Answer:
a) 19.85% probability that a total of two people are struck by lightning during first four months of the year.
b) 22.68% probability that the year has 5 good and 7 bad months
Step-by-step explanation:
We are going to use the Poisson distribution and the binomial distribition to solve this question.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given time interval.
Binomial distribution:
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
a.Find the probability that a total of two people are struck by lightning during first four months of the year.
10 people during a year(12 months).
In 4 months, the mean is ![\mu = \frac{10*4}{12} = 3.33](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B10%2A4%7D%7B12%7D%20%3D%203.33)
This is P(X = 2).
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 2) = \frac{e^{-3.33}*(3.33)^{2}}{(2)!} = 0.1985](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-3.33%7D%2A%283.33%29%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.1985)
19.85% probability that a total of two people are struck by lightning during first four months of the year.
b.Say that a month is good is no one is struck by lightning, and bad otherwise. Find the probability that the year has 5 good and 7 bad months.
Probability that a month is good.
P(X = 0), Poisson
The mean is ![\mu = \frac{10*1}{12} = 0.8333](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B10%2A1%7D%7B12%7D%20%3D%200.8333)
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-0.8333}*(0.8333)^{0}}{(0)!} = 0.4346](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-0.8333%7D%2A%280.8333%29%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.4346)
Find the probability that the year has 5 good and 7 bad months.
Now we use the binomial distribution, we want P(X = 5) when n = 12, p = 0.4346. So
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 5) = C_{12,5}.(0.4346)^{5}.(0.5654)^{7} = 0.2268](https://tex.z-dn.net/?f=P%28X%20%3D%205%29%20%3D%20C_%7B12%2C5%7D.%280.4346%29%5E%7B5%7D.%280.5654%29%5E%7B7%7D%20%3D%200.2268)
22.68% probability that the year has 5 good and 7 bad months