Hola,
Sea $$X_i=\begin{cases}0 & \text{si no toca en el sorteo}~ i\\ 1 & \text{si sí toca en el sorteo}~ i\end{cases}$$
$$X_i\sim \text{Be}(p=10^{-5})$$
Sea $$X=\sum_i^n X_i$$ número de veces que toca tras $$n$$ sorteos. $$X\sim \text{Bi}(n,p)$$.
$$E(X)=np$$
$$E(X)>1\Longrightarrow n>10^5~ \text{sorteos}\cdot\dfrac{1~\text{año}}{1~ \text{sorteo}}\cdot\dfrac{1~ \text{vida}}{80~ \text{años}}=1250~ \text{vidas}$$
[attachment id=0 msg=458733]
clear, clc
y=@(p) log(1-p)/log(0.99999);
x1=0:0.001:1;
figure(1)
close(1)
figure(1)
hold on
x=0:0.1:1;
for k=1:length(x)
plot([x(k) x(k)],[0 y(x(k))],'k')
plot([0 x(k)],[y(x(k)) y(x(k))],'k')
plot(x(k),y(x(k)),'bo','MarkerFaceColor','b','MarkerSize',3)
text(-0.15,y(x(k)),[num2str(ceil(y(x(k)))) '\rightarrow'],'color','b')
end
plot(x1,y(x1),'b')
axis([0 1 0 3e5])
set(gca,'ytick',[])
xlabel('$p^*$','interpreter','latex')
title('$P(X\ge 1)=1-(0.99999)^n\ge p^*$','interpreter','latex')
text(-0.1,3e5,'$n$','interpreter','latex')
text(-0.13,3e5-1e4,'(sorteos)','FontSize',7)
