Question

an aircraft has three independent computer systems all doing the same critical task. each system has a probability of failing on a flight as follows: 0.02 for system 1, 0.1 for system 2, and 0.2 for system 3. what is the probability that all systems will fail on a flight?

Explanation:

Step1: Use the multiplication rule for independent events

For independent events \(A\), \(B\), and \(C\), the probability that \(A\), \(B\), and \(C\) all occur is \(P(A\cap B\cap C)=P(A)\times P(B)\times P(C)\)
Let \(P_1 = 0.02\) (probability of System 1 failing), \(P_2=0.1\) (probability of System 2 failing), \(P_3 = 0.2\) (probability of System 3 failing)
The formula for the probability that all systems fail is \(P = P_1\times P_2\times P_3\)

Step2: Substitute the values into the formula

Substitute \(P_1 = 0.02\), \(P_2=0.1\), \(P_3 = 0.2\) into the formula:
\(P=0.02\times0.1\times0.2\)
First, calculate \(0.02\times0.1=\frac{2}{100}\times\frac{1}{10}=\frac{2}{1000} = 0.002\)
Then, \(0.002\times0.2=\frac{2}{1000}\times\frac{2}{10}=\frac{4}{10000}=0.0004\)

Answer:

\(0.0004\)