# Determine percentage of crow hybrids

We had in class following question which I had no idea how to get to the correct answer:

The carrion crow and the hooded crow are fertile together, but their reproductive success is reduced by 50%. In a certain region exist two populations of both species of roughly the same size. Thus, mixed couples occur in about 10% of all cases. What is the percentage of hybrids in the F1-generation?

(Translated from German.)

I thought it would be something like 5/95 = 5,3% but apparently the answer is 1%. Why?

• why do you think it would be 5/95? and how, according to you, can the answer be 1%? please show some research effort before asking question here. – another 'Homo sapien' Feb 6 '17 at 12:02
• 1% is the answer according to the solution. 5/95 because 90% of the couples have 100% success, and 10% have 50% sucess. @another'Homosapien' – Adrian Feb 6 '17 at 12:11
• Homework questions are off-topic here unless you show some research effort. Tell us what research effort you have put on the question, lest this question is likely to be closed. – another 'Homo sapien' Feb 6 '17 at 12:17
• Well I told you what I thought was the correct approach. I did not find any hints in our material. I did further research before posting here. I didnt find any source dealing with mathematical discussions of this topic. Further, to me my solution seems correct. @another'Homosapien' – Adrian Feb 6 '17 at 12:34

In order to make your reasoning clearer, you should use more formal notations and explain your thinking step by step. Here's a proposition.

Let's use the following notations :

• $C$: the total number of couples (mixed and not mixed)
• $r$: the reproductive success
• $F1_h$: the number of hybrids in the F1 generation
• $F1_{nh}$: the number of non hybrids in the F1 generation

You are looking for the percentage of hydrids in the F1 generation, which is: $x = \frac{F1_h}{F1_h + F1_{nh}}$

You know that $10~\%$ of the couples are mixed couples and that their reproductive success is reduced by $50~\%$. This can be written: $\begin{cases} 0.9\cdot C \cdot r = F1_{nh} \\ 0.1\cdot C\cdot \frac{r}{2} = F1_h \end{cases}$

Thus: $\displaystyle x = \frac{F1_h}{F1_h + F1_{nh}} = \frac{0.05\cdot C\cdot r}{C\cdot r\cdot 0.95} = \frac{0.05}{0.95}$

This is indeed the result you suggested. So the correction you were given might not be correct. Or maybe there was some more information in your homework that you ignored...