Answer:
A home security system is designed to have a 99% reliability rate. Suppose that twelve homes equipped with this system experience an attempted burglary. Find the probability of these events:
a. At least one of the alarms is triggered.
b. More than ten of the alarms are triggered.
c. Eleven or fewer alarms are triggered.
The answers to the question are
(a) 1
(b) 0.9937
(c) 0.1137
Step-by-step explanation:
It is first important to note that the question is a Binomial experiment, therefore
n = Number of burglary attempt = 12
p = the probability of an alarm being triggered = 99 % = 0.99
(a) The probability of at least one is given by
at k = 0, P(X=0) = C₀¹²×0.99×(1-0.99)¹²⁻⁰ ≈ 0
By the complement rule of probability, the probability of at least one is the complement of the probability that non were triggered
Hence P(X ≥ 1) = 1 - P(X=0) = 1 - 0 = 1
(b) The probability that 11 or 12 re triggered k = 11, 12
P(X = 11) = C¹²₁₁×0.99¹¹ ×(1-0.99)¹²⁻¹¹ ≈ 12×0.99¹¹×0.01¹ = 0.1074
P(X = 12) = C¹²₁₂×0.99¹² ×(1-0.99)¹²⁻¹² ≈ 1×0.99¹²×0.01⁰ = 0.8863
P(X > 10) = P(X = 11) + P(X = 12) = 0.1074 + 0.8863 = 0.9937
(c) The probability that 11 or fewer alarms were triggered
P(X = 12) = C¹²₁₂ × 0.99¹² ×(1-0.99)¹²⁻¹² ≈ 1×0.99¹²×0.01⁰ = 0.8863
we use the complement rule and we get
P(X ≤ 11) = 1 - P(X = 12)
= 1 - 0.8863 = 0.1137
