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Probability

Events, complements, independence, the addition and multiplication rules.

Probability measures how likely an event is, from 0 to 1. For independent events multiply; for mutually exclusive events add; for anything else, P(A or B) = P(A) + P(B) − P(A and B). Two dice showing double six is (1/6)(1/6) = 1/36.

Terminology

Probability is a measure that is associated with how certain we are of results, or outcomes, of a particular activity. When the activity is a planned operation carried out under controlled conditions, it is called an experiment. If the result is not predetermined, then the experiment is said to be a chance experiment. Each time the experiment is attempted is called a trial.

Examples of chance experiments include the following:

  • flipping a fair coin,
  • spinning a spinner,
  • drawing a marble at random from a bag, and
  • rolling a pair of dice.

A result of an experiment is called an outcome. The sample space of an experiment is the set, or collection, of all possible outcomes.

There are four main ways to represent a sample space:

Flipping a Fair CoinFlipping Two Fair Coins
Systematic List of Outcomesheads (H)
tails (T)		HH
HT
TH
TT
Tree Diagram*
Venn Diagram*
Set Notation\(S=\ \{H,\ T\}\)\(S=\ \{HH,\ HT,\ TH,\ TT\}\)

*We will investigate tree diagrams and Venn diagrams in Section 3.5.

Note—when represented as a set, the sample space is denoted with an uppercase S.

An event is any combination of outcomes. It is a subset of the sample space, so uppercase letters like A and B are commonly used to represent events. For example, if the experiment is to flip three fair coins, event A might be getting at most one head.

\[P(A)=\frac{\text{Number of outcomes in event A}}{\text{Total number of possible outcomes.}}\]\[P(A)=\frac{\text{Number of times event A occurs.}}{\text{Total number of trials}}\]
Example

Try it.

A fair, six-sided die is rolled. The sample space, S, is {1, 2, 3, 4, 5, 6}. Describe each event and calculate its probability.

  1. Event T = the outcome is two.
  2. Event A = the outcome is an even number.
  3. Event B = the outcome is less than four.
  4. The complement of A
  5. A GIVEN B
  6. B GIVEN A
  7. A AND B
  8. A OR B
  9. A OR B′
  10. Event N = the outcome is a prime number.
  11. Event I = the outcome is seven.
Solution
  1. T = {2}, P(T) = \(\frac{\text{number of outcomes in }\ T}{\text{number of outcomes in }\ S}\) = \(\frac{1}{6}\)
  2. A = {2, 4, 6}, P(A) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
  3. B = {1, 2, 3}, P(B) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
  4. A′ = {1, 3, 5}, P(A′) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
  5. A|B = {2}, There are three outcomes in B, and only 1 of these lies in A, so P(A|B) = \(\frac{1}{3}\)
  6. B|A = {2}, There are three outcomes in A, and only 1 of these lies in B, so P(B|A) = \(\frac{1}{3}\)
  7. A AND B = {2}, P(A AND B) = \(\frac{1}{6}\)
  8. A OR B = {1, 2, 3, 4, 6}, P(A OR B) = \(\frac{5}{6}\)
  9. A OR B′ = {2, 4, 5, 6}, P(A OR B′) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
  10. N = {2, 3, 5}, P(N) = \(\frac{1}{2}\)
  11. It is impossible to roll a die and get an outcome of 7, so P(7) = 0.

Condensed — the full section is in OpenStax Statistics.

Terminology

Use the following information to answer the next four exercises. A box is filled with several party favors. It contains 12 hats, 15 noisemakers, 10 finger traps, and five bags of confetti.
Let H = the event of getting a hat.
Let N = the event of getting a noisemaker.
Let F = the event of getting a finger trap.
Let C = the event of getting a bag of confetti.


Use the following information to answer the next six exercises. A jar of 150 jelly beans contains 22 red jelly beans, 38 yellow, 20 green, 28 purple, 26 blue, and the rest are orange.
Let B = the event of getting a blue jelly bean
Let G = the event of getting a green jelly bean.
Let O = the event of getting an orange jelly bean.
Let P = the event of getting a purple jelly bean.
Let R = the event of getting a red jelly bean.
Let Y = the event of getting a yellow jelly bean.


Use the following information to answer the next six exercises. There are 23 countries in North America, 12 countries in South America, 47 countries in Europe, 44 countries in Asia, 54 countries in Africa, and 14 countries in Oceania (Pacific Ocean region).
Let A = the event that a country is in Asia.
Let E = the event that a country is in Europe.
Let F = the event that a country is in Africa.
Let N = the event that a country is in North America.
Let O = the event that a country is in Oceania.
Let S = the event that a country is in South America.


Use the following information to answer the next two exercises. You see a game at a local fair. You have to throw a dart at a color wheel. Each section on the color wheel is equal in area.
Let B = the event of landing on blue.
Let R = the event of landing on red.
Let G = the event of landing on green.
Let Y = the event of landing on yellow.


Use the following information to answer the next 10 exercises. On a baseball team, there are infielders and outfielders. Some players are great hitters, and some players are not great hitters.
Let I = the event that a player in an infielder.
Let O = the event that a player is an outfielder.
Let H = the event that a player is a great hitter.
Let N = the event that a player is not a great hitter.


Use the following information to answer the next two exercises. You are rolling a fair, six-sided number cube. Let E = the event that it lands on an even number. Let M = the event that it lands on a multiple of three.

Independent Events

Two events are independent if the following are true:

  • P(A|B) = P(A)
  • P(B|A) = P(B)
  • P(A AND B) = P(A)P(B)

Two events A and B are independent events if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are not independent, then we say that they are dependent events.

Sampling may be done with replacement or without replacement.

  • 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.

A bag contains four blue and three white marbles. James draws one marble from the bag at random, records the color, and replaces the marble. The probability of drawing blue is \(\frac{4}{7}\). When James draws a marble from the bag a second time, the probability of drawing blue is still \(\frac{4}{7}\). James replaced the marble after the first draw, so there are still four blue and three white marbles.

  • 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 or not independent.

The bag still contains four blue and three white marbles. Maria draws one marble from the bag at random, records the color, and sets the marble aside. The probability of drawing blue on the first draw is \(\frac{4}{7}\). Suppose Maria draws a blue marble and sets it aside. When she draws a marble from the bag a second time, there are now three blue and three white marbles. So, the probability of drawing blue is now \(\frac{3}{6}=\frac{1}{2}\). Removing the first marble without replacing it influences the probabilities on the second draw.

If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise.

Condensed — the full section is in OpenStax Statistics.

Mutually Exclusive Events

A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P(A AND B) = 0.

For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}. A AND B = {4, 5}. P(A AND B) = \(\frac{2}{10}\) and is not equal to zero. Therefore, A and B are not mutually exclusive.
A and C do not have any numbers in common so P(A AND C) = 0. Therefore, A and C are mutually exclusive.

If it is not known whether A and B are mutually exclusive, assume they are not until you can show otherwise. The following examples illustrate these definitions and terms.

Example

Flip two fair coins. This is an experiment.

The sample space is {HH, HT, TH, TT}, where T = tails and H = heads. The outcomes are HH, HT, TH, and TT. The outcomes HT and TH are different. The HT means that the first coin showed heads and the second coin showed tails. The TH means that the first coin showed tails and the second coin showed heads.

  • Let A = the event of getting at most one tail. At most one tail means zero or one tail. Then A can be written as {HH, HT, TH}. The outcome HH shows zero tails. HT and TH each show one tail.
  • Let B = the event of getting all tails. B can be written as {TT}. B is the complement event of A, so B = A′. Also, P(A) + P(B) = P(A) + P(A′) = 1.
  • The probabilities for A and for B are P(A) = \(\frac{3}{4}\) and P(B) = \(\frac{1}{4}\).
  • Let C = the event of getting all heads. C = {HH}. Since B = {TT}, P(B AND C) = 0. B and C are mutually exclusive. (B and C have no members in common because you cannot have all tails and all heads at the same time.)
  • Let D = event of getting more than one tail. D = {TT}. P(D) = \(\frac{1}{4}\text{.}\)
  • Let E = event of getting a head on the first roll. This implies you can get either a head or tail on the second roll. E = {HT, HH}. P(E) = \(\frac{2}{4}\text{.}\)
  • Find the probability of getting at least one (one or two) tail in two flips. Let F = event of getting at least one tail in two flips. F = {HT, TH, TT}. P(F) = \(\frac{3}{4}\text{.}\)

Condensed — the full section is in OpenStax Statistics.

Two Basic Rules of Probability

In calculating probability, there are two rules to consider when you are determining if two events are independent or dependent and if they are mutually exclusive or not.

The Multiplication Rule

If A and B are two events defined on a sample space, then P(A AND B) = P(B)P(A|B).

This equation can be rewritten as P(A AND B) = P(B)P(A|B), the multiplication rule.

If A and B are independent, then P(A|B) = P(A). In this special case, P(A AND B) = P(A|B)P(B) becomes P(A AND B) = P(A)P(B).

A bag contains four green marbles, three red marbles, and two yellow marbles. Mark draws two marbles from the bag without replacement. The probability that he draws a yellow marble and then a green marble is

\[\begin{array}{l}P(\text{yellow and green})=P(\text{yellow})⋅P(\text{green | yellow}) \\ \\ =\frac{2}{9}⋅\frac{4}{8} \\ =\frac{1}{9}\end{array}\]

Notice that \(P(\text{green | yellow})=\frac{4}{8}\). After the yellow marble is drawn, there are four green marbles in the bag and eight marbles in all.

The Addition Rule

If A and B are defined on a sample space, then P(A OR B) = P(A) + P(B) − P(A AND B).

Draw one card from a standard deck of playing cards. Let H = the card is a heart, and let J = the card is a jack. These events are not mutually exclusive because a card can be both a heart and a jack.

\[\begin{array}{l}P(H\text{ or }J)=P(H)+P(J)-P(H\text{ and }J) \\ \\ =\frac{13}{52}+\frac{4}{52}-\frac{1}{52} \\ =\frac{16}{52} \\ =\frac{4}{13} \\ \approx .3077\end{array}\]

If A and B are mutually exclusive, then P(A AND B) = 0. Then P(A OR B) = P(A) + P(B) − P(A AND B) becomes
P(A OR B) = P(A) + P(B).

Draw one card from a standard deck of playing cards. Let H = the card is a heart and S = the card is a spade. These events are mutually exclusive because a card cannot be a heart and a spade at the same time. The probability that the card is a heart or a spade is

\[\begin{array}{l}P(H\text{ or }S)=P(H)+P(S) \\ \\ =\frac{13}{52}+\frac{13}{52} \\ =\frac{26}{52} \\ =\frac{1}{2} \\ =.5\end{array}\]
Example

Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska.

  • Klaus can only afford one vacation. The probability that he chooses A is P(A) = .6 and the probability that he chooses B is P(B) = .35.
  • P(A AND B) = 0 because Klaus can only afford to take one vacation.
  • Therefore, the probability that he chooses either New Zealand or Alaska is P(A OR B) = P(A) + P(B) = .6 + .35 = .95. Note that the probability that he does not choose to go anywhere on vacation must be .05.

Condensed — the full section is in OpenStax Statistics.

Two Basic Rules of Probability

Use the following information to answer the next 10 exercises. Forty-eight percent of all voters of a certain state prefer life in prison without parole over the death penalty for a person convicted of first-degree murder. Among Latino registered voters in this state, 55 percent prefer life in prison without parole over the death penalty for a person convicted of first-degree murder. Of all citizens in this state, 37.6 percent are Latino.

In this problem, let

  • C = citizens of a certain state (registered voters) preferring life in prison without parole over the death penalty for a person convicted of first-degree murder.
  • L = registered voters of the state who are Latino.

Suppose that one citizen is randomly selected.

Contingency Tables

A two-way table provides a way of portraying data that can facilitate calculating probabilities. When used to calculate probabilities, a two-way table is often called a contingency table. The table helps in determining conditional probabilities quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. We used two-way tables in Chapters 1 and 2 to calculate marginal and conditional distributions. These tables organize data in a way that supports the calculation of relative frequency and, therefore, experimental (empirical) probability. Later on, we will use contingency tables again, but in another manner.

Condensed — the full section is in OpenStax Statistics.

Contingency Tables

Use the following information to answer the next four exercises. shows a random sample of musicians and how they learned to play their instruments.

GenderSelf-TaughtStudied in SchoolPrivate InstructionTotal
Female12382272
Male19241558
Total316237130

Tree and Venn Diagrams

Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities.

Tree Diagrams

A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of branches that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram:

Condensed — the full section is in OpenStax Statistics.

Venn Diagram

A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events.

Example

Suppose an experiment has the outcomes 1, 2, 3, . . . , 12 where each outcome has an equal chance of occurring. Let event A = {1, 2, 3, 4, 5, 6} and event B = {6, 7, 8, 9}. Then A AND B = {6} and A OR B = {1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is as follows:

Example

Flip two fair coins. Let A = tails on the first coin. Let B = tails on the second coin. Then A = {TT, TH} and B = {TT, HT}. Therefore, A AND B = {TT}. A OR B = {TH, TT, HT}.

The sample space when you flip two fair coins is X = {HH, HT, TH, TT}. The outcome HH is in NEITHER A NOR B. The Venn diagram is as follows:

Condensed — the full section is in OpenStax Statistics.

Робочий приклад: 1/6 * 1/6

Arithmetic · Arithmetic

Evaluate 1/36

\frac{1}{36}

Крок за кроком

  1. 1 \cdot \frac{1}{6} \cdot 1 \cdot \frac{1}{6} = \frac{1}{36}

    Evaluate.

Показати відповідь
\frac{1}{36}

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. The sample space S is the whole numbers starting at one and less than 20.

    1. S = ________

      Let event A = the even numbers and event B = numbers greater than 13.

    2. A = ________, B = ________
    3. P(A) = ________, P(B) = ________
    4. A AND B = ________, A OR B = ________
    5. P(A AND B) = ________, P(A OR B) = ________
    6. A′ = ________, P(A′) = ________
    7. P(A) + P(A′) = ________
    8. P(A|B) = ________, P(B|A) = ________; are the probabilities equal?
    Показати відповідь
    1. S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}
    2. A = {2, 4, 6, 8, 10, 12, 14, 16, 18}, B = {14, 15, 16, 17, 18, 19}
    3. P(A) = \(\frac{\text{number of outcomes in }\ A}{\text{number of outcomes in }\ S}\) = \(\frac{9}{19}\), P(B) = \(\frac{\text{number of outcomes in }\ B}{\text{number of outcomes in }\ S}\) = \(\frac{6}{19}\)
    4. The set A AND B contains all outcomes that lie in both sets A and B, so A AND B = {14,16,18}, The set A OR B contains all outcomes that lie either of the sets A or B, so A OR B = {2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19}.
    5. P(A AND B) = \(\frac{3}{19}\text{,}\) P(A OR B) = \(\frac{12}{19}\)
    6. A' consists of all outcomes in the sample space, S, that DO NOT lie in A, so A′ = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19; P(A′) = \(\frac{10}{19}\).
    7. P(A) + P(A′) = \(\frac{9}{19}\) + \(\frac{10}{19}\) = 1
    8. P(A|B) = \(\frac{P(A\text{ AND }B)}{P(B)}\) \(\ =\ \frac{\frac{3}{19}}{\frac{6}{19}}\) = \(\frac{3}{6}\), P(B|A) = \(\frac{P(A\text{ AND }B)}{P(A)}\) \(\ =\ \frac{\frac{3}{19}}{\frac{9}{19}}\) = \(\frac{3}{9}\), No, the probabilities are not equal.

  2. The sample space S is all the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)).

    1. S = ________
      Let event A = the sum is even and event B = the first number is prime.
    2. A = ________, B = ________
    3. P(A) = ________, P(B) = ________
    4. A AND B = ________, A OR B = ________
    5. P(A AND B) = ________, P(A OR B) = ________
    6. B′ = ________, P(B′) = ________
    7. P(A) + P(A′) = ________
    8. P(A|B) = ________, P(B|A) = ________; are the probabilities equal?
    Показати відповідь
    1. S = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)}
    2. A = {(1,1), (1,3), (2,2), (2,4), (3,1), (3,3)}
      B = {(2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)}
    3. P(A) = \(\frac{1}{2}\), P(B) = \(\frac{2}{3}\)
    4. A AND B = {(2,2), (2,4), (3,1), (3,3)}
      A OR B = {(1,1), (1,3), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)}
    5. P(A AND B) = \(\frac{1}{3}\), P(A OR B) = \(\frac{5}{6}\)
    6. B′ = {(1,1), (1,2), (1,3), (1,4)}, P(B′) = \(\frac{1}{3}\)
    7. P(B) + P(B′) = 1
    8. P(A|B) = \(\frac{P\text{(}A\text{ AND }B\text{)}}{P\text{(}B\text{)}}\text{ = }\frac{1}{2}\), P(B|A) = \(\frac{P\text{(}A\text{ AND }B\text{)}}{P\text{(}B\text{)}}\text{ = }\frac{2}{3}\), No.

  3. A fair, six-sided die is rolled. The sample space, S, is {1, 2, 3, 4, 5, 6}. Describe each event and calculate its probability.

    1. Event T = the outcome is two.
    2. Event A = the outcome is an even number.
    3. Event B = the outcome is less than four.
    4. The complement of A
    5. A GIVEN B
    6. B GIVEN A
    7. A AND B
    8. A OR B
    9. A OR B′
    10. Event N = the outcome is a prime number.
    11. Event I = the outcome is seven.
    Показати відповідь
    1. T = {2}, P(T) = \(\frac{\text{number of outcomes in }\ T}{\text{number of outcomes in }\ S}\) = \(\frac{1}{6}\)
    2. A = {2, 4, 6}, P(A) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
    3. B = {1, 2, 3}, P(B) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
    4. A′ = {1, 3, 5}, P(A′) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
    5. A|B = {2}, There are three outcomes in B, and only 1 of these lies in A, so P(A|B) = \(\frac{1}{3}\)
    6. B|A = {2}, There are three outcomes in A, and only 1 of these lies in B, so P(B|A) = \(\frac{1}{3}\)
    7. A AND B = {2}, P(A AND B) = \(\frac{1}{6}\)
    8. A OR B = {1, 2, 3, 4, 6}, P(A OR B) = \(\frac{5}{6}\)
    9. A OR B′ = {2, 4, 5, 6}, P(A OR B′) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
    10. N = {2, 3, 5}, P(N) = \(\frac{1}{2}\)
    11. It is impossible to roll a die and get an outcome of 7, so P(7) = 0.

  4. Let’s denote the events M = the subject is male, F = the subject is female, R = the subject is right-handed, L = the subject is left-handed. Compute the following probabilities:

    1. P(M)
    2. P(F)
    3. P(R)
    4. P(L)
    5. P(M AND R)
    6. P(F AND L)
    7. P(M OR F)
    8. P(M OR R)
    9. P(F OR L)
    10. P(M')
    11. P(R|M)
    12. P(F|L)
    13. P(L|F)
    Показати відповідь
    1. \(P(M)=\frac{\text{number of males}}{\text{total number of subjects}}=\frac{43+9}{43+9+44+4}=\frac{52}{100}=.52\)
    2. \(P(F)=\frac{\text{number of females}}{\text{total number of subjects}}=\frac{44+4}{43+9+44+4}=\frac{48}{100}=.48\)
    3. \(P(R)=\frac{\text{number of right-handed subjects}}{\text{total number of subjects}}=\frac{43+44}{43+9+44+4}=\frac{87}{100}=.87\)
    4. \(P(L)=\frac{\text{number of left-handed subjects}}{\text{total number of subjects}}=\frac{9+4}{43+9+44+4}=\frac{13}{100}=.13\)
    5. \(P(M\text{ and }R)=\frac{\text{number of male, right-handed subjects}}{\text{total number of subjects}}=\frac{43}{100}=.43\)
    6. \(P(F\text{ and }L)=\frac{\text{number of female, left-handed subjects}}{\text{total number of subjects}}=\frac{4}{100}=.04\)
    7. \(P(M\text{ or }F)=\frac{\text{number of subjects that are male or female}}{\text{total number of subjects}}=\frac{52+48}{100}=\frac{100}{100}=1\)
    8. \[P(M\text{ or }R)=\frac{\text{number of subjects that are male or right-handed}}{\text{total number of subjects}}=\frac{43+9+44}{100}=\frac{96}{100}=.96\]
    9. \[P(F\text{ or }L)=\frac{\text{number of subjects that are female or left-handed}}{\text{total number of subjects}}=\frac{44+4+9}{100}=\frac{57}{100}=.57\]
    10. \(P({M}^{'})=\frac{\text{number of subjects who are not male}}{\text{total number of subjects}}=\frac{44+4}{43+9+44+4}=\frac{48}{100}=.48\)
    11. \[P(R|M)=\frac{P(R\text{ and }M)}{P(M)}=\frac{0.43}{0.52}˜=.8269\] (rounded to four decimal places)
    12. \[P(F|L)=\frac{P(F\text{ and }L)}{P(L)}=\frac{0.04}{0.13}˜=.3077\] (rounded to four decimal places)
    13. \[P(L|F)=\frac{P(L\text{ and }F)}{P(F)}=\frac{0.04}{0.48}˜=.0833\] (rounded to four decimal places)

  5. Find P(H).

  6. Find P(N).

    Показати відповідь

    P(N) = \(\frac{15}{42}\) = \(\frac{5}{14}\) = .36

  7. Find P(F).

  8. Find P(C).

    Показати відповідь

    P(C) = \(\frac{5}{42}\) = .12

  9. Find P(B).

  10. Find P(G).

    Показати відповідь

    P(G) = \(\frac{20}{150}\) = \(\frac{2}{15}\) = .13

  11. Find P(P).

  12. Find P(R).

    Показати відповідь

    P(R) = \(\frac{22}{150}\) = \(\frac{11}{75}\) = .15

  13. Find P(Y).

  14. Find P(O).

    Показати відповідь

    P(O) = \(\frac{150-22-38-20-28-26}{150}\) = \(\frac{16}{150}\) = \(\frac{8}{75}\) = .11

  15. Find P(A).

  16. Find P(E).

    Показати відповідь

    P(E) = \(\frac{47}{194}\) = .24

  17. Find P(F).

  18. Find P(N).

    Показати відповідь

    P(N) = \(\frac{23}{194}\) = .12

  19. Find P(O).

  20. Find P(S).

    Показати відповідь

    P(S) = \(\frac{12}{194}\) = \(\frac{6}{97}\) = .06

  21. What is the probability of drawing a red card in a standard deck of 52 cards?

  22. What is the probability of drawing a club in a standard deck of 52 cards?

    Показати відповідь

    \(\frac{13}{52}\) = \(\frac{1}{4}\) = .25

  23. What is the probability of rolling an even number of dots with a fair, six-sided die numbered one through six?

  24. What is the probability of rolling a prime number of dots with a fair, six-sided die numbered one through six?

    Показати відповідь

    \(\frac{3}{6}\) = \(\frac{1}{2}\) = .5

  25. If you land on Y, you get the biggest prize. Find P(Y).

  26. If you land on red, you don’t get a prize. What is P(R)?

    Показати відповідь

    \(P(R)=\frac{4}{8}=.5\)

  27. Write the symbols for the probability that a player is not an outfielder.

  28. Write the symbols for the probability that a player is an outfielder or is a great hitter.

    Показати відповідь

    P(O OR H)

  29. Write the symbols for the probability that a player is an infielder and is not a great hitter.

  30. Write the symbols for the probability that a player is a great hitter, given that the player is an infielder.

    Показати відповідь

    P(H|I)

  31. Write the symbols for the probability that a player is an infielder, given that the player is a great hitter.

  32. Write the symbols for the probability that of all the outfielders, a player is not a great hitter.

    Показати відповідь

    P(N|O)

  33. Write the symbols for the probability that of all the great hitters, a player is an outfielder.

  34. Write the symbols for the probability that a player is an infielder or is not a great hitter.

    Показати відповідь

    P(I OR N)

  35. Write the symbols for the probability that a player is an outfielder and is a great hitter.

  36. Write the symbols for the probability that a player is an infielder.

    Показати відповідь

    P(I)

  37. What is the word for the set of all possible outcomes?

  38. What is conditional probability?

    Показати відповідь

    The likelihood that an event will occur given that another event has already occurred.

  39. A shelf holds 12 books. Eight are fiction and the rest are nonfiction. Each is a different book with a unique title. The fiction books are numbered one to eight. The nonfiction books are numbered one to four. Randomly select one book
    Let F = event that book is fiction
    Let N = event that book is nonfiction
    What is the sample space?

  40. What is the sum of the probabilities of an event and its complement?

    Показати відповідь

    1

Спробуйте власну

Parts of this page are adapted from OpenStax Statistics (CC BY 4.0). Condensed and re-explained here; errors are ours.

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Sampling and dataDescribing data with graphsMean, median and modeCounting: permutations and combinationsDiscrete random variablesContinuous random variablesThe normal distributionThe central limit theoremConfidence intervalsHypothesis testingComparing two samplesChi-square testsLinear regression and correlationANOVA and the F distribution