1.1: Independent Events
Learning Objectives
- Know the definition and the notion for independent events.
- Use the rules for addition, multiplication, and complementation to solve for probabilities of particular events in finite sample spaces.
What’s in a Word?
The words dependent and independent are used by students and teachers on a daily basis. In fact, they are probably used quite frequently. You may tell your parent or guardian that you are independent enough to go to the movies on your own with your friends. You could say that when you bake a cake or make a cup of hot chocolate, the taste of these are dependent on what ingredients you use. In the English language, the term dependent means to be unable to do without, whereas independent means to be free from any outside influence.
What about in mathematics? What do the terms dependent and independent actually mean? This lesson will explore the mathematics of independence and dependence.
What are Venn Diagrams and Why are They Used?
In probability, a Venn diagram is a graphic organizer that shows a visual representation for all possible outcomes of an experiment and the events of the experiment in ovals. Normally, in probability, the Venn diagram will be a box with overlapping ovals inside. Look at the diagram below:
The \begin{align*}S\end{align*} represents all of the possible outcomes of an experiment. It is called the sample space. The ovals \begin{align*}A\end{align*} and \begin{align*}B\end{align*} represent the outcomes of the events that occur in the sample space. Let’s look at an example.
Let’s say our sample space is the numbers from 1 to 10. Event \begin{align*}A\end{align*} is randomly choosing one of the odd numbers from 1 to 10, and event \begin{align*}B\end{align*} is randomly choosing one of the prime numbers from 1 to 10. Remember that a prime number is a number whose only factors are 1 and itself. Now let’s draw the Venn diagram to represent this example.
We know that:
\begin{align*}S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\!\\ A = \{1, 3, 5, 7, 9\}\!\\ B = \{2, 3, 5, 7\}\end{align*}
Notice that 3 of the prime numbers are part of both sets and are, therefore, in the overlapping part of the Venn diagram. The numbers 4, 6, 8, and 10 are the numbers not part of \begin{align*}A\end{align*} or \begin{align*}B\end{align*}, but they are still members of the sample space. Now you try.
Example 1
2 coins are tossed one after the other. Event \begin{align*}A\end{align*} consists of the outcomes when tossing heads on the first toss. Event \begin{align*}B\end{align*} consists of the outcomes when tossing heads on the second toss. Draw a Venn diagram to represent this example.
Solution:
We know that:
\begin{align*}S = \{HH,HT, TH, TT\}\!\\ A = \{HH,HT\}\!\\ B = \{HH,TH\}\end{align*}
Notice that event \begin{align*}A\end{align*} and event \begin{align*}B\end{align*} share the Heads + Heads outcome and that the sample space contains Tails + Tails, which is neither in event \begin{align*}A\end{align*} nor event \begin{align*}B\end{align*}.
Example 2
Event \begin{align*}A\end{align*} represents randomly choosing a student from \begin{align*}ABC\end{align*} High School who holds a part-time job. Event \begin{align*}B\end{align*} represents randomly choosing a student from \begin{align*}ABC\end{align*} High School who is on the honor roll. Draw a Venn diagram to represent this example.
Solution:
We know that:
\begin{align*}S =\end{align*} {students in \begin{align*}ABC\end{align*} High School}
\begin{align*}A =\end{align*} {students holding a part-time job}
\begin{align*}B =\end{align*} {students on the honor roll}
Notice that the overlapping oval for \begin{align*}A\end{align*} and \begin{align*}B\end{align*} represents the students who have a part-time job and are on the honor roll. The sample space, \begin{align*}S\end{align*}, outside the ovals represents students neither holding a part-time job nor on the honor roll.
In a Venn diagram, when events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} occur, the symbol used is \begin{align*}\cap\end{align*}. Therefore, \begin{align*}A \cap B\end{align*} is the intersection of events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} and can be used to find the probability of both events occurring. If, in a Venn diagram, either \begin{align*}A\end{align*} or \begin{align*}B\end{align*} occurs, the symbol is \begin{align*}\cup\end{align*}. This symbol would represent the union of events \begin{align*}A\end{align*} and \begin{align*}B\end{align*}, where the outcome would be in either \begin{align*}A\end{align*} or \begin{align*}B\end{align*}.
Example 3
You are asked to roll a die. Event \begin{align*}A\end{align*} is the event of rolling a 1, 2, or a 3. Event \begin{align*}B\end{align*} is the event of rolling a 3, 4, or a 5. Draw a Venn diagram to represent this example. What is \begin{align*}A \cap B\end{align*}? What is \begin{align*}A \cup B\end{align*}?
Solution:
We know that:
\begin{align*}S = \{1, 2, 3, 4, 5, 6\}\!\\ A = \{1, 2, 3\}\!\\ B = \{3, 4, 5\}\end{align*}
\begin{align*}A \cap B = \{3\}\!\\ A \cup B = \{1, 2, 3, 4, 5\}\end{align*}
Independent Events
In mathematics, the term independent means to have one event not dependent on the other. It is similar to the English definition. Suppose you are trying to convince your parent/guardian to let you go to the movies on your own. Your parent/guardian is thinking that if you go, you will not have time to finish your homework. For this reason, you have to convince him/her that you are independent enough to go to the movies and finish your homework. Therefore, you are trying to convince your parent/guardian that the 2 events, going to the movies and finishing your homework, are independent. This is similar to the mathematical definition. Say you were asked to pick a particular card from a deck of cards and roll a 6 on a die. It does not matter if you choose the card first and roll a 6 second, or vice versa. The probability of rolling the 6 would remain the same, as would the probability of choosing the card.
Going back to our Venn diagrams, independent events are represented as those events that occur in both sets. If we look just at Example 2, event \begin{align*}A\end{align*} is randomly choosing a student holding a part-time job, and event \begin{align*}B\end{align*} is randomly choosing a student on the honor roll. These 2 events are independent of each other. In other words, whether you hold a part-time job is not dependent on your being on the honor roll, or vice versa. The outcome of one event is not dependent on the outcome of the second event. To calculate the probability, you would look at the overlapping part of the diagram. The region representing \begin{align*}A\end{align*} and \begin{align*}B\end{align*} is the probability of both events occurring. Let’s look at a specific example.
In \begin{align*}ABC\end{align*} High School, 30 percent of the students have a part-time job, and 25 percent of the students from the high school are on the honor roll. Event \begin{align*}A\end{align*} represents randomly choosing a student holding a part-time job. Event \begin{align*}B\end{align*} represents randomly choosing a student on the honor roll. What is the probability of both events occurring?
Event \begin{align*}A\end{align*} is randomly choosing a student holding a part-time job, and event \begin{align*}B\end{align*} is randomly choosing a student on the honor roll. These 2 events are independent of each other. In other words, whether you hold a part-time job is not dependent on your being on the honor roll, or vice versa. The outcome of one event is not dependent on the outcome of the second event. To calculate the probability, you would look at the overlapping part of the Venn diagram. The region representing \begin{align*}A\end{align*} and \begin{align*}B\end{align*} is the probability of both events occurring. Let’s look at the probability calculation, which is done with the Multiplication Rule:
\begin{align*}P(A) &= 30\% \ \text{or} \ 0.30\\ P(B) &= 25\% \ \text{or} \ 0.25\\ P(A \ \text{and} \ B) &= P(A) \times P(B)\\ P(A \ \text{and} \ B) &= 0.30 \times 0.25\\ P(A \ \text{and} \ B) &= 0.075\end{align*}
In other words, 7.5% of the students of \begin{align*}ABC\end{align*} high school are both on the honor roll and have a part-time job.
In Example 1, 2 coins are tossed one after the other. Remember that event \begin{align*}A\end{align*} consists of the outcomes when getting heads on the first toss, and event \begin{align*}B\end{align*} consists of the outcomes when getting heads on the second toss. What would be the probability of tossing the coins and getting a head on both the first coin and the second coin? We know that the probability of getting a head on a coin toss is \begin{align*}\frac{1}{2}\end{align*}, or 50%. In other words, we have a 50% chance of getting a head on a toss of a fair coin and a 50% chance of getting a tail.
\begin{align*}P(A) &= 50\% \ \text{or} \ 0.50\\ P(B) &= 50\% \ \text{or} \ 0.50\\ \\ P(A \ \text{and} \ B) &= P(A) \times P(B)\\ P(A \ \text{and} \ B) &= 0.50 \times 0.50\\ P(A \ \text{and} \ B) &= 0.25\end{align*}
Therefore, there is a 25% chance of getting 2 heads when tossing 2 fair coins.
Example 4
2 cards are chosen from a deck of cards. The first card is replaced before choosing the second card. What is the probability that they both will be sevens?
Solution:
Let \begin{align*}A = 1^{\text{st}}\end{align*} seven chosen.
Let \begin{align*}B = 2^{\text{nd}}\end{align*} seven chosen.
A little note about a deck of cards
A deck of cards consists of 52 cards.
Each deck has 4 parts (suits) with 13 cards in them.
Each suit has 3 face cards.
\begin{align*}& 4 \ \text{suits} \qquad 1 \ \text{seven} \ \text{per suit}\\ & \ \searrow \qquad \swarrow\\ \text{The total number of sevens in the deck} &= 4 \times 1=4.\end{align*}
Since the card was replaced, these events are independent:
\begin{align*}P(A) &= \frac{4}{52}\\ \\ & \qquad \qquad \ \ \text{Note: The total number of cards is}\\ P(B) &= \frac{4}{52} \ \swarrow \ \text{52 after choosing the first card,}\\ & \qquad \qquad \ \ \text{because the first card is replaced.}\\ \\ P(A \ \text{and} \ B) &= \frac{4}{52} \times \frac{4}{52} \ \text{or} \ P(A \cap B)=\frac{4}{52} \times \frac{4}{52}\\ \\ P(A \cap B) &= \frac{16}{2704}\\ \\ P(A \cap B) &= \frac{1}{169}\end{align*}
Example 5
The following table represents data collected from a grade 12 class in DEF High School.
Gender | University | Community College | Total |
---|---|---|---|
Males | 28 | 56 | 84 |
Females | 43 | 37 | 80 |
Total | 71 | 93 | 164 |
Suppose 1 student was chosen at random from the grade 12 class.
(a) What is the probability that the student is female?
(b) What is the probability that the student is going to university?
Now suppose 2 people both randomly chose 1 student from the grade 12 class. Assume that it's possible for them to choose the same student.
(c) What is the probability that the first person chooses a student who is female and the second person chooses a student who is going to university?
Solution:
\begin{align*}\text{Probabilities:} \ P(\text{female}) &= \frac{80}{164} \swarrow \fbox{164 \ \text{total students}}\\ P(\text{female}) &= \frac{20}{41}\\ P(\text{going to university}) &= \frac{71}{164}\\ \\ P(\text{female}) \times P(\text{going to university})&= \frac{20}{41} \times \frac{71}{164}\\ &= \frac{1420}{6724}\\ &= \frac{355}{1681}\\ &= 0.211\end{align*}
Therefore, there is a 21.1% probability that the first person chooses a student who is female and the second person chooses a student who is going to university.
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