What are the Properties of Probability?
Impossible and Guaranteed Events
What is an Impossible Event?
The Empty Set
The empty set, denoted as \( \emptyset \), is a set that contains no elements. The probability of the empty set is always \[ P(\emptyset) = 0. \]
The Impossible Event
The impossible event is an event \(E\) such that \(P(E)=0\). In other works, \(E=\emptyset\), i.e., the event has no outcomes.
What is a Sure Thing in Probability?
A sure thing (or certain event) is an event that includes all possible outcomes in the sample space \( S \). The probability of a sure thing \(E\) is always \[P(E)=P(S) = 1.\]
Example 1
A space mission is set to launch only if the weather is clear. The probability of a successful launch is 0.9 if conditions are favorable, but if there is a major storm, the probability of launching is 0.
- Part A: What is the probability that the rocket launches during a major storm?
- Part B: Explain why this is an example of an impossible event.
Solution
- Part A: Since the probability of launching during a major storm is 0, we write: \[ P(\text{Launch in Storm}) = 0 \]
- Part B: This is an example of an impossible event because it never occurs under the given conditions. In probability terms, an event with \( P(E) = 0 \) is an impossible event.
$$\tag*{\(\blacksquare\)}$$
Example 2
Every school day, students at Riverside High have a morning announcement. The principal starts the announcement promptly at 8:00 AM, and this has happened every school day for the past five years.
- Part A: What is the probability that a morning announcement occurs on a school day?
- Part B: Explain why this is an example of a sure thing.
Solution
- Part A: Since the announcement always happens on a school day, we write: \[ P(\text{Morning Announcement}) = 1 \]
- Part B: This is an example of a sure thing because it always occurs under the given conditions. In probability terms, an event with \( P(E) = 1 \) is a certain event.
$$\tag*{\(\blacksquare\)}$$
Probability Properties
What are the Properties?
- The probability of any event \( E \) must be between or equal to 0 and 1.
- The sum of all probabilities in a sample space \( S \) must be 1.
- If two events \( E \) and \( F \) do not have any outcomes in common, then the probability of either event occuring is the sum of their individual probabilities.
What is the Notation for these Properties?
- For any event \(E\), \(0\leq P(E)\leq 1\).
- For all outcomes \(x\) in a sample space \(S\), \(\Sigma P(x)=P(S)=1\).
- If \(E\text{ and }F=\emptyset\), then \(P(E\text{ or }F)=P(E)+P(F)\).
Examples
Example 3
A student calculates the probabilities of different events in a game and gets the following: \[ P(A) = 0.4\quad P(B) = 0.5\quad P(C) = 1.1\quad P(D) = -0.2 \] Which of these probabilities are valid? Explain your reasoning based on the properties of probability.
Solution
- \( P(A) = 0.4 \) and \( P(B) = 0.5 \) are valid since they are between 0 and 1.
- \( P(C) = 1.1 \) is invalid because probabilities cannot be greater than 1.
- \( P(D) = -0.2 \) is invalid because probabilities cannot be negative.
$$\tag*{\(\blacksquare\)}$$
Example 4
A fair six-sided die is rolled. The probability of rolling each number is: \[\begin{align*} P(1) &= \frac{{1}}{{6}} & P(2) &= \frac{{1}}{{6}} & P(3) &= \frac{{1}}{{6}} \\\\P(4) &= \frac{{1}}{{6}} & P(5) &= \frac{{1}}{{6}} & P(6) &= \frac{{1}}{{6}}\end{align*} \] Find the sum of the six dice roles and explain why the probability you got makes sense.
Solution
Adding up all probabilities, we get: \[ \begin{align*}P(1) + P(2) + P(3) + P(4) + P(5) + P(6) &= \dfrac{{1}}{{6}} + \dfrac{{1}}{{6}} + \dfrac{{1}}{{6}} + \dfrac{{1}}{{6}} + \dfrac{{1}}{{6}} + \dfrac{{1}}{{6}}\\\\& = 1\end{align*} \] This makes sense since we added up the probabilities of every possible role in the sample space, and the sample space always have a probability of 1.
$$\tag*{\(\blacksquare\)}$$
Example 5
A spinning wheel has three sections: Red, Blue, and Green. The probabilities are: \[ P(\text{{Red}}) = 0.3 \quad P(\text{{Blue}}) = 0.5 \quad P(\text{{Green}}) = 0.2 \] What is the probability that the wheel lands on Red or Green?
Solution
Since a region cannot be both Red and Green, we can add their probabilities:\[\begin{align*} P(\text{Red or Green}) &= P(\text{{Red}}) + P(\text{{Green}})\\ &= 0.3 + 0.2 \\&= 0.5\end{align*} \]
$$\tag*{\(\blacksquare\)}$$
Example 6
A factory produces three types of parts: Type A, Type B, and Type C. The probabilities that a randomly chosen part is of a particular type is known for Types A and B: \[ P(A) = 0.4 \qquad P(B) = 0.35 \] Using this information, determine the probability that a randomly chosen part will be Type C.
Solution
Since all the probabilities must sum to 1, we have that \[P(A)+P(B)+P(C)=1.\] Since we know that \( P(A) = 0.4\) and \(P(B) = 0.35\), we can substitute them into this equation and solve for \(P(C)\) like a variable: \[\begin{align*}P(A)+P(B)+P(C)&=1\\\\0.4+0.35+P(C)&=1\\\\0.75+P(C)&=1\\\\ P(C)&=1-0.75\\\\P(C)&=0.25\end{align*}\] Therefore, a randomly chosen part will be Type C 25% of the time.
$$\tag*{\(\blacksquare\)}$$