probability that he is NOT killed in $20$ years Question

$\text{suppose that the probability of being killed in a single flight is 
 }P_{c}=\frac{10^{-6}}{4} \text{based}$
  $\text{ on available statistics. Assume that different flights are independent. If a businessman}$ $\text{takes 20 flights per year, what is the probability that he is killed in a plane crash within the}$ $\text{next 20 years? (Let's assume that he will not die because of another }$
  $\text{reason within the next 20 years.)}?$

My Approach

$$\text{probability of NOT  being killed in a single flight is 
 }P_{c}=1-\frac{10^{-6}}{4}$$

There are $20$ flights per year .so Probability that he will not be dying in a year
$$=\left(1-\frac{10^{-6}}{4}\right)^{20}$$

probability that he is killed in a plane crash within the next $20$ years=probability that he is NOT killed in $20$ years and will be dying in next upcoming $20$years.
  $$\left(\left(1-\frac{10^{-6}}{4}\right)^{20}\right)^{20} \times \left(\frac{10^{-6}}{4}\right)^{20}$$

Am i correct?
Bit confused.please help me
Thanks!
 A: The total number of flights that he will take during the next $20$ years is $N=20\times20=400$.
Let $p_s$ be the probability that he survives a given single flight. Then we have
$$p_s=1-p_c$$
Since these flights are independent, the probability that he will survive all $N=400$ flights is $$P(\mbox{Survive N flights })=p_s\times p_s\times......\times p_s=p_s^{N}=(1-p_c)^N$$Let $A$ be the event that the businessman is killed in a plane crash within the next $20$ years. Then$$P(A)=1-(1-p_c)^N=9.9995\times10^{-5}\approx \dfrac{1}{10000}$$
A: There is another approach to this problem.
Consider $P(n)$ be the probability that the businessman dies in his $n$th flight.
 $p_c$ is the probability of the person being killed in any flight.
Now, $P(i) = {(1-p_c)}^{i-1} \dot p_c$ i.e. he survived $i-1$ flights and died in $i$th flight.
He will take total $20 \times 20$ flights in 20 years. The probability that he will die in any of these flights is $\sum_{i=1}^{20 \times 20}{P(i)}$.  
Therefore, the required probability is $1 - \sum_{i=1}^{20 \times 20}{P(i)}$
