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
Let’s say our sample space is the numbers from 1 to 10. Event
We know that:
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
Example 1
2 coins are tossed one after the other. Event
Solution:
We know that:
Notice that event
Example 2
Event
Solution:
We know that:
Notice that the overlapping oval for
In a Venn diagram, when events
Example 3
You are asked to roll a die. Event
Solution:
We know that:
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
In
Event \begin{align*}A\end{align*}
\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*}
In Example 1, 2 coins are tossed one after the other. Remember that event \begin{align*}A\end{align*}
\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*}
Let \begin{align*}B = 2^{\text{nd}}\end{align*}
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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