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there are seven geologic faults (and possibly more) capable of generati…

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 decimal rounded to two decimal places as needed.)

Explanation:

Step1: Calculate non - eruption probabilities

Convert the given eruption probabilities to non - eruption probabilities.
The non - eruption probabilities are:
For 24% eruption probability, non - eruption probability is $1 - 0.24=0.76$.
For 21% eruption probability, non - eruption probability is $1 - 0.21 = 0.79$.
For 11% eruption probability, non - eruption probability is $1 - 0.11=0.89$.
For 10% eruption probability, non - eruption probability is $1 - 0.10 = 0.90$.
For 4% eruption probability, non - eruption probability is $1 - 0.04=0.96$.
For 3% eruption probability (twice), non - eruption probabilities are both $1 - 0.03 = 0.97$.

Step2: Calculate the probability that none of the faults erupt

Since the events are independent, the probability that none of the faults erupt is the product of all non - eruption probabilities.
$P(\text{none})=0.76\times0.79\times0.89\times0.90\times0.96\times0.97\times0.97$
$P(\text{none})\approx0.41$

Step3: Calculate the probability that at least one fault erupts

The probability that at least one fault erupts is the complement of the probability that none of the faults erupt.
$P(\text{at least one}) = 1 - P(\text{none})$
$P(\text{at least one})=1 - 0.41 = 0.59$

Answer:

$0.59$