Respuesta :
Using the binomial distribution, we find that:
a) There is a 0.648 = 64.8% probability that there will be at least two hurricanes from which the home will experience no damage before the second hurricane from which the home will experience damage.
b) There is a 0.1728 = 17.28% probability that the home will experience its second damage from the fourth hurricane, given that the home has not experienced two damages from the first two hurricanes.
c) The mode is of 7.5 hurricanes.
For each hurricane, there are only two possible outcomes. Either it causes damage, or it does not. The probability of the home experiencing damage from a hurricane is independent of any other hurricane, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
It gives the probability of exactly x successes on n trials, with p probability.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The mode of the number of trials needed for q successes is given by:
[tex]M = \frac{q}{p}}[/tex]
In this problem, there is a 0.4 probability of experiencing damage from a hurricane, thus [tex]p = 0.4[/tex].
Item a:
This is the probability of at most 1 damage in 3 hurricanes, which is:
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
With [tex]n = 3[/tex].
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{3,0}.(0.4)^{0}.(0.6)^{3} = 0.216[/tex]
[tex]P(X = 1) = C_{3,1}.(0.4)^{1}.(0.6)^{2} = 0.432[/tex]
Then
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.216 + 0.432 = 0.648[/tex]
0.648 = 64.8% probability that there will be at least two hurricanes from which the home will experience no damage before the second hurricane from which the home will experience damage.
Item b:
- 1 in the first three, with 0.432 probability.
- Damage in the fourth, with 0.4 probability.
Then
[tex]0.432(0.4) = 0.1728[/tex]
0.1728 = 17.28% probability that the home will experience its second damage from the fourth hurricane, given that the home has not experienced two damages from the first two hurricanes.
Item c:
- Damage from 3 hurricanes, thus q = 3.
Then:
[tex]M = \frac{3}{0.4}} = 7.5[/tex]
The mode is of 7.5 hurricanes.
A similar problem is given at https://brainly.com/question/24923631
