Answer
No, it would not be unusual.
Work Step by Step
$P(no~more~than~3)=P(X\leq3)=P(0)+P(1)+P(2)+P(3)={}_{30}C_{0}~0.05^0~(0.95)^{30}+{}_{30}C_{1}~0.05^1~(0.95)^{29}+{}_{30}C_{2}~0.05^2~(0.95)^{28}+{}_{30}C_{3}~0.05^3~(0.95)^{27}=\frac{30!}{0!\times(30-0)!}\times1\times0.95^{30}+\frac{30!}{1!\times(30-1)!}\times0.05\times(0.95)^{29}+\frac{30!}{2!\times(30-2)!}\times0.0025\times(0.95)^{28}+\frac{30!}{3!\times(30-3)!}\times0.000125\times(0.95)^{27}=1\times1\times0.95^{30}+30\times0.05\times(0.95)^{29}+435\times0.0025\times(0.95)^{28}+4060\times0.000125\times(0.95)^{27}=0.9392$
The probability that more than 3 died is the complement of the probability that no more than 3 died.
Using the Complement Rule (see page 275):
$P(more~than~3)=1-P(no~more~than~3)=1-0.9392=0.0608$
$P(more~than~3)\gt0.05$. So, it is not an unusual event.
