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 probabilities of these events. (Round your answers to three decimal places.)

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