Question

there are seven geologic faults (and possibly more) capable of generating a magnitude 6.7 earthquake in the region around a certain city. their probabilities of rupturing by the year 2032 are 24%, 21%, 11%, 10%, 4%, 3%, and 3%. complete parts (a) and (b)
(a) calculate the probability that at least one of these faults erupts by the year 2032, assuming that these are independent events.
the probability is
(type an integer or

× try again.

first find the probability that no faults rupture. recall that if two events, e and f, are independent, then p(e ∩ f)=p(e)·p(f). the event that at least one fault ruptures is the complement of no faults rupturing

ok

Explanation:

Step1: Find probabilities of non - rupture

The probabilities of the faults rupturing are \(24\%,21\%,11\%,10\%,4\%,3\%,3\%\). So the probabilities of non - rupture are \(1 - 0.24=0.76\), \(1 - 0.21 = 0.79\), \(1 - 0.11=0.89\), \(1 - 0.10 = 0.90\), \(1 - 0.04=0.96\), \(1 - 0.03=0.97\), \(1 - 0.03=0.97\) respectively.

Step2: Calculate probability of no faults rupturing

Since the events are independent, the probability that no faults rupture is \(P(\text{no rupture})=0.76\times0.79\times0.89\times0.90\times0.96\times0.97\times0.97\)
\[

$$\begin{align*} P(\text{no rupture})&=0.76\times0.79\times0.89\times0.90\times0.96\times0.97\times0.97\\ &\approx0.417 \end{align*}$$

\]

Step3: Calculate probability of at least one fault rupturing

The probability that at least one fault ruptures is the complement of the event that no faults rupture. So \(P(\text{at least one rupture}) = 1 - P(\text{no rupture})\)
\[

$$\begin{align*} P(\text{at least one rupture})&=1- 0.417\\ &=0.583 \end{align*}$$

\]

Answer:

\(0.583\)