Question

In Sweden, 10 people are struck by lightning every year, on average. Answer thefollowing questions about a given year in Sweden using a relevant approximation. (Assumeperson-to-person and day-to-day independence, and assume all months have the same lengthand the same chance of lightning.)a.Find the probability that a total of two people are struck by lightning during first fourmonths 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.

Answer 1

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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ 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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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 $\mu = \frac{10*4}{12} = 3.33$

This is P(X = 2).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-3.33}*(3.33)^{2}}{(2)!} = 0.1985$

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 $\mu = \frac{10*1}{12} = 0.8333$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.8333}*(0.8333)^{0}}{(0)!} = 0.4346$

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

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{12,5}.(0.4346)^{5}.(0.5654)^{7} = 0.2268$

22.68% probability that the year has 5 good and 7 bad months