1.1: Independent Events
Learning Objectives
 Know the definition of the notion of independent events.
 Use the rules for addition, multiplication, and complementation to solve for probabilities of particular events in finite sample spaces.
What is Probability?
The simplest definition of probability is the likelihood of an event. If, for example, you were asked what the probability is that the sun will rise in the east, your likely response would be 100%. We all know that the sun rises in the east and sets in the west. Therefore, the likelihood that the sun will rise in the east is 100% (or all the time). If, however, you were asked the likelihood that you were going to eat carrots for lunch, the probability of this happening is not as easy to answer.
Sometimes probabilities can be calculated or even logically deduced. For example, if you were to flip a coin, you have a \begin{align*}\frac{50}{50}\end{align*} chance of landing on heads so the probability of getting heads is 50%. The likelihood of landing on heads (rather than tails) is 50% or \begin{align*}\frac{1}{2}\end{align*}. This is easily figured out more so than the probability of eating carrots at lunch.
Probability and Weather Forecasting
Meteorologists use probability to determine the weather. In Manhattan on a day in February, the probability of precipitation (P.O.P.) was projected to be 0.30 or 30%. When meteorologists say the P.O.P. is 0.30 or 30%, they are saying that there is a 30% chance that somewhere in your area there will be snow (in cold weather) or rain (in warm weather) or a mixture of both. If you were planning on going to the beach and the P.O.P. was 0.75, would you go? Would you go if the P.O.P. was 0.25?
However, probability isn’t just used for weather forecasting. We use it everywhere. When you roll a die you can calculate the probability of rolling a six (or a three), when you draw a card from a deck of cards, you can calculate the probability of drawing a spade (or a face card), when you play the lottery, when you read market studies they quote probabilities. Yes, probabilities affect us in many ways.
Bias and Probability
A. Eric Hawkins is taking science, math, and English, this semester. There are 30 people in each of his classes. Of these 30 people, 25 passed the science midsemester test, 24 passed the midsemester math test, and 28 passed the midsemester English test. He found out that 4 students passed both math and science tests. Eric found out he passed all three tests.
(a) Draw a VENN DIAGRAM to represent the students who passed and failed each test.
(b) If a student’s chance of passing math is 70%, and passing science is 60%, and passing both is 40%, what is the probability that a student, chosen at random, will pass math or science.
At the end of the lesson, you should be able to answer this question. Let’s begin.
Probability and Odds
The probability of something occurring is not the same as the odds of an event occurring. Look at the two formulas below.
\begin{align*}\text{Probability}\ (success) = \frac{number\ of\ ways\ to\ get\ success}{total\ number\ of\ possible\ outcomes}\\ \\ \text{Odds}\ (success) = \frac{number\ of\ ways\ to\ get\ success}{number\ of\ ways\ to\ not\ get\ success}\end{align*}
What do you see as the difference between the two formulas? Let’s look at an example.
Example 1: Imagine you are rolling a die.
(a) Calculate the probability of rolling a “5.”
(b) Calculate the odds of rolling a “5.”
Solution:
(a) \begin{align*}\text{Probability}\ (success) = \frac{number\ of\ ways\ to\ get\ success}{total\ number\ of\ possible\ outcomes}\end{align*}
(b) \begin{align*}\text{Odds}\ (success) = \frac{number\ of\ ways\ to\ get\ success}{number\ of\ ways\ to\ not\ get\ success}\end{align*}
So now we can calculate the probability and we know the difference between probability and odds. Let’s move one step further. Imagine now you were rolling a die and tossing a coin. What is the probability of rolling a 5 and flipping the coin to get heads?
Solution:
\begin{align*}\text{Probability}\ (success) = \frac{number\ of\ ways\ to\ get\ success}{total\ number\ of\ possible\ outcomes}\end{align*}
Die: \begin{align*}P (5) = \frac{1}{6}\end{align*}
Coin: \begin{align*}P (H) = \frac{1}{2}\end{align*}
Die and Coin:
\begin{align*}P(5\ AND\ H) & = \frac{1}{6} \times \frac{1}{2} \\ P(5\ AND\ H) & = \frac{1}{12}\end{align*}
The previous question is an example of an INDEPENDENT EVENT. When two events occur in such a way that the probability of one is independent of the probability of the other, the two are said to be independent. Can you think of some examples of independent events? Roll two dice. If one die roll was a six (6), does this mean the other die rolled cannot be a six? Of course not! The two dies are independent. Rolling one die is independent of the roll of the second die. The same is true if you choose a red candy from a candy dish and flip a coin to get heads. The probability of these two events occurring is also independent.
We often represent an independent event in a VENN DIAGRAM. Look at the diagrams below. \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are two events in a sample space.
For independent events, the VENN DIAGRAM will show that all the events belong to sets \begin{align*}A\end{align*} AND \begin{align*}B\end{align*}.
Example 2: Two cards are chosen from a deck of cards. What is the probability that they both will be face cards?
Solution:
Let \begin{align*}A = 1^{st}\end{align*} Face card chosen
Let \begin{align*}B = 2^{nd}\end{align*} Face card chosen
A little note about a deck of cards
A deck of cards \begin{align*}= 52\end{align*} cards
Each deck has four parts (suits) with 13 cards in them.
\begin{align*}& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad 4\ \text{suits} \qquad \quad 3\ \ \text{face cards per suit} \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \searrow \quad \ \swarrow \\ & \text{Therefore, the total number of face cards in the deck} = 4 \times 3 = 12\end{align*}
\begin{align*}P(A) & = \frac{12}{52} \\ P(B) & = \frac{11}{51} \\ P(A\ AND\ B) & = \frac{12}{52} \times \frac{11}{51}\ \text{or}\ P(A \cap B) = \frac{12}{52} \times \frac{11}{51} = \frac{33}{663} \\ P(A \cap B) & = \frac{11}{221}\end{align*}
Example 3: You have different pairs of gloves of the following colors: blue, brown, red, white and black. Each pair is folded together in matching pairs and put away in your closet. You reach into the closet and choose a pair of gloves. The first pair you pull out is blue. You replace this pair and choose another pair. What is the probability that you will choose the blue pair of gloves twice?
Solution:
\begin{align*}P(\text{blue and blue}) & = P(\text{blue} \cap \text{blue}) = P(\text{blue}) \times P(\text{blue}) \\ & = \frac{1}{5} \times \frac{1}{5} \\ & = \frac{1}{25}\end{align*}
What if you were to choose a blue pair of gloves or a red pair of gloves? How would this change the probability? The word OR changes our view of probability. We have, up until now worked with the word AND. Going back to our VENN DIAGRAM, we can see that the sample space increases for \begin{align*}A\end{align*} or \begin{align*}B\end{align*}.
Example 4: You have different pairs of gloves of the following colors: blue, brown, red, white and black. Each pair is folded together in matching pairs and put away in your closet. You reach into the closet and choose a pair of gloves. What is the probability that you will choose the blue pair of gloves or a red pair of gloves?
Solution:
\begin{align*}P(\text{blue or red}) & = P(\text{blue} \cup \text{red}) = P(\text{blue}) + P(\text{red}) \\ & = \frac{1}{5} + \frac{1}{5} \\ & = \frac{2}{5}\end{align*}
We have one more set of terms to look at before we finish of our first look at independent and events in probability. These terms are MUTUALLY INCLUSIVE and MUTUALLY EXCLUSIVE. Mutually exclusive events cannot occur in a single event or at the same time. For example, a number cannot be both even and odd or you cannot have picked a single card from a deck of cards that is both a ten and a jack. Mutually inclusive events can occur at the same time. For example a number can be both less than 5 and even or you can pick a card from a deck of cards that can be a club and a ten. The addition principle accounts for this “double counting.”
Addition Principle
\begin{align*}P(A \cup B) & = P(A) + P(B)  P(A \cap B) \\ P(A \cap B) & = 0\ \text{for mutually exclusive events}\end{align*}
Example 5: Two cards are drawn from a deck of cards. The first card is replaced before choosing the second card.
A: \begin{align*}1^{st}\end{align*} card is a club
B: \begin{align*}1^{st}\end{align*} card is a 7
C: \begin{align*}2^{nd}\end{align*} card is a heart
Find the following probabilities:
(a) \begin{align*}P(A\ \text{or}\ B)\end{align*}
(b) \begin{align*}P (B\ \text{or}\ A)\end{align*}
(c) \begin{align*}P (A\ \text{and}\ C)\end{align*}
Solution:
(a) \begin{align*}P(A\ or\ B) & = \frac{13}{52} + \frac{4}{52}  \frac{1}{52} \\ P(A\ or\ B) & = \frac{16}{52} \\ P(A\ or\ B) & = \frac{4}{13}\end{align*} (b) \begin{align*}P(B\ or\ A) & = \frac{4}{52} + \frac{13}{52}  \frac{1}{52} \\ P(B\ or\ A) & = \frac{16}{52} \\ P(B\ or\ A) & = \frac{4}{13}\end{align*}(c) \begin{align*}P(A\ and\ C) & = \frac{13}{52} \times \frac{13}{52} \\ P(A\ and\ C) & = \frac{169}{2704} \\ P(A\ and\ C) & = \frac{1}{16}\end{align*}
Let’s go back to our original problem now and see if we can solve it.
Bias and Probability
B. Eric Hawkins is taking science, math, and English, this semester. There are 30 people in each of his classes. 25 passed the science midsemester test, 24 passed the mid semester math test, and 28 passed the midsemester English test. He found out that 4 students passed both math and science tests. Eric found out he passed all three tests.
(c) Draw a VENN DIAGRAM to represent the students who passed and failed each test.
(d) If a student’s chance of passing math is 70%, and passing science is 60%, and passing both is 40%, what is the probability that a student, chosen at random, will pass math or science.
(a)
(b) Let \begin{align*}M =\end{align*} Math test
Let \begin{align*}S =\end{align*} Science test
\begin{align*}P(M\ or\ S) & = 0.70 + 0.60  0.40 \\ P(M\ or\ S) & = 0.90 \\ P(M\ or\ S) & = 90\%\end{align*}
Lesson Summary
Probability and odds are two important terms that must be identified and kept clear in our minds. The fact remains that probability affects almost every part of our lives. In order to determine probability mathematically, we need to consider other definitions such as the difference between independent and dependent events, as well as the difference between a mutually exclusive event and a mutually inclusive event. The calculations involved in probability are dependent on the distinction between these (no pun intended!). For mutually inclusive events, it is important to remember the addition rule so that we do not double count in our calculations.
Points to Consider
 Why is the term probability more useful than the term odds?
 Are VENN DIAGRAMS a useful tool for visualizing probability events?
Vocabulary
 Dependent Events
 Two or more events whose outcomes affect each other. The probability of occurrence of one event depends on the occurrence of the other.
 Independent Events
 Two or more events whose outcomes do not affect each other.
 Mutually Exclusive Events
 Two outcomes or events are mutually exclusive when they cannot both occur simultaneously.
 Mutually Inclusive Events
 Two outcomes or events are mutually exclusive when they can both occur simultaneously.
 Outcome
 A possible result of one trial of a probability experiment.
 Probability
 The chance that something will happen.
 Random Sample
 A sample in which everyone in a population has an equal chance of being selected; not only is each person or thing equally likely, but all groups of persons or things are also equally likely.
 Venn Diagram
 A diagram of overlapping circles that shows the relationships among members of different sets.
Review Questions
Jack is looking for a new car to drive. He goes to the lot and finds a number to choose from. There are three conditions he is looking for: price, gas mileage, and safety record. He decides to draw a VENN DIAGRAM to organize all of the vehicles he has found to help him determine what car to pick. Look at the following VENN DIAGRAM to answer each of the questions 1 through 9.
 What is the sample space for Price and Gas Mileage?
 What is the sample space for Price and Safety Record?
 What is the sample space for Gas Mileage and Safety Record?
 What is the sample space for Price or Gas Mileage?
 What is the sample space for Price or Safety Record?
 What is the sample space for Gas Mileage or Safety Record?
 What is the sample space for Price and Gas Mileage and Safety Record?
 What is the sample space for Price or Gas Mileage or Safety Record?
 Did Jack find the car he was looking for? How can you tell?
 If a die is tossed twice, what is the probability of rolling a 4 followed by a 5?
 A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a jack and an eight?
 Two cards are drawn from a deck of cards. Determine the probability of each of the following events:
 P(heart) or P(club)
 P(heart) and P(club)
 P(jack) or P(heart)
 P(red) or P(ten)
 A box contains 5 purple and 8 yellow marbles. What is the probability of successfully drawing, in order, a purple marble and then a yellow marble? {Hint: in order means they are not replaced}
 A bag contains 4 yellow, 5 red, and 6 blue marbles. What is the probability of drawing, in order, 2 red, 1 blue, and 2 yellow marbles?
 Fifteen airmen are in the line crew. They must take care of the coffee mess and line shack cleanup. They put slips numbered 1 through 15 in a hat and decide that anyone who draws a number divisible by 5 will be assigned the coffee mess and anyone who draws a number divisible by 4 will be assigned cleanup. The first person draws a 4, the second a 3, and the third and 11. What is the probability that the fourth person to draw will be assigned:
 the coffee mess?
 the cleanup?
Review Answers
 5
 6
 4
 31
 35
 32
 1
 49
 Yes he did find his car because the answer to question 8 is “1” meaning he found only one car with all three of his conditions.
 \begin{align*}\frac{1}{36}\end{align*}

\begin{align*}\frac{1}{169}\end{align*}
 \begin{align*}\frac{1}{2}\end{align*}
 \begin{align*}\frac{1}{16}\end{align*}
 \begin{align*}\frac{4}{13}\end{align*}
 \begin{align*}\frac{7}{13}\end{align*}
 \begin{align*}\frac{10}{39}\end{align*}

\begin{align*}\frac{4}{1001}\end{align*}
 \begin{align*}\frac{1}{4}\end{align*}
 \begin{align*}\frac{1}{6}\end{align*}
Answer Key for Review Questions (Even Numbers)
Jack is looking for a new car to drive. He goes to the lot and finds a number to choose from. There are three conditions he is looking for: price, gas mileage, and safety record. He decides to draw a VENN DIAGRAM to organize all of the vehicles he has found to help him determine what car to pick. Look at the following VENN DIAGRAM to answer each of the questions 1 through 9.
2. 6
4. 31
6. 32
8. 49
10. \begin{align*}\frac{1}{36}\end{align*}
12. (a) \begin{align*}\frac{1}{2}\end{align*}
(b) \begin{align*}\frac{1}{16}\end{align*}
(c) \begin{align*}\frac{4}{13}\end{align*}
(d) \begin{align*}\frac{7}{13}\end{align*}
14. \begin{align*}\frac{4}{1001}\end{align*}
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