Answer:
6 alarms to be 99% certain that a burglar trying to enter is detected by at least one alarm
Step-by-step explanation:
For each alarm, there are only two possible outcomes. Either a theft is detected, or is not. The probability of an alarm detecting a theft is independent of others alarms. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
0.55 that a single alarm will detect a burglar.
This means that ![p = 0.55](https://tex.z-dn.net/?f=p%20%3D%200.55)
(a) How many such alarms should be used to be 99% certain that a burglar trying to enter is detected by at least one alarm?
The probability that no alarms detect a burglar is P(X = 0). Either no alarms detect, or at least one does. We need to have a 99% probability that at least one does. So we need P(X = 0) = 1 - 0.99 = 0.01.
We do by trial and error, find n until P(X = 0) <= 0.01.
n = 1
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{1,0}.(0.55)^{0}.(0.45)^{1} = 0.45](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B1%2C0%7D.%280.55%29%5E%7B0%7D.%280.45%29%5E%7B1%7D%20%3D%200.45)
n = 2
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{2,0}.(0.55)^{0}.(0.45)^{2} = 0.2025](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B2%2C0%7D.%280.55%29%5E%7B0%7D.%280.45%29%5E%7B2%7D%20%3D%200.2025)
n = 3
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{3,0}.(0.55)^{0}.(0.45)^{3} = 0.091125](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B3%2C0%7D.%280.55%29%5E%7B0%7D.%280.45%29%5E%7B3%7D%20%3D%200.091125)
n = 4
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{4,0}.(0.55)^{0}.(0.45)^{4} = 0.0410](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B4%2C0%7D.%280.55%29%5E%7B0%7D.%280.45%29%5E%7B4%7D%20%3D%200.0410)
n = 5
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{5,0}.(0.55)^{0}.(0.45)^{5} = 0.0185](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B5%2C0%7D.%280.55%29%5E%7B0%7D.%280.45%29%5E%7B5%7D%20%3D%200.0185)
n = 6
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{6,0}.(0.55)^{0}.(0.45)^{6} = 0.0083](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B6%2C0%7D.%280.55%29%5E%7B0%7D.%280.45%29%5E%7B6%7D%20%3D%200.0083)
So we need 6 alarms to be 99% certain that a burglar trying to enter is detected by at least one alarm