Let’s take a look at this scenario: \(84\%\) of graduates from the University of Illinois either found a job or began graduate school within six months of graduating. You select \(15\) random graduates. What is the probability that exactly \(12\) of them will have either found a job or began graduate school?
There are several pieces to consider here since we need to include the number of ways that we could select \(12\) students out of the \(15\) and we need to consider the probability for each student falling into that \(84\%\) category. The process for this will feel very similar to what you used for expanding binomials with the Binomial Theorem.
The number of ways to select \(12\) students out of \(15\) will involve combinations and then we’ll multiply that number by the probability of \(12\) students having found a job/started grad school and \(3\) students not having done that.
\(_{15}C_{12}\left(.84\right)^{12}\left(.16\right)^3=.2300\)
This is called Binomial Probability and we can use is when the following conditions are met:
1. There are \(n\) independent trials.
2. Each trial only has two outcomes…we usually refer to one of the outcomes as a success and one as a failure.
3. The probability of an outcome remains the same for each trial.
If those are met, then the probability of \(k\) success in \(n\) trials where the probability of a successful outcome is \(p\) is: \(_nC_k\left(p\right)^k\left(1-p\right)^{n-k}\).
Example 1: \(47\%\) of students at a local high school will attend college out of state.
a) What is the probability that out of \(8\) randomly selected students, exactly \(5\) of them will attend college out of state?
b) What is the probability that out of \(8\) randomly selected students, no more than \(3\) of them will attend college out of state?
There are several pieces to consider here since we need to include the number of ways that we could select \(12\) students out of the \(15\) and we need to consider the probability for each student falling into that \(84\%\) category. The process for this will feel very similar to what you used for expanding binomials with the Binomial Theorem.
The number of ways to select \(12\) students out of \(15\) will involve combinations and then we’ll multiply that number by the probability of \(12\) students having found a job/started grad school and \(3\) students not having done that.
\(_{15}C_{12}\left(.84\right)^{12}\left(.16\right)^3=.2300\)
This is called Binomial Probability and we can use is when the following conditions are met:
1. There are \(n\) independent trials.
2. Each trial only has two outcomes…we usually refer to one of the outcomes as a success and one as a failure.
3. The probability of an outcome remains the same for each trial.
If those are met, then the probability of \(k\) success in \(n\) trials where the probability of a successful outcome is \(p\) is: \(_nC_k\left(p\right)^k\left(1-p\right)^{n-k}\).
Example 1: \(47\%\) of students at a local high school will attend college out of state.
a) What is the probability that out of \(8\) randomly selected students, exactly \(5\) of them will attend college out of state?
b) What is the probability that out of \(8\) randomly selected students, no more than \(3\) of them will attend college out of state?
We can use a histogram to display the probabilities of all of the possible outcomes that can occur in a given situation.
Example 2: A survey found that about \(65\%\) of households donate to at least one charity throughout the year. You ask \(5\) randomly chosen households if they have donated to at least one charity in the last year.
a) Create a binomial probability distribution and graph the results.
b) What is the probability that at least one household will have donated in the last year?
c) What is the probability that no more than three households will have donated in the last year?