## 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.

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
$\frac{3- \sqrt{5}}{4}$
.