Example: Given a fair coin, the probability of a flip turning up heads is \(0.5\) (or 50%). Similarly, given a fair six-sided die, the probability of a roll resulting in a number greater than four is \(1/3\) (or about 33.3%) because there are two outcomes satifying the criteria (rolling a 5 or rolling a 6) out of the six total possible outcomes.

Try It! Now it is your turn. Try the next few problems. Be sure to note any questions you have as you work through them.

Good work on that last set of questions. In those problems you could find the probability by counting the number of ways the desired outcome could occur and then dividing that number by the total number of outcomes possible. In the last question, there were simply more ways to roll a five (four ways to do it) than to roll a two (just one way). What if we try doing something a bit more complicated? Say we wanted to know the probability of rolling at least a two on a single roll of a die and then flipping a “tails” on a single flip of a coin?

Probability and Independent Events: If \(A\) and \(B\) are independent events (that is, the probability that \(B\) occurs does not depend on whether or not \(A\) occured and vice-versa), then the probability of \(A\) and \(B\) occurring is the product of the probability of \(A\) occurring and the probability of \(B\) occurring. Mathematically, we write: \(\mathbb{P}\left[A~\text{and}~B\right] = \mathbb{P}\left[A\right]\cdot\mathbb{P}\left[B\right]\).

Good work so far. Let’s say you forgot to study for your chemistry quiz today. It is a four question multiple choice quiz with options \(a)\) - \(e)\). You decide that your best option is to guess randomly on each of the questions. Answer the following, using the empty code block below to carry out any necessary computations.

Binomial Distribution: Let \(X\) be the number of successes resulting from a binomial experiment with \(n\) trials. We can compute the following probabilities:

• The probability of exactly \(k\) successes is given by \(\displaystyle{\mathbb{P}\left[X = k\right] = \binom{n}{k}\cdot p^k\left(1 - p\right)^{n-k} \approx \tt{dbinom(k, n, p)}}\)
• The probability of at most \(k\) successes is given by \(\displaystyle{\mathbb{P}\left[X \leq k\right] = \sum_{i=0}^{k}{\binom{n}{i}\cdot p^i\left(1 - p\right)^{n-i}} \approx \tt{pbinom(k, n, p)}}\)

In the equations above, \(\binom{n}{k} = \frac{n!}{k!\left(n-k\right)!}\) counts the number of ways to arrange the \(k\) successes amongst the \(n\) trials. That being said, the R functionality, dbinom() and pbinom() allow us to bypass the messy formulas – but you’ll still need to know what these functions do in order to use them correctly!

Tip: We need to use the binomial distribution to find probabilities associated with numbers of successful (or failing) outcomes in which we do not know for certain the trials on which the successes (or failures) occur.

The code block below is set up to find the probability of exactly two flips of a coin landing heads-up out of seven total flips. Edit the code block that that it finds the probability that you got exactly two of the four questions on your chemistry quiz from earlier correct. As a reminder, there were five answer options for each question and you were guessing randomly.

dbinom(2, 7, 0.5)

[1] 0.1640625

Good work. Now you’ll get to try a few more problems! As you work through the next set of questions, you may want to check out this example and solution. Note that in that document, I mention that drawing a simple picture for each problem will help you decide which function(s) you might use and whether you might need to make multiple computations. This is a really important strategy that will help you in developing a strategy to solve each problem.

Practice: For each of the following, consider a scenario in which a random sample of 18 students is asked (in private) whether they’ve failed to hand in at least one assignment this semester. We assume that about 34% of students fail to hand in at least one assignment.

1. Find the probability that exactly 7 of the 18 students have failed to hand in at least one assignment.

Example: Find the probability that at most 15 of the 18 students have failed to hand in at least one assignment. That is,

\(\displaystyle{\mathbb{P}\left[X \leq 15\right] = \sum_{i=0}^{15}{\binom{18}{i}\cdot 0.34^{i}\left(1 - 0.34\right)^{18-i}} \approx \tt{pbinom(15, 18, 0.34)}})\)

pbinom(15, 18, 0.34)

## [1] 0.9999977

2. Find the probability that at most 9 of the 18 students have failed to hand in at least one assignment.

Example: Find the probability that at least 5 of the 18 students have failed to hand in at least one assignment. That is,

\(\displaystyle{\mathbb{P}\left[X \ge 5\right] = 1 - \mathbb{P}\left[X \le 4\right] = \tt{1 - pbinom(4, 18, 0.34)}}\)

1 - pbinom(4, 18, 0.34)

## [1] 0.7865908

3. Find the probability that at least 11 of the 18 students have failed to hand in at least one assignment.

Example: Find the probability that at least 10 and at most 15 students have failed to hand in at least one assignment. That is,

\(\displaystyle{\mathbb{P}\left[10 \le X \le 15\right] = \mathbb{P}\left[X \le 15\right]-\mathbb{P}\left[X \le 9\right] = \tt{pbinom(15, 18, 0.34) - pbinom(9, 18, 0.34)}}\)

pbinom(15, 18, 0.34) - pbinom(9, 18, 0.34)

## [1] 0.04942319

4. Find the probability that between a minimum of 6 and a maximum of 12 out of the 18 students have failed to hand in at least one assignment.

5. The expected number of successes in a binomial experiment is sometimes denoted by \(\mathbb{E}\left[X\right]\) or \(\mu\), and it can be computed as \(\mu = \mathbb{E}\left[X\right] = n\cdot p\), where \(n\) denotes the number of trials run and \(p\) denotes the probability of success on a single trial. Sometimes it is convenient to think of the expected number of successes as “the mean”. Use the code block below to compute the expected number of students who have failed to hand in at least one assignment:
6. The standard deviation in the number of successes for a binomial experiment can also be computed. The quantity \(\displaystyle{\sigma = \sqrt{n\cdot p\left(1 - p\right)}}\), where \(n\) denotes the number of trials run and \(p\) denotes the probability of success on a single trial is the standard deviation in number of successes. Use the code block below to compute the standard deviation in number of students who have failed to hand in at least one assignment from random samples of 18 students:

Be sure to write down what questions you had as you worked through these problems and to have a teacher, colleague, or tutor help clarify things for you.