Question

Maggie is practicing her penalty kicks for her upcoming soccer game. During the practice, she attempts 10 penalty kicks. If each attempt at the penalty kick is independent of the other attempts and if she scores 65% of the time, historically, what is the probability that she scores at least eight goals

Answer 1

`P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)`

And we can find the individual probabilities like this:

`P(X=8)=(10C8)(0.65)^0 (1-0.65)^(10-8)=0.176`

`P(X=9)=(10C9)(0.65)^1 (1-0.65)^(10-9)=0.072`

`P(X=10)=(10C10)(0.65)^2 (1-0.65)^(10-10)=0.013`

And adding we got:

`P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)= 0.176+0.072+0.013=0.262`

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".

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

`X \sim Binom(n=10, p=0.65)`

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!)`

Solution to the problem

For this case we want to find this probability:

`P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)`

And we can find the individual probabilities like this:

`P(X=8)=(10C8)(0.65)^0 (1-0.65)^(10-8)=0.176`

`P(X=9)=(10C9)(0.65)^1 (1-0.65)^(10-9)=0.072`

`P(X=10)=(10C10)(0.65)^2 (1-0.65)^(10-10)=0.013`

And adding we got:

`P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)= 0.176+0.072+0.013=0.262`