The probability of at least two persons of $6$ boys and $8$ girls live for $80$ years I have this small problem, that has been preventing me from proceeding in my studies. I'd be very grateful if someone could help me out.
"$29.4$% of boys and $55.1$% of girls are expected to live for $80$ years. Hence find the probability that at least two people out of a group of $6$ boys and $8$ girls live for $80$ years."
Thanks!
-José
 A: Let $a$ be the probability none of them reaches age $80$. Let $b$ be the probability that exactly $1$ of them reaches age $80$. Our required probability is $1-(a+b)$.  We show how to compute $a$ and $b$.  
Computing $a$: The probability all are dead is the probability all the boys are, times the probability all the girls are. The probability all $6$ boys are dead before age $80$ is $(0.706)^6$. Find a similar expression for the girls, and multiply.
Computing $b$: Exactly $1$ person can reach age $80$ in two ways. We find the probability of each and add. Let's compute the probability $1$ boy is alive and all the girls are dead.
The probability exactly $1$ boy is alive is $\binom{6}{1}(0.294)(0.706)^5$. Multiply this by the probability all the girls are dead. You know this if you completed the calculation of $a$. 
(You are probably familiar with the above binomial distribution calculation. The live one can be chosen in $\binom{6}{1}$ ways. We multiplied this by the probability he is alive and the other $5$ are dead.)
For  $1$ of the girls alive and the boys all dead, do something similar. 
A: \begin{align}
P(N\geq 2)&=1-P(N\leq1)=1-P(N=0)-P(N=1)\\
P(N=0)&=P(B=0,G=0)\\&\stackrel{ind.}{=}P(B=0)\times P(G=0)\\&=(1-0.294)^6\times(1-0.551)^8\\
&\approx 0.02\%\\
P(N=1)&=P(B=1,G=0)+P(B=0,G=1)\\&=6\times 0.294\times (1-0.294)^5\times (1-0.551)^8+8\times 0.551\times (1-0.551)^7\times (1-0.294)^6\\
&\approx 0.25\%\\
P(N\geq 2)&\approx 99.72\%
\end{align}
