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Answer:
[tex]0.4138[/tex]
Step-by-step explanation:
Given
[tex]x \to storm[/tex]
[tex]\mu_x = 1.0[/tex]
[tex]y \to fire[/tex]
[tex]\mu_y = 1.5[/tex]
[tex]z \to theft[/tex]
[tex]\mu_z = 2.4[/tex]
Let the event that the above three factors is greater than 3 be represented as:
[tex]P(A > 3)[/tex]
Using complement rule, we have:
[tex]P(A > 3) = 1 - P(A \le 3)[/tex]
This gives:
[tex]P(A > 3) = 1 - P(\{x \le 3\}\ n\ \{y \le 3\}\ n \{z \le 3\}\)[/tex]
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The exponential distribution formula of each is:
[tex]P(x \le k) = 1 - e^{-\frac{k}{\mu}}[/tex]
So, we have:
[tex]k = 3; \mu_x = 1[/tex]
[tex]P(x \le 3) = 1 - e^{-\frac{3}{1}} = 1 - e^{-3} = 0.9502[/tex]
[tex]k=3; \mu_y = 1.5[/tex]
[tex]P(y \le 3) = 1 - e^{-\frac{3}{1.5}} = 1 - e^{-2} = 0.8647[/tex]
[tex]k = 3; \mu_z = 2.4[/tex]
[tex]P(z \le 3) = 1 - e^{-\frac{3}{2.4}} = 1 - e^{-1.25} = 0.7135[/tex]
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[tex]P(A > 3) = 1 - P(\{x \le 3\}\ n\ \{y \le 3\}\ n \{z \le 3\}\)[/tex]
[tex]P(A > 3) = 1 - (0.9502 * 0.8647 *0.7135)[/tex]
[tex]P(A > 3) = 1 - 0.5862[/tex]
[tex]P(A > 3) = 0.4138[/tex]