Respuesta :
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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
0.55 that a single alarm will detect a burglar.
This means that [tex]p = 0.55[/tex]
(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
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{1,0}.(0.55)^{0}.(0.45)^{1} = 0.45[/tex]
n = 2
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{2,0}.(0.55)^{0}.(0.45)^{2} = 0.2025[/tex]
n = 3
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{3,0}.(0.55)^{0}.(0.45)^{3} = 0.091125[/tex]
n = 4
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{4,0}.(0.55)^{0}.(0.45)^{4} = 0.0410[/tex]
n = 5
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{5,0}.(0.55)^{0}.(0.45)^{5} = 0.0185[/tex]
n = 6
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{6,0}.(0.55)^{0}.(0.45)^{6} = 0.0083[/tex]
So we need 6 alarms to be 99% certain that a burglar trying to enter is detected by at least one alarm
