Question

Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places.

P(X>1),  n=4,  p=0.6.

Answer 1

`P(X>1)= 1-P(X \leq 1)= 1- [P(X=0) +P(X=1)]`

And if we use the probability mass function we got:

`P(X=0)=(4C0)(0.6)^0 (1-0.6)^(4-0)=0.0256`

`P(X=1)=(4C1)(0.6)^1 (1-0.6)^(4-1)=0.1536`

And replacing we got:

`P(X>1) =1- [0.0256 +0.1536]= 0.8208`

Explanation:

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

`X \sim Binom(n=4, p=0.6)`

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!)`

We want to find the following probability:

`P(X >1)`

And for this case we can use the complement rule and we got:

`P(X>1)= 1-P(X \leq 1)= 1- [P(X=0) +P(X=1)]`

And if we use the probability mass function we got:

`P(X=0)=(4C0)(0.6)^0 (1-0.6)^(4-0)=0.0256`

`P(X=1)=(4C1)(0.6)^1 (1-0.6)^(4-1)=0.1536`

And replacing we got:

`P(X>1) =1- [0.0256 +0.1536]= 0.8208`