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

Answer:

a) $P(X \geq 1) = 1-P(X$

$P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}$

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:

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

Step-by-step 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:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want to find this probability:

$P(X \geq 1) = 1-P(X$

$P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}$

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:

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