Binomial probability is used in various scenarios, like quality control and genetics. By understanding binomial probability, we can make informed predictions and decisions based on repeated experiments. Use this resource to learn how to apply the concepts of binomial probability.
Binomial probability ("bi" meaning "two" and "nomos" meaning "parts") helps us find the chance of getting a certain number of successes in a series of tries. It's like flipping a coin and wanting to know how often it lands on heads after several flips.
This helps us predict outcomes when we have repeated experiments or tests, making it useful for planning and decisions.
We can use binomial probability to determine the likelihood of tossing a head compared to a tail. Image by AlLes, via Pixabay
Binomial experiments
In a statistical experiment, a random variable may be described as having a binomial distribution when there are:
a number of repeated trials
only two possible outcomes on each trial: success or failure.
If it produces a binomial distribution or binomial data, it involves binomial probability.
Some examples of such experiments include:
finding the probability of obtaining \(5\) heads in \(6\) tosses of a coin, with six repeated trials, each with two possible outcomes: head or tail
finding the probability that in a randomly selected group of \(30\) people, none of them will have a particular disease, with \(30\) repeated "trials", each with two possible outcomes: have disease or do not have disease
finding the probability that in a sample of \(100\) manufactured components, no more than \(2\) will be defective, with \(100\) repeated trials, each with two possible outcomes: defective or not defective.
Example – finding probability in binomial experiments
In a particular family, the probability that a child will have red hair is \(\dfrac{1}{4}\). If the parents have three children, find the probability that:
all three will have red hair
none will have red hair
at least one child will have red hair
only the first child will have red hair
exactly one child will have red hair.
This is a binomial probability experiment because it involves three "events" (having three children) and two possible outcomes for each event (having red hair and not having red hair). Let's let \(R\) be the outcome of having red hair and \(R'\) be the outcome of not having red hair.
For all three to have red hair, we are looking at \(\Pr(R_{1}\cap R_{2}\cap R_{3})\).
\[\begin{align*} \Pr(R_{1}\cap R_{2}\cap R_{3}) & = \frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}\\
& = \frac{1}{64}
\end{align*}\]
For none with red hair, we are looking at \(\Pr(R'_{1}\cap R'_{2}\cap R'_{3})\).
\[\begin{align*} \Pr(R'_{1}\cap R'_{2}\cap R'_{3}) & = \frac{3}{4}\times\frac{3}{4}\times\frac{3}{4}\\
& = \frac{27}{64}
\end{align*}\]
The outcome where at least one child with red hair complements the event where no children have red hair, so we can use the formula:
\[\begin{align*} 1-\Pr(R'_{1}\cap R'_{2}\cap R'_{3}) & = 1-\frac{27}{64}\\
& = \frac{37}{64}
\end{align*}\]
For only the first child with red hair, we are looking at:
\[\begin{align*} \Pr(R_{1}\cap R'_{2}\cap R'_{3}) & = \frac{1}{4}\times\frac{3}{4}\times\frac{3}{4}\\
& = \frac{9}{24}
\end{align*}\]
There are many ways in which only one child will have red hair. It could be the first child only, the second child only, or the first child only. Because of this, we must sum the probabilities of each of these.
In Example – finding probability on binomial experiments, we used the basic probability rules but there are specific rules we can use when we know we are dealing with binomial probability.
If "\(n\)" is the number of trials and "\(p\)" is the probability of the outcome of interest, then the probability of "\(x\)" outcomes is given by the formula:
where \(^{n}C_{x}\) is the number of ways of selecting \(x\) items from a population of \(n\) items, without regard to order, and \(^{n}C_{x}=\dfrac{n!}{x!(n-x)!}\).
The exclamation mark (\(!\)) means that we are looking at factorials. For \(n!\), we multiply \(n\) by all of the integers counting from \(n\) to \(1\). This means \(3!\) would be \(3\times2\times1=6\) and \(6!\) would be \(6\times5\times4\times3\times2\times1=720\).
These formulas may look a bit scary, but looking at an example will help!
Example – using the binomial probability formula
One in every hundred items a machine produces is defective. What is the probability that, in a sample of five items produced by this machine:
exactly \(3\) are defective?
none are defective?
at least \(1\) is defective?
The binomial probability formula uses \(n\) which is the number of items, \(p\) is the probability of the outcome and \(x\) is the number of times the outcome is achieved.
\(n=5\)
\(p=\dfrac{1}{100}\)
\(x\) depends on the question being asked: a, b or c.
For a, \(x=3\). Let's find \(^{n}C_{x}\) first.
\[\begin{align*} ^{n}C_{x} & = \frac{n!}{x!(n-x)!}\\
^{5}C_{3} & = \frac{5\times4\times3\times2\times1}{3\times2\times1(2\times1)}\\
& = 10
\end{align*}\]
Now, we can substitute this into the probability formula:
\[\begin{align*} \Pr(X=3) & = ^{n}C_{x}\times p^{x}\times(1-p)^{n-x}\\
& = 10\times\left(\frac{1}{100}\right)^{3}\times\left(1-\frac{1}{100}\right)^{5-3}\\
& = 10\times\left(\frac{1}{1000000}\right)\times\left(\frac{99}{100}\right)^{2}\\
& \approx 0.00001
\end{align*}\]
For c, at least \(1\) defective is complementary to all defective, so we can calculate the probability easily.
\[\begin{align*} \Pr(X\geq1) & = 1-\Pr(X=0)\\
& = 1-0.95\\
& = 0.05
\end{align*}\]
Binomial probability distribution
A list of all possible outcomes of an event and their associated probabilities is called a probability distribution.
This is useful for calculating values like:
mean (\(\mu\) or \(E(X)\)), which tells us the expected number of successes or expected result in a given number of trials
\[\mu=E(X)=np\]
standard deviation (\(\sigma\) or \(SD(x)\), which tells us how much the actual number of successes might vary from the average.
\[\sigma=SD(x)=\sqrt{np(1-p)}\]
Example – summarising data from binomial probability experiments
One in every hundred items a machine produces is defective. The probability distribution table shown is for the event where the item is defective \(X\), to five decimal places.
\(\mathbf{x}\) (Number of defective items)
\(0\)
\(1\)
\(2\)
\(3\)
\(4\)
\(5\)
\(\mathbf{\Pr(X=x)}\)
\(0.95099\)
\(0.04803\)
\(0.00097\)
\(0.00001\)
\(0.00000\)
\(0.00000\)
Find the expected number of defectives in a batch of \(1000\) items, including the standard deviation.
\[\begin{align*} \mu & = np\\
& = 1000\times\frac{1}{100}\\
& = 10
\end{align*}\] \[\begin{align*} \sigma & = \sqrt{np(1-p)}\\
& = \sqrt{1000\times\frac{1}{100}\left(1-\frac{99}{100}\right)}\\
& = 3.15
\end{align*}\]
This means that in a batch of \(1000\), we would expect to get \(10\) defectives and the number of defectives will deviate from this amount by an "average" of \(3.15\).
We would expect most batches to have between: \(10-3.15\approx7\) and \(10+3.15\approx13\) defective items.
Exercise – finding binomial probability
The probability that an archer will hit a bullseye is \(0.7\). If he is allowed ten attempts, find the probability that he:
hits it every time
misses each time
scores two bulls-eyes
scores at least two bulls-eyes
How many bulls-eyes would you expect the archer from the previous question to score?
