Answer
$P(1)=0.315$
If we randomly select 10 visitors to the ER 1000 times, we would expect about 315 trials result in exactly 1 of 10 will die within 1 year.
Work Step by Step
$P(x)={}_nC_{x}~p^x~(1-p)^{n-x}=$
n = 10
x = 1
p = 0.05
1 - p = 1 - 0.05 = 0.95
n - x = 10 - 1 = 9
$P(1)={}_{10}C_{1}\times0.05^1\times(0.95)^{9}=\frac{10!}{1!(10-1)!}\times0.05\times(0.95)^9=\frac{10!}{9!}\times0.05\times(0.95)^9$
But, $10!=10\times9\times8\times7\times6\times5\times4\times3\times2\times1=10\times9!$
$P(1)=\frac{10\times9!}{9!}\times0.05\times(0.95)^9=10\times0.05\times(0.95)^9=0.315$
