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Mathematics
24.04.2021 11:00
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A home security system is designed to have a 99% reliability rate. Suppose that

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.
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sandrafina2004
sandrafina2004
4,8(63 marks)

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(X8) = 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(X8) = 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

willia343
willia343
4,9(38 marks)

0.8 or 4/5


Step-by-step explanation:1/5 x 4 = 4/5

And to check your work  

4/5 / 4 = 1/5

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