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

Answer:

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

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:

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

Step-by-step 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:  

$nCx=\frac{n!}{(n-x)! x!}$  

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:

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