Question

a string of christmas lights contains 20 lights. the lights are wired in series, so that if any light fails, the whole string will go dark. each light has probability 0.98 of working for a 3 - year period. the lights fail independently of each other. find the probability that the string of lights will remain bright for 3 years. (round to 3 decimal places. leave your answer in decimal form.)

Explanation:

Step1: Use the multiplication rule for independent events

Since the lights fail independently, and the string remains bright only if all 20 lights work. The probability that one light works is \(p = 0.98\). For \(n=20\) independent lights, the probability that all work is \(P=\prod_{i = 1}^{20}p_i\). Because \(p_1=p_2=\cdots=p_{20}=0.98\), we have \(P = 0.98^{20}\).

Step2: Calculate \(0.98^{20}\)

We know that \(a^n=e^{n\ln(a)}\). So, \(n = 20\) and \(a=0.98\). First, \(\ln(0.98)\approx- 0.0202\). Then \(n\ln(a)=20\times(-0.0202)=-0.404\). And \(e^{-0.404}\approx0.668\).

Answer:

\(0.668\)