Multiplication Rule & Independent Events (College Board AP® Statistics): Study Guide

Look out for scenarios that involve sampling without replacement. These could involve conditional probabilities.

Chad has 25 comic books, 12 of which involve the superhero Dr Data. Chad chooses two different comic books at random to take on vacation with him.

Calculate the probability that both of the chosen comic books involve Dr Data.

Answer:

Calculate the probability that the first chosen comic book involves Dr Data

There are 25 choices and 12 of them involve Dr Data

$P\left(\mathrm{first}\mathrm{is}DrData\right)=\frac{12}{25}$

The first comic book is not replaced once chosen, therefore the probability for the second comic book is a conditional probability

Calculate the probability that the second comic book involves Dr Data given that the first one does

If the first chosen comic involves Dr Data then there are 24 comics remaining and 11 of them involve Dr Data

$P\left(\mathrm{second}\mathrm{is}DrData|\mathrm{first}\mathrm{is}DrData\right)=\frac{11}{24}$

Find the probability that both comics involve Dr Data by using the multiplication rule

$P\left(\mathrm{both}\mathrm{are}DrData\right)=P\left(\mathrm{first}\mathrm{is}DrData\right)·P\left(\mathrm{second}\mathrm{is}DrData|\mathrm{first}\mathrm{is}DrData\right)$

$\begin{array}{rcl}P\left(\mathrm{both}\mathrm{are}DrData\right)& =& \frac{12}{25}·\frac{11}{24}\\ & =& \frac{11}{50}\end{array}$

The probability that both of the chosen comic books involve Dr Data is $\frac{11}{50}$ (or 0.22)

Do not assume two events are independent unless you are told in the question!

A college lecturer surveys a large group of first-year students about their accommodation. Students are asked whether they live on campus or live elsewhere. They are also asked whether they have a private or shared bathroom. The relative frequency of each category is shown in the table.

For the students surveyed, are the events of living on campus and having a private bathroom independent? Justify your answer.

Answer:

Identify the relevant probabilities

$\begin{array}{rcl}P\left(\mathrm{campus}\right)& =& 0.648\\ P\left(\mathrm{private}\right)& =& 0.330\\ P\left(\mathrm{campus}\mathrm{and}\mathrm{private}\right)& =& 0.232\end{array}$

There are three methods to check whether two events are independent

Method 1: Checking $P\left(A|B\right)=P\left(A\right)$

$P\left(\mathrm{campus}|\mathrm{private}\right)=P\left(\mathrm{campus}\right)$

Calculate the probability that a student lives on campus given that they have a private bathroom

Use the formula $P\left(\mathrm{campus}|\mathrm{private}\right)=\frac{P\left(\mathrm{campus}\mathrm{and}\mathrm{private}\right)}{P\left(\mathrm{private}\right)}$

$\begin{array}{rcl}P\left(\mathrm{campus}|\mathrm{private}\right)& =& \frac{0.232}{0.330}\\ & =& 0.703...\end{array}$

Compare this to the probability that a student lives on campus

$0.703...\ne 0.648$

The events of living on campus and having a private bathroom are not independent because $P\left(\mathrm{campus}|\mathrm{private}\right)\ne P\left(\mathrm{campus}\right)$

Method 2: Checking $P\left(B|A\right)=P\left(B\right)$

$P\left(\mathrm{private}|\mathrm{campus}\right)=P\left(\mathrm{private}\right)$

Calculate the probability that a student has a private bathroom given that they live on campus

Use the formula $P\left(\mathrm{private}|\mathrm{campus}\right)=\frac{P\left(\mathrm{private}\mathrm{and}\mathrm{campus}\right)}{P\left(\mathrm{campus}\right)}$

$\begin{array}{rcl}P\left(\mathrm{private}|\mathrm{campus}\right)& =& \frac{0.232}{0.648}\\ & =& 0.358...\end{array}$

Compare this to the probability that a student has a private bathroom

$0.358...\ne 0.330$

The events of living on campus and having a private bathroom are not independent because $P\left(\mathrm{private}|\mathrm{campus}\right)\ne P\left(\mathrm{private}\right)$

Method 3: Checking $P\left(A\cap B\right)=P\left(A\right)·P\left(B\right)$

$P\left(\mathrm{campus}\mathrm{and}\mathrm{private}\right)=P\left(\mathrm{campus}\right)·P\left(\mathrm{private}\right)$

Multiply together the probability that a student lives on campus and the probability that a student has a private bathroom

$\begin{array}{rcl}P\left(\mathrm{campus}\right)·P\left(\mathrm{private}\right)& =& 0.648·0.330\\ & =& 0.21384\end{array}$

Compare this to the probability that a student lives on campus and has a private bathroom

$0.21384\ne 0.232$

The events of living on campus and having a private bathroom are not independent because $P\left(\mathrm{campus}\mathrm{and}\mathrm{private}\right)\ne P\left(\mathrm{campus}\right)·P\left(\mathrm{private}\right)$