# 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.

• @TooTone Yes, I was thinking of the intersecting event too.
– Hawk
Commented Feb 13, 2014 at 16:19

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.

• On second thought, I don't think my first equation is correct. I have to rethink it. Commented Feb 14, 2014 at 4:29
• OK, I have created 2 cases depending on the value of $\alpha.$ Try it out. Commented Feb 17, 2014 at 18:52
• Can you be more specific? Commented Feb 17, 2014 at 18:59
• You seem to have misunderstood the question. Your result (when $\alpha \ne \frac12$) is independent of $\alpha$, but that can't possibly be true. Consider for instance the two extremes $\alpha = 0$ and $\alpha = 1$. Commented Feb 17, 2014 at 18:59
• Yes, I was assuming $\alpha > 0$ and $\alpha < 1.$ I've added that in to my solution. Commented Feb 17, 2014 at 19:02

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.

• thanks for the attribution -;) Commented Feb 17, 2014 at 20:22
• As far as I can tell, this produces the same answers as my solution. What am I missing in my solution? Commented Feb 17, 2014 at 21:19
• Right back at you! Nice try, but I can reduce your answer to mine with algebra. You think yours is dependent on $\alpha,$ but it is not! Commented Feb 17, 2014 at 23:01
• Well I'm blowed. You are absolutely right. I apologise profusely! Commented Feb 17, 2014 at 23:46
• No problem. This one's been beating me up for several days. Commented Feb 18, 2014 at 0:05

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.

• Hang on $-$ how can you deduce $\frac{b}{g}$ from an equation of the form $rb = \frac12 - sg$? I think the question is flawed. Commented Feb 17, 2014 at 19:06
• Your final equation assumes that the first term on the right-hand side of the previous equation is $\frac12g$, not $\frac12$. Commented Feb 17, 2014 at 19:10
• Aha $-$ we can solve this using the condition $b+g=1$. See my answer. Commented Feb 17, 2014 at 19:40
• @TonyK doh! I despair of my algebra sometimes!!! I'm gonna need to fix that... Commented Feb 17, 2014 at 20:08