Question

Suppose that the probability that a person is killed by lightning, in a year is, independently of other people, 1/(500 million). Assume that the U.S. population is 300 million. Write down the exact expression for the probability P(3 or more people will be killed by lightning in the U.S. next year). Evaluate this expression to 6 decimal places. Write down a relevant approximate expression for the probability from (a). Justify briefly the approximation. Evaluate this expression to 6 decimal places.

Answer 1

a)
`P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)]`

The individual pprobabilities are:

`P(X=0)=(300000000C0)((1)/(500000000))^3 (1-(1)/(500000000))^(300000000-0)=0.548812`

`P(X=1)=(300000000C0)((1)/(500000000))^3 (1-(1)/(500000000))^(300000000-0)=0.329287`

`P(X=2)=(300000000C0)((1)/(500000000))^3 (1-(1)/(500000000))^(300000000-0)=0.098786`

And if we replace we got:

`P(X \geq 3) = 1- [0.548812+0.329287+0.098786]=0.023115`

b)
`P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)]`

And we can calculate this with the following excel formula:

`=1-BINOM.DIST(2;300000000;(1/500000000);TRUE)`

And we got:

`P(X\geq 3) = 0.023115`

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

`X \sim Binom(n=300000000, p=(1)/(500000000))`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

(a) Write down the exact expression for the probability P(3 or more people will be killed by lightning in the U.S. next year). Evaluate this expression to 6 decimal places.

For this case we want this probability, we are going to use the complement rule:

`P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)]`

The individual probabilities are:

`P(X=0)=(300000000C0)((1)/(500000000))^3 (1-(1)/(500000000))^(300000000-0)=0.548812`

`P(X=1)=(300000000C0)((1)/(500000000))^3 (1-(1)/(500000000))^(300000000-0)=0.329287`

`P(X=2)=(300000000C0)((1)/(500000000))^3 (1-(1)/(500000000))^(300000000-0)=0.098786`

And if we replace we got:

`P(X \geq 3) = 1- [0.548812+0.329287+0.098786]=0.023115`

(b) Write down a relevant approximate expression for the probability from (a). Justify briefly the approximation. Evaluate this expression to 6 decimal places.

For this case we want this probability, we are going to use the complement rule:

`P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)]`

And we can calculate this with the following excel formula:

`=1-BINOM.DIST(2;300000000;(1/500000000);TRUE)`

And we got:

`P(X\geq 3) = 0.023115`