2.1.3 - Probability Rules

2.1.3 - Probability Rules

The probability rules covered in this lesson can be found in section P.1 of the Lock5 textbook.

Earlier in this lesson you were introduced to proportions. We used the notation: \(Proportion=\frac{Number\;in\;the\;category}{Total\;number}\).

When we discuss probabilities, we will use the notation below where \(P(A)\) is the probability of event \(A\) occurring. Probabilities are typically written in decimal form but may also be translated to percentages. 

Note that this is the same formula that you learned earlier in Lesson 2.1.1 for a proportion.

Probability of Event A
\(P(A)=\dfrac{Number\;in\;group\;A}{Total\;number}\)

Example: Spades

What is the probability that a randomly selected card from a standard 52-card deck will be a spade? There are 13 spades in the deck of 52.

\(P(spade)=\dfrac{13}{52}=0.25\)

The probability of pulling a spade is 0.25. We could also say that there is a 25% chance of pulling a spade.

Example: Odd Numbers

If you roll a six-sided die, what is the probability of getting an odd number? There are three odd numbers on the die (1, 3, 5).

\(P(odd)=\dfrac{3}{6}=0.50\)

The probability of rolling an odd number is 0.50. We could also say that there 50% chance of rolling an odd number.

Example: Raffle

There are a total of 500 raffle tickets and you have purchased 10. What is the probability that one of your tickets will be randomly selected to win the raffle?

\(P(winning)=\dfrac{10}{500}=0.02\)

The probability of you winning is 0.02. We could also say that there is a 2% chance that you will win.


2.1.3.1 - Range of Probabilities

2.1.3.1 - Range of Probabilities

The probability of an impossible event is 0 and the probability of a certain event is 1. The range of possible probabilities is: \(0 \leq P(A) \leq 1\). It is not possible to have a probability less than 0 or greater than 1. 

Example: Rolling an 8

It is impossible to roll an eight on a six-sided die.

\(P(rolling\; 8)= \dfrac{0}{6} = 0\)

Example: Blue Cards

In a standard 52-card deck all cards are black or red. There are no blue cards.

\(P(blue)=\dfrac{0}{52}=0\)

Example: Rolling a Value Between 1 and 6

A six-sided die contains the values 1, 2, 3, 4, 5, and 6. All rolls will result in a value between 1 and 6.

\(P(rolling \;1 \;to\; 6)=\dfrac{6}{6}=1.00\)


2.1.3.2 - Combinations of Events

2.1.3.2 - Combinations of Events

In situations with two or more categorical variables there are a number of different ways that combinations of events can be described: intersections, unions, complements, and conditional probabilities. Each of these combinations of events is covered in your textbook. However, note that your textbook does not use the symbols that are most commonly used when discussing these combinations of events. The symbols that we will be using are in the table below. In this section, you will also learn about disjoint events and independent events

Combination of Events
Combination Symbol Definition
Intersection \(P(A\cap B)\) Probability of A and B
Union \(P(A\cup B)\)

Probability of A or B

Note: This includes the possibility of A and B

Complement \(P(A^C)\) The probability of NOT A
Conditional \(P(A\mid B)\) The probability of A given B

2.1.3.2.1 - Disjoint & Independent Events

2.1.3.2.1 - Disjoint & Independent Events

Note that disjoint events and independent events are different. Events are considered disjoint if they never occur at the same time; these are also known as mutually exclusive events. Events are considered independent if they are unrelated.

Disjoint Events

Two events that do not occur at the same time. These are also known as mutually exclusive events

In the Venn diagram below event A and event B are disjoint events because the two do not overlap.

Mutually Exclusive
Venn diagram
A visual representation in which the sample space is depicted as a box and events are represented as circles within the sample space.
Independent Events
Unrelated events. The outcome of one event does not impact the outcome of the other event.

Example: Freshmen & Sophomores

Let's consider undergraduate class status. A student can be classified as a freshman, sophomore, junior, or senior.

Being a freshman and being a sophomore are disjoint events because an individual cannot be classified as both at the same time. 

Being a freshman is not independent of being a sophomore. If I know that an individual is a freshman then the probability that they are a sophomore is 0; knowing that the student was a freshman provided information that influenced my prediction of them being a sophomore. 

Example: Class Status & Gender

Assume that there is no relationship between gender and class status. This means that within each class (freshmen, sophomores, juniors, seniors) the proportion of students who are men is consistent. It also means that within each gender the proportion of students who are freshmen, sophomores, juniors, and seniors is consistent.

In this case, we could say that the events of being a man and being a senior are independent events. Knowing that a student is a man does not influence the likelihood of him being a senior. Knowing that a student is a senior does not change the likelihood of them being a man.

There are some men who are seniors, so these events are not disjoint. 


2.1.3.2.2 - Intersections

2.1.3.2.2 - Intersections
Intersection

The overlap of two or more events is symbolized by the character \(\cap\). 

\(P(A \cap B)\) is read as "the probability of A and B."

Intersection of A and B

Example: Red King

What is the probability of randomly selecting a card from a standard 52-card deck that is a red card and a king?

There are 2 kings that are red cards: the king of hearts and the king of diamonds.

\(P(red \cap king)=\dfrac{2}{52}=.0385\)

Example: Female Undergraduate Students

The two-way table below displays the World Campus enrollment from Fall 2015 in terms of level (undergraduate and graduate) and biological sex. What proportion of World Campus students were female and undergraduate students?

  Female Male Total
Undergraduate 3814 3428 7242
Graduate 2213 2787 5000
Total 6027 6215 12242

There are 3814 students who are females and undergraduates out of a total of 12242 students.

\(P(F \cap U)=\dfrac{3814}{12242}=0.312\)


2.1.3.2.3 - Unions

2.1.3.2.3 - Unions
Union

A union contains the area in A or B and is symbolized by \(\cup\). Note that this also includes the overlap of A and B (i.e., the intersection).

\(P(A \cup B)\) is read as "the probability of A or B."

Union of A and B
Union
\(P(A\cup B) = P(A)+P(B)-P(A\cap B)\)

Example: Hearts or Spades

What is the probability of randomly selecting a card from a standard 52-card deck that is a heart or spade?

There are 13 cards that are hearts, 13 cards that are spades, and no cards that are both a heart and a spade.

\(P(heart \cup spade)=\dfrac{13}{52}+\dfrac{13}{52}-\dfrac{0}{52}= \dfrac {26}{52}=0.5\)

Example: Hearts or Aces

What is the probability of randomly selecting a card from a standard 52-card deck that is a heart or an ace?

There are 13 cards that are hearts and 4 cards that are aces. There is one ace of hearts, so one of those 4 aces has already been counted.

\(P(heart \cup ace)=\dfrac{13}{52}+\dfrac{4}{52}-\dfrac{1}{52}=\dfrac{16}{52}=0.308\)

Example: Female or Undergraduate

The two-way table below displays the World Campus enrollment from Fall 2015 in terms of level (undergraduate and graduate) and biological sex. What proportion of World Campus students were female or undergraduate students?

  Female Male Total
Undergraduate 3814 3428 7242
Graduate 2213 2787 5000
Total 6027 6215 12242

When we have a contingency table we can take the appropriate values from the table as opposed to using the formula given above. There are 3814 female undergraduate students, 3428 male undergraduate students, 2213 female graduate students, and a total of 12242 students.

\(P(F \cup U)=\dfrac{3814+3428+2213}{12242}=\dfrac{9455}{12242}=0.772\)

Note that the final answer would be the same if we had used the formula:

\(P(F \cup U) = \dfrac{6027}{12242}+\dfrac{7242}{12242}-\dfrac{3814}{12242}= \dfrac{9455}{12242}=0.772\)


2.1.3.2.4 - Complements

2.1.3.2.4 - Complements
Complement

The probability that the event does not occur. The complement of \(P(A)\) is \(P(A^C)\). This may also be written as \(P(A')\).

In the diagram below we can see that \(A^{C}\) is everything in the sample space that is not A.

Complement of A
Complement of A
\(P(A^{C})=1−P(A)\)

Example: Coin Flip

When flipping a coin, one can flip heads or tails. Thus, \(P(Tails^{C})=P(Heads)\) and \(P(Heads^{C})=P(Tails)\)

Example: Hearts

If you randomly select a card from a standard 52-card deck, you could pull a heart, diamond, spade, or club. The complement of pulling a heart is the probability of pulling a diamond, spade, or club. In other words: \(P(Heart^{C})=P(Diamond,\; Spade,\;\;Club)\)

The complement of any outcome is equal to one minus the outcome. In other words: \(P(A^{C})=1-P(A)\)

It is also true then that: \(P(A)=1-P(A^{C})\)

Example: Rain

Light Rain Showers

According to the weather report, there is a 30% chance of rain today: \(P(Rain) = .30\) 

Raining and not raining are complements.

\(P(Not \:rain)=P(Rain^{C})=1-P(Rain)=1-.30=.70\)

There is a 70% chance that it will not rain today.

Example: Winning

The probability that your team will win their next game is calculated to be .45, in other words:

\(P(Winning)=.45\)

Winning and losing are complements of one another. Therefore the probability that they will lose is:

\(P(Losing)=P(Winning^{C})=1-.45=.55\)

The sum of all of the probabilities for possible events is equal to 1.

Example: Cards

In a standard 52-card deck there are 26 black cards and 26 red cards. All cards are either black or red.

\(P(red)+P(black)=\frac{26}{52}+\frac{26}{52}=1\)

Example: Dominant Hand

Of individuals with two hands, it is possible to be right-handed, left-handed, or ambidextrous. Assuming that these are the only three possibilities and that there is no overlap between any of these possibilities:

\(P(right\;handed)+P(left\;handed)+P(ambidextrous) = 1\)


2.1.3.2.5 - Conditional Probability

2.1.3.2.5 - Conditional Probability
Conditional Probability

The probability of one event occurring given that it is known that a second event has occurred. This is communicated using the symbol \(\mid\) which is read as "given."

For example, \(P(A\mid B)\) is read as "Probability of A given B."

Example: PA Resident given Undergraduate

The two-way table below displays the World Campus enrollment from Fall 2019 in terms of level (undergraduate and graduate) and residency (Pennsylvania and non-Pennsylvania). Given that an individual is an undergraduate student, what is the probability that the student is a Pennsylvania resident?

  Pennsylvania Non-Pennsylvania Total
Undergraduate 3757 4603 8360
Graduate 2253 4074 6327
Total 6010 8677 14687

We know that the individual is an undergraduate student so we will only look at the 8360 undergraduate students. Of those 8360 undergraduate students, 3757 were Pennsylvania residents.

\(P(PA \mid Undergrad) = \dfrac{3757}{8360}=0.449\)


2.1.3.2.5.1 - Advanced Conditional Probability Applications

2.1.3.2.5.1 - Advanced Conditional Probability Applications

Advanced Formulas

Conditional probabilities can also be computed using the following formulas. Note that these two formulas are identical, but A and B are switched. Again, if the contingency table is available it is usually most efficient to take the appropriate values from the table, as shown above, as opposed to using these formulas.

Conditional Probability of A Given B
\(P(A\mid B)=\dfrac{P(A \: \cap\: B)}{P(B)}\)
Conditional Probability of B Given A
\(P(B\mid A)=\dfrac{P(A \: \cap\: B)}{P(A)}\)

Example: Clubs

In a standard 52-card deck, there are 26 black cards including 13 clubs. All clubs are black, therefore there are 13 black clubs.

What is the probability that a randomly selected card is a club given that it is a black card?

We are given that \(P(club)=\frac{13}{52}=0.25\), \(P(black)=\frac{26}{52}=0.50\), and  \(P(club \: \cap\: black)=\frac{13}{52}0.25\)

\(P(club\mid black)=\dfrac{P(club \: \cap\: black)}{P(black)}=\dfrac{0.25}{0.50}=0.50\)

Given that a randomly selected card is black, there is a 50% chance that it's a club.

Independent Events Written as Conditional Probabilities

If events A and B are independent then \(P(A) = P(A \mid B)\). In other words, whether or not event B occurs does not change the probability of event A occurring.

Example: Checking for Independence, Aces and Hearts

A card is randomly drawn from a 52-card deck. Are the events of drawing an ace and drawing a heart independent?

In a standard 52-card deck, there are 4 aces and 13 hearts. Therefore \(P(ace)=\frac{4}{52}\) and \(P(heart)=\frac{13}{52}\). Out of 13 hearts, 1 is an ace, which translates to \(P(ace \mid heart) = \frac{1}{13}\).

To determine if these two events are independent we can compare \(P(A)\) to \(P(A\mid B)\). If we call being an ace event A and being a heart event B, then we're comparing \(P(ace)\) to \(P(ace \mid heart)\).

\(P(ace)=\frac{4}{52}=0.0769\)

\(P(ace \mid heart) = \frac{1}{13}=0.0769\)

These values are identical, therefore we can conclude that the events of drawing an ace and drawing a heart are independent. 

 


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