Question

with the salary cap in the NFL, it is said that on "any given Sunday" any team could beat any other team. If we assume every week a team has a 50% chance of winning, what is the probability that a team will have at least 1 win

Answer 1

Answer:

$P(X \geq 1) = 1-P(X<1) = 1-P(X=0)$

And we can find the individual probability like this:

$P(X=0) = 16C0 (0.5)^0 (1-0.5)^{16-0} = 0.0000153$

And replacing we got:

$P(X \geq 1) = 1-P(X<1) = 1-P(X=0)=1- 0.0000153=0.99998$

Step-by-step explanation:

Assuming the following question: With the salary cap in the NFL, it is said that on any given Sunday any team could beat any other team. If we assume every week of the 16 week season a team has a 50% chance of winning, what is the probability that a team will have at least 1 win?

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:  

$X \sim Binom(n=16, p=0.5)$  

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:  

$nCx=\frac{n!}{(n-x)! x!}$  

And we want this probability:

$P(X \geq 1)$

And using the complement rule we got:

$P(X \geq 1) = 1-P(X<1) = 1-P(X=0)$

And we can find the individual probability like this:

$P(X=0) = 16C0 (0.5)^0 (1-0.5)^{16-0} = 0.0000153$

And replacing we got:

$P(X \geq 1) = 1-P(X<1) = 1-P(X=0)=1- 0.0000153=0.99998$