In a quiz contest,probabilities of teachers,boy students and girl students answering the question correctly are $\alpha,\beta$ and $\gamma$... In a quiz contest,probabilities of teacher,boy students and girl students answering the question correctly are $\alpha,\beta$ and $\gamma$ respectively.The probability of teacher and student agreeing to the same answer is $\dfrac12$.Find the ratio of the number of boy students to the girl students.  
My work:
There arise two cases when they agree-
(1) The teacher and the student are correct.
(2) The teacher and the student are wrong.  
Case (1) has two subcases-
(i) The teacher and the male student are correct.
(ii) The teacher and the female student are correct.  
Case (2) has two subcases-
(i) The teacher and the male student are wrong.
(ii) The teacher and the female student are wrong.  
From this we can get,
$\alpha \beta+(1-\alpha)(1-\beta)+\alpha \gamma+(1-\alpha)(1-\gamma)=\dfrac12$  
Now, I consider that there are total $n$ students and I consider there are $k$ male students. I need to find out the ratio $k:n-k$. But I cannot proceed further. Please help.
 A: Assume $0 < \alpha < 1.$ Let the number of boys be $b,$ the number of girls be $g,$ and the desired ratio be $x={b/g}.$ If $\alpha$ is unequal to ${1 \over 2},$ the restriction that the teacher agrees with the students with probability $1 \over 2$ can be expressed as $${1 \over 2}= { {\beta b + \gamma g} \over {b+g}}$$ Now divide both numerator and denominator of the right hand side by $g.$ Then we have $${1 \over 2}= { {\beta x + \gamma} \over {x+1} }$$ Solving for $x,$ we find $$x = { {2 \gamma - 1} \over {1- 2 \beta} } $$ In the case for $\alpha = {1 \over 2},$ then the ratio of boys to girls could be anything.
A: 
In a quiz contest,probabilities of teacher,boy students and girl students answering the question correctly are α, β and γ respectively.

I think the only way this question is tractable is if you assume independence between teacher and students getting the answer correct.

The probability of teacher and student agreeing to the same answer is $\frac12$.

I interpret this as meaning choosing any student randomly, the probability that that the teacher and student agree is $\frac12$.

Find the ratio of the number of boy students to the girl students.

Let $b,g$ be the probabilities of being a boy or a girl.
By the theorem of total probability, $\mathbb{P}$(agree) = $\mathbb{P}$(agree|student is boy)$\;b$ + $\mathbb{P}$(agree|student is girl)$\;g$.
And as you wrote, the probability of agreeing is in each case (again by the theorem of total probability) the probability of both being correct + the probability of both being wrong. Hence
$$\begin{align}
(\alpha\beta + (1-\alpha)(1-\beta))b + (\alpha\gamma + (1-\alpha)(1-\gamma))g &= \frac{1}{2} \\\\
(\alpha\beta + (1-\alpha)(1-\beta))b &= \frac{1}{2} - (\alpha\gamma + (1-\alpha)(1-\gamma))g \\\\
(\alpha\beta + (1-\alpha)(1-\beta))b &= \frac{1 -2(\alpha\gamma + (1-\alpha)(1-\gamma))g}{2} \\\\
\end{align}$$
This can be solved by using the fact that $b = (1-g)$: see TonyK's answer for the rest of the story.
A: Starting from (i.e. stealing) TooTone's equation:
$$(\alpha\beta + (1-\alpha)(1-\beta))b = \frac{1}{2} - (\alpha\gamma + (1-\alpha)(1-\gamma))g$$
we can write it as $rb=\frac12 - sg$, and together with the further condition $b+g=1$, we get $rb=\frac12 - s(1-b)$, or
$$b = \frac{1-2s}{2(r-s)}$$
Similarly,
$$g = \frac{1-2r}{2(s-r)}$$
Hence
$$\frac{b}{g} = \frac{1-2s}{2r-1}$$
Writing this all out:
$$\frac{b}{g}=\frac{1-2(\alpha\gamma + (1-\alpha)(1-\gamma))}{2(\alpha\beta + (1-\alpha)(1-\beta))-1}$$
This is undefined if $r=s=\frac12$. In that case, the boy/girl ratio is undetermined.
