Construct sample spaces and calculate probabilities for simple and/or compound events.
Use appropriate rules of probability for compound events.
An experiment is a procedure that can be infinitely repeated and has a well-defined set of outcomes.
An outcome is the result of a single trial (individual repetition) of an experiment.
A chance experiment (or random experiment) is an experiment that has more than one possible outcome and whose outcomes cannot be predicted with certainty.
The sample space of a chance experiment is the set of all possible outcomes.
An event is a subset of the sample space.
The complement \(E^c\) of event \(E\) is the set of all outcomes in a sample space that are NOT included in event \(E\).
The intersection \(A\cap B\) of two events \(A\) and \(B\) is the set of all outcomes in the sample space that are shared by \(A\) and \(B\).
The union \(A\cup B\) of two events \(A\) and \(B\) is the set of all outcomes in the sample space that are either in \(A\) or \(B\).
Two events \(A\) and \(B\) are mutually exclusive if their intersection \(A\cap B\) is empty.
A probability \(P(E)\) is the measures of how likely an outcome in the event \(E\) will occur in a chance experiment.
Equally likely means that each outcome of an experiment occurs with equal chance.
When the outcomes in the sample space of a chance experiment are equally likely, the probability of an event \(E\) is $$P(E)=\dfrac{\text{number of outcomes in }E}{\text{number of outcomes in }S}$$
Chance experiment that involves tossing fair coins, rolling fair dice and drawing a card from a well-mixed deck of cards have equally likely outcomes.
Note that many chance experiments do not have equally likely outcomes. For example, the majors of students in a class are not equally likely outcomes.
Imagine flipping one fair coin (which means the chance of a head and the chance of a tail are the same). What is the probability of getting the head.
Solution: There are two possible outcomes: Head or Tail.
So the sample space is the set $$S = \{\text{Head}, \text{Tail}\}.$$
The event \(E\) of getting a head is the subset $$E=\{\text{Head}\}.$$
The probability \(P(E)=\dfrac{1}{2}=0.5\).
An empirical (or a statistical) probability is the relative frequency of occurrence of outcomes from observations in repeated experiments: $$ \begin{aligned} P(E)=&\dfrac{\text{number of occurrence of event } E}{\text{total number of observations}}\\[0.5em] =&\dfrac{\text{frequency in }E}{\text{total frequency}}. \end{aligned} $$
A statistics class has 5 math majors and 20 other majors. If a students was randomly select from the class, what’s the probability that the selected students is a math major?
Solution: The sample space is the set of all students in the statistics class. The event is the set of the 5 math majors. Then the probability is $$ P(E)=\dfrac{\text{frequency of math majors}}{\text{total frequency of students}}=\dfrac{5}{25}=0.2. $$
Theoretical probability is an expected value that can be calculated by mathematical theory and assumptions.
When all outcomes in the sample space are equally likely, the probability of a desired event \(E\), known as a theoretical probability, is calculated by $$ P(E)=\dfrac{\text{number of desired outcomes for event }E}{\text{number of all possible outcomes}}. $$
Tree diagrams are often used for counting all possible outcomes.
Find the probability of getting two heads when flipping a fair coins twice.
Solution: In the first time, the coin has two possible outcomes. The second time, the coin still has two possible outcomes. By the fundamental counting principle, we know that the sample space \(S\) contains \(2\cdot 2=4\) possible outcomes.
The event \(E\) of getting 2 heads contains only one outcome: head and head. So the probability of getting two head when flipping a fair coin twice is \(P(E)=\frac{1}{4}.\)
The empirical probability of an event is an “estimate” that based upon observed data from an experiment.
The theoretical probability of an event is an “expected” probability based upon counting rules.
Law of Large Numbers: As an experiment is repeated over and over, that is the number of trials getting larger and larger, the empirical probability of an event approaches the theoretical probability of the event. (Wiki: Law of large numbers.)
By the law of large number, we can say that the probability of any event is the long-term relative frequency of that event.
Flipping a fair coin twice, find the probabilities of getting exactly one head.
When an event \(E\) consists of infinitely many outcomes, the right-hand side of the equality in Property 3 will be an infinite sum.
Easy consequence 1: If events \(A\) and \(B\) are mutually exclusive, then $$P(A\cup B)=P(A)+P(B).$$
Easy consequence 2: The probability \(P(E)\) of an event \(E\) and the probability \(P(E^c)\) of the complement event \(E^c\) satisfies the identity: $$P(E)+P(E^c)=1.$$
Equivalently, $$ P(E^c)=1-P(E)\quad\text{or}\quad P(E)=1-P(E^c). $$
A six-sided fair die is rolled. Denote by \(E\) the event of getting a number less than \(3\).
Solution: The sample space of the six-sided die is \(\{1,2,3,4,5,6\}\).
The event \(E\) consists of 2 numbers: \(E=\{1, 2\}\).
The probability is $$ P(E)=\frac26=\frac13. $$
The complement \(E^c=\{3, 4, 5, 6\}\) and the probability is $$ P(E^c)=\frac46=\frac23. $$
Then \(P(E)+P(E^c)=\frac13+\frac23=1.\)
Two six-sided fair dice were rolled. Find the probability of getting two numbers whose sum is at least 4.
Solution: The sample space contains \(6\cdot 6=36\) possible out comes.
Source of image: https://sasandr.wordpress.com/2012/05/04/rolling-the-dice-ii/
Let \(E\) be the event of the sum is at least 4. Then the complement \(E^c\) consists pairs of numbers whose sum are at most 3. There are 3 such pairs: $$E^c=\{(1, 1), (1, 2), (2, 1)\}.$$
Therefore, $$ \begin{aligned} P(E^c)=&P((1,1))+P((1,2))+P((2,1))\\ =&\frac16\cdot\frac16+\frac16\cdot\frac16+\frac16\cdot\frac16=\frac{1}{12}. \end{aligned} $$
Apply the complement rule, we find $$P(E)=1-P(E^c)=\frac{11}{12}.$$
When outcomes in the sample spaces are equally likely,
the probability of the intersection of two events is $$ P(A\cap B)=\dfrac{\text{numbers of elements in } A\cap B}{\text{number of elements in the sample space }S}. $$
the probability of the union of two events is $$ P(A\cup B)=\dfrac{\text{numbers of elements in } A\cup B}{\text{number of elements in the sample space }S}. $$
In general, the probability of the union of two events from a chance experiment is defined by the basic rules and the addition rule.
A card was randomly drawn from a deck of 52 cards. Find the probability of getting each of the following card or combination.
Image source: Wikipedia: Standard 52-card deck
Solution: The sample space \(S\) consists of 52 cards. There are 13 hearts and 12 faces. $$P(\text{heart})=\frac{13}{52}=\frac14,\quad P(\text{face})=\frac{12}{52}=\frac3{13}.$$ There are 3 heart face cards. Then $$ P(\text{heart and face})=\frac{3}{52}=\frac{1}{14}. $$ There are \(22\) cards that are either hearts or faces. Then $$ P(\text{heart or face})=\frac{22}{52}=\frac{11}{26}. $$ Since there is no card which is club and spade, we have $$ P(\text{club and spade})=0. $$
Note: From the addition formula, \(P(\text{heart or face})=P(\text{heart})+P(\text{face})-P(\text{heart and face})=\frac{13}{52}+\frac{12}{52}-\frac{3}{52}=\frac{22}{52}=\frac{11}{26}.\)
The conditional probability of \(A\) given \(B\), written as \(P(A\mid B)\), is the probability that event \(A\) will occur given that the event \(B\) has already occurred.
In the case that the chance experiment has equally likely outcomes, the conditional probability is, $$ P(A\mid B)=\dfrac{\text{numbers of elements in }A\cap B}{\text{number of elements in }B}. $$
In general, we may use fundamental rules of probability and the multiplication rule to calculate the conditional probability.
Multiplication Rule: the probability of the intersection of two events \(A\) and \(B\) satisfies the following equality $$ P(A\cap B)=P(B)P(A\mid B)=P(A)P(B\mid A). $$
The multiplication rule gives a formula for conditional probability: $$ P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}\qquad\qquad P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}. $$
Two events \(A\) and \(B\) are independent if $$ P(A\mid B)=P(A)\quad \text{ or } \quad P(B)=P(B\mid A). $$ Equivalently, $$ P(A\cap B)=P(A)P(B). $$
Fundamental Counting Principle: if there are \(m\) ways of doing something and \(n\) ways of doing another thing independently, then there are \(m\cdot n\) ways of performing both actions in order.
See https://online.stat.psu.edu/stat414/lesson/3 for more information and examples on counting technique.
A fair six-sided die is rolled.
Find the probability that the number rolled is a two, given that it is even.
Solution: Let \(A\) be the event of all possible even outcomes and \(B\) the event consisting of the outcome 2. Then \(A=\{2, 4, 6\}\) and \(B=\{2\}\). The intersection event \(A\cap B\) consists of the number \(2\). By the definition of conditional probability, $$P(B|A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{1}{3}.$$
Note: Since there are 6 possible outcomes in total, \(P(A)=\frac36=\frac12\), \(P(A\cap B)=\frac16\), and then \(P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac16/\frac12=\frac13\).
Consider flipping a fair coin and rolling a fair six-sided die together.
Solution: By the fundamental counting principle, the sample space consists of \(2\times 6=12\) elements. Let \(H\) and \(D\) be the event of getting a head and getting a number bigger than 4 respectively. Then $$ H=\{H1, H2, H3, H4, H5, H6\},\quad D=\{H5,H6,T5,T6\},\quad D\cap H=\{H5, H6\}. $$ Therefore, \(P(H)=\frac{6}{12}=\frac12\) and \(P(D)=\frac{4}{12}=\frac13\)
Given that a head shows, the change of getting a number bigger than 4 is $$P(D\mid H)=\frac{2}{6}=\frac13.$$
This can also be obtained by the multiplication rule, $$P(D\cap H)=P(H)P(D\mid H)=\frac12\cdot\frac13=\frac16.$$
Since $$P(D)=\frac13=P(D\mid H)\quad\text{and then}\quad P(D\cap H)=P(H)P(D|H)=P(H)P(D),$$ the events \(H\) and \(D\) are independent.
The probability that a student borrows a statistics book from the library is 0.3. The probability that a student borrows a biology book is 0.4. Given that a student borrowed a biology book, the probability that he/she borrows a statistics book is 0.6.
Solution: Denote by \(S\) the event that a student borrows a statistics book, and \(B\) the event that the student borrows a biology book.
From the given conditions, we know that \(P(S)=0.3\), \(P(B)=0.4\) and \(P(S\mid B)=0.6\).
By the multiplication rule, we know $$ P(S\cap B)=P(B)P(S\mid B)=0.4\cdot 0.6=0.24. $$
By the addition rule, we get $$ \begin{aligned} P(S\cup B)=&P(S)+P(B)-P(S\cap B)\\ =&0.3+0.4-0.24=0.46. \end{aligned} $$
With replacement: If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.
Without replacement: When sampling is done without replacement, each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent.
Two cards were randomly drawn from a standard deck of 52 cards with replacement. Find the probability of getting exactly one club card.
Solution: There are two different pairs with exactly one club card: (club, not club), (not club, club). When drawing with replacement, the events are considered to be independent. Therefore, the probability in those two situations are $$P(\text{(club, not club)})=P(\text{club})\cdot P(\text{not club})=\frac{13}{52}\cdot\frac{39}{52},$$ $$P(\text{(not club, club)})=P(\text{not club})\cdot P(\text{club})=\frac{39}{52}\cdot\frac{13}{52}.$$ Then the probability of getting exactly one club is $$P(\text{exactly one club})=\frac{13}{52}\cdot\frac{39}{52}+\frac{39}{52}\cdot\frac{13}{52}=\frac{3}{8}.$$
Two cards were randomly drawn from a standard deck of 52 cards without replacement, which means the first card will not be put back.
Solution: Let \(S1\) be the event of getting a spade in the first drawing and \(S2\mid S1\) be the event of getting the second spade given the first card is a spade. Then \(P(S1)=\frac{13}{52}=\frac14\), \(P(S2\mid S1)=\frac{12}{51}\), and $$ P(S1 \text{ and } S2)=P(S1)P(S2\mid S1)=\frac{1}{4}\cdot\frac{12}{51}=\frac{4}{51}. $$
Let \(NS1\) and \(NS2\) be events of not getting a spade card in first and second drawing respectively. Then $$ P(S1 \text{ and } NS2)+P(NS1\text{ and } S2)= \frac{13}{52}\cdot \frac{39}{51} + \frac{39}{52}\cdot\frac{13}{51}=\frac{39}{102}. $$
A special deck of 16 cards has 4 that are blue, 4 yellow, 4 green, and 4 red. The four cards of each color are numbered from one to four. A single card is drawn at random. Find the following probabilities.
A box contains 10 pens, 6 black and 4 red. Two pens are drawn without replacement, which means that the first one is not put back.