Independent Marksmen Problem

Suppose two marksmen are aiming at a target. Each marksman aims independently, but player A is twice as likely to hit the target as player B. If the probability the the target is hit is 1/2, what is the probability that A hits the target?
Let the probability that B hits the target be  
\[x\]
, then the probability that A hits the target is  
\[2x\]
.
Since A and B aim independently the probability they both hit the target is  
\[x \times 2x=2x^2\]
. The probability that A but not B hits the target is  
\[2x-2x^2\]
.
The probability that B but not A hits the target is  
\[x-2x^2\]
.
The Venn diagram is shown below.

marksmen problem

Then  
\[(2x-2x^2)+2x^2+(x-2x^2)=\frac{1}{2} \rightarrow 4x^2-6x+1=0\]
.
The solution is  
\[x= \frac{6 \pm \sqrt{6"-4 \times 4 \times 1}}{2 \times 4}=\frac{3 \pm \sqrt{5}}{4}\]
.
\[x \lt 1\]
  so  
\[\frac{3- \sqrt{5}}{4}\]
.