variable giving the number of the answer selected by student $i$, where there are $m$ students and $n$ possible answers.  We assume
these random variables are independent, and that if $c$ is the correct answer
$P(A_i = c) = p$ while for $a \ne c$, $P(A_i = a) = (1-p)/(n-1)$ (i.e. an incorrect answer is equally likely to be any of the incorrect alternatives).
Suppose $x > m/2$ (so at most one answer can be selected by $x$ of the students).
The probability that exactly $x$ of the $m$ students select the correct answer
is $${m \choose x} p^x (1-p)^{m-x}$$  The probability that exactly $x$ of the $m$ students select a particular incorrect answer is $${m \choose x} \left(\dfrac{1-p}{n-1}\right)^x \left(1 - \dfrac{1-p}{n-1}\right)^{m-x}$$
The total probability that exactly $x$ of the $m$ students select the same answer is then 
$$ {m \choose x} p^x (1-p)^{m-x} + (n-1) {m \choose x} \left(\dfrac{1-p}{n-1}\right)^x \left(1 - \dfrac{1-p}{n-1}\right)^{m-x}$$
Thus the conditional probability that an answer is correct, given that it was selected by exactly $x$ of the $m$ students, is 
$$ \dfrac{p^x (1-p)^{m-x}}{p^x (1-p)^{m-x} + (n-1)  \left(\dfrac{1-p}{n-1}\right)^x \left(1 - \dfrac{1-p}{n-1}\right)^{m-x}}$$
