A certain component of an electronic device has a probability of $0.1$ of failing. If there are $6$ such components in a circuit. What is the probability that at least one fails?
The Answer is $0.47$.
My Solution:
At least $1$ means more than $1$ failures
$P(1\, \text{fail}) = 0.1 \\ P(2\, \text{fails})=0.1\times0.1 \\P(3\, \text{fails})=0.1^{3}\\P(4\, \text{fails})=0.1^{4}\\P(5\, \text{fails})=0.1^{5}\\P(6\, \text{fails})=0.1^6\\ P(\text{Total})=P(1) +P(2)+...+P(6)=0.111111$
Where did I get wrong?