# A home has installed 4 security systems to detect attempted break-ins. What is the probability that at least one of the 4 security systems will detect it?

## They operate independently, and each has a .40 probability of detecting a break-in.

Jan 19, 2017

color(green)(87.04%)

#### Explanation:

Probability of at least one system detecting a break-in
$\textcolor{w h i t e}{\text{XXX}} = 1 -$ (probability of none detecting a break-in)

If each system has a probability of $0.40$ of detecting a break-in
then each system has a probability of $1 - 0.40 = 0.60$ of not detecting a break-in.

The first system will fail to detect a break-in $0.60$ of the time.
Of that $0.60$ of the time that the first system fails,
$\textcolor{w h i t e}{\text{XXX}}$the second system will fail $0.60$ of the time.
Therefore both of the first 2 systems will fail $0.60 \times 0.60 = 0.36$ of the time.

Similarly the first 3 systems will fail $0.36 \times 0.60 = 0.216$ of the time;

and all 4 systems will fail $0.216 \times 0.60 = 0.1296$ of the time.

Therefore at least one of the systems will succeed 1-0.1296=0.8704=87.04% of the time 