Question

A home security system is designed to have a 95% reliability rate. Suppose that eleven homes equipped with this system experience an attempted burglary. Find the probabilities of these events. (Round your answers to three decimal places.)

Answer 1

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

And we can find:

`P(X=0) = (11C0) (0.95)^0 (1-0.95)^(11-0)= 4.88x10^(-15)`

So then:

`P(X \geq 1)= 1- 4.88x10^(-15)\approx 1`

b)
`P(X >7) = P(X \geq 8) = P(X=8) +P(X=9) +P(X=10) + P(X=11)`

`P(X=8) = (11C8) (0.95)^8 (1-0.95)^(11-8)=0.0137`

`P(X=9) = (11C9) (0.95)^9 (1-0.95)^(11-9)= 0.0867`

`P(X=10) = (11C10) (0.95)^(10) (1-0.95)^(11-10)= 0.3293`

`P(X=11) = (11C11) (0.95)^(11) (1-0.95)^(11-11)= 0.5688`

Adding the values we got:

`P(X >7) = P(X \geq 8) = P(X=8) +P(X=9) +P(X=10) + P(X=11)=0.9984`

c)
`P(X\leq 8)`

`P(X \leq 8) = 1- P(X>8) = 1- [P(X=9) + P(X=10) + P(X=11)]`

`P(X=9) = (11C9) (0.95)^9 (1-0.95)^(11-9)= 0.0867`

`P(X=10) = (11C10) (0.95)^(10) (1-0.95)^(11-10)= 0.3293`

`P(X=11) = (11C11) (0.95)^(11) (1-0.95)^(11-11)= 0.5688`

`P(X \leq 8) = 1- P(X>8) = 1- [0.0867+ 0.3293 + 0.5688]=0.0152`

Explanation:

Assuming the following questions:

a. At least one of the alarms is triggered.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

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!)`

For this case we define X as " the number of alarms that are triggered" and we define the random variable with the following distribution:

`X \sim Bin (n =11, p = 0.95)`

For this case we want this probability:

`P(X \geq 1)`

And we can use the complement rule and we got:

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

And we can find:

`P(X=0) = (11C0) (0.95)^0 (1-0.95)^(11-0)= 4.88x10^(-15)`

So then:

`P(X \geq 1)= 1- 4.88x10^(-15)\approx 1`

b. More than seven of the alarms are triggered.

For this case we want this probability:

`P(X >7) = P(X \geq 8) = P(X=8) +P(X=9) +P(X=10) + P(X=11)`

We find the individual proababilities and we got:

`P(X=8) = (11C8) (0.95)^8 (1-0.95)^(11-8)=0.0137`

`P(X=9) = (11C9) (0.95)^9 (1-0.95)^(11-9)= 0.0867`

`P(X=10) = (11C10) (0.95)^(10) (1-0.95)^(11-10)= 0.3293`

`P(X=11) = (11C11) (0.95)^(11) (1-0.95)^(11-11)= 0.5688`

Adding the values we got:

`P(X >7) = P(X \geq 8) = P(X=8) +P(X=9) +P(X=10) + P(X=11)=0.9984`

c. Eight or fewer alarms are triggered.

We want this probability:

`P(X\leq 8)`

Using the complement rule we have:

`P(X \leq 8) = 1- P(X>8) = 1- [P(X=9) + P(X=10) + P(X=11)]`

`P(X=9) = (11C9) (0.95)^9 (1-0.95)^(11-9)= 0.0867`

`P(X=10) = (11C10) (0.95)^(10) (1-0.95)^(11-10)= 0.3293`

`P(X=11) = (11C11) (0.95)^(11) (1-0.95)^(11-11)= 0.5688`

`P(X \leq 8) = 1- P(X>8) = 1- [0.0867+ 0.3293 + 0.5688]=0.0152`