Question

A home security system is designed to have a 99% reliability rate. Suppose that nine homes equipped with this system experience an attempted burglary. Find the probabilities of these events:_________.

A. At least one alarm is triggered.
B. More than seven alarms are triggered.
C. Eight or fewer alarms are triggered.

Answer 1

a)
`P(X \geq 1) = 1-P(X<1) = 1-P(X=0)`

`P(X=0)=(9C0)(0.99)^0 (1-0.99)^(9-0)=1x10^(-18)`

And replacing we got:

`P(X \geq 1) =1 -1x10^(-18) \approx 1`

b)
`P(X=7)=(9C7)(0.99)^7 (1-0.99)^(9-7)=0.003355`

`P(X=8)=(9C8)(0.99)^8 (1-0.99)^(9-8)=0.083047`

`P(X=9)=(9C9)(0.99)^9 (1-0.99)^(9-9)=0.913517`

And adding we got:

`P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992`

c)
`P(X \leq 8) =1 -P(X>8) = 1-P(X=9)`

`P(X=9)=(9C9)(0.99)^9 (1-0.99)^(9-9)=0.913517`

And replacing we got:

`P(X \leq 8)= 1-0.913517=0.086483`

Explanation:

Let X the random variable of interest "numebr of times that an alarm is triggered", on this case we now that:

`X \sim Binom(n=9, p=0.99)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

Part a

We want to find this probability:

`P(X \geq 1) = 1-P(X<1) = 1-P(X=0)`

`P(X=0)=(9C0)(0.99)^0 (1-0.99)^(9-0)=1x10^(-18)`

And replacing we got:

`P(X \geq 1) =1 -1x10^(-18) \approx 1`

Part b

`P(X \geq 7)= P(X=7) +P(X=8)+ P(X=9)`

`P(X=7)=(9C7)(0.99)^7 (1-0.99)^(9-7)=0.003355`

`P(X=8)=(9C8)(0.99)^8 (1-0.99)^(9-8)=0.083047`

`P(X=9)=(9C9)(0.99)^9 (1-0.99)^(9-9)=0.913517`

And adding we got:

`P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992`

Part c

`P(X \leq 8) =1 -P(X>8) = 1-P(X=9)`

`P(X=9)=(9C9)(0.99)^9 (1-0.99)^(9-9)=0.913517`

And replacing we got:

`P(X \leq 8)= 1-0.913517=0.086483`