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Independent and mutually exclusive do not mean the same thing. Independent EventsTwo events are independent if the following are true:
Two events \(\text{A}\) and \(\text{B}\) are independent 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. Sampling a population Sampling may be done with replacement or without replacement (Figure \(\PageIndex{1}\)):
If it is not known whether \(\text{A}\) and \(\text{B}\) are independent or dependent, assume they are dependent until you can show otherwise. Example \(\PageIndex{1}\): Sampling with and without replacement You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \(\text{J}\) (jack), \(\text{Q}\) (queen), \(\text{K}\) (king) of that suit. a. Sampling with replacement: Suppose you pick three cards with replacement. The first card you pick out of the 52 cards is the \(\text{Q}\) of spades. You put this card back, reshuffle the cards and pick a second card from the 52-card deck. It is the ten of clubs. You put this card back, reshuffle the cards and pick a third card from the 52-card deck. This time, the card is the \(\text{Q}\) of spades again. Your picks are {\(\text{Q}\) of spades, ten of clubs, \(\text{Q}\) of spades}. You have picked the \(\text{Q}\) of spades twice. You pick each card from the 52-card deck. b. Sampling without replacement: Suppose you pick three cards without replacement. The first card you pick out of the 52 cards is the \(\text{K}\) of hearts. You put this card aside and pick the second card from the 51 cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining 50 cards in the deck. The third card is the \(\text{J}\) of spades. Your picks are {\(\text{K}\) of hearts, three of diamonds, \(\text{J}\) of spades}. Because you have picked the cards without replacement, you cannot pick the same card twice. Exercise \(\PageIndex{1}\) You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \(\text{J}\) (jack), \(\text{Q}\) (queen), \(\text{K}\) (king) of that suit. Three cards are picked at random.
With replacement Answer bNo Example \(\PageIndex{2}\) You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \(\text{J}\) (jack), \(\text{Q}\) (queen), and \(\text{K}\) (king) of that suit. \(\text{S} =\) spades, \(\text{H} =\) Hearts, \(\text{D} =\) Diamonds, \(\text{C} =\) Clubs.
Which of a. or b. did you sample with replacement and which did you sample without replacement? Answer aWithout replacement Answer bWith replacement Exercise \(\PageIndex{2}\) You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \(\text{J}\) (jack), \(\text{Q}\) (queen), and \(\text{K}\) (king) of that suit. \(\text{S} =\) spades, \(\text{H} =\) Hearts, \(\text{D} =\) Diamonds, \(\text{C} =\) Clubs. Suppose that you sample four cards without replacement. Which of the following outcomes are possible? Answer the same question for sampling with replacement.
a. Possible; b. Impossible, c. Possible Answer - with replacementa. Possible; c. Possible, c. Possible Mutually Exclusive Events\(\text{A}\) and \(\text{B}\) are mutually exclusive events if they cannot occur at the same time. This means that \(\text{A}\) and \(\text{B}\) do not share any outcomes and \(P(\text{A AND B}) = 0\). For example, suppose the sample space \[S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}. \nonumber\] Let \(\text{A} = \{1, 2, 3, 4, 5\}, \text{B} = \{4, 5, 6, 7, 8\}\), and \(\text{C} = \{7, 9\}\). \(\text{A AND B} = \{4, 5\}\). \[P(\text{A AND B}) = \dfrac{2}{10} \nonumber\] and is not equal to zero. Therefore, \(\text{A}\) and \(\text{B}\) are not mutually exclusive. \(\text{A}\) and \(\text{C}\) do not have any numbers in common so \(P(\text{A AND C}) = 0\). Therefore, \(\text{A}\) and \(\text{C}\) are mutually exclusive. If it is not known whether \(\text{A}\) and \(\text{B}\) are mutually exclusive, assume they are not until you can show otherwise. The following examples illustrate these definitions and terms. Example \(\PageIndex{3}\) Flip two fair coins. 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.
Exercise \(\PageIndex{3}\) Draw two cards from a standard 52-card deck with replacement. Find the probability of getting at least one black card. AnswerThe sample space of drawing two cards with replacement from a standard 52-card deck with respect to color is \(\{BB, BR, RB, RR\}\). Event \(A =\) Getting at least one black card \(= \{BB, BR, RB\}\) \(P(\text{A}) = \dfrac{3}{4} = 0.75\) Example \(\PageIndex{4}\) Flip two fair coins. Find the probabilities of the events.
Solution Look at the sample space in Example \(\PageIndex{3}\).
\(\text{J}\) and \(\text{H}\) have nothing in common so \(P(\text{J AND H}) = 0\). \(\text{J}\) and \(\text{H}\) are mutually exclusive. Exercise \(\PageIndex{4}\) A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events:
Answer
Example \(\PageIndex{5}\) Roll one fair, six-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event \(\text{A} =\) a face is odd. Then \(\text{A} = \{1, 3, 5\}\). Let event \(\text{B} =\) a face is even. Then \(\text{B} = \{2, 4, 6\}\).
Are \(\text{C}\) and \(\text{E}\) mutually exclusive events? (Answer yes or no.) Why or why not? Answer No. \(\text{C} = \{3, 5\}\) and \(\text{E} = \{1, 2, 3, 4\}\). \(P(\text{C AND E}) = \dfrac{1}{6}\). To be mutually exclusive, \(P(\text{C AND E})\) must be zero.
Exercise \(\PageIndex{5}\) Let event \(\text{A} =\) learning Spanish. Let event \(\text{B}\) = learning German. Then \(\text{A AND B}\) = learning Spanish and German. Suppose \(P(\text{A}) = 0.4\) and \(P(\text{B}) = 0.2\). \(P(\text{A AND B}) = 0.08\). Are events \(\text{A}\) and \(\text{B}\) independent? Hint: You must show ONE of the following:
Answer \[P(\text{A|B}) = \dfrac{\text{P(A AND B)}}{P(\text{B})} = \dfrac{0.08}{0.2} = 0.4 = P(\text{A})\] The events are independent because \(P(\text{A|B}) = P(\text{A})\). Example \(\PageIndex{6}\) Let event \(\text{G} =\) taking a math class. Let event \(\text{H} =\) taking a science class. Then, \(\text{G AND H} =\) taking a math class and a science class. Suppose \(P(\text{G}) = 0.6\), \(P(\text{H}) = 0.5\), and \(P(\text{G AND H}) = 0.3\). Are \(\text{G}\) and \(\text{H}\) independent? If \(\text{G}\) and \(\text{H}\) are independent, then you must show ONE of the following:
The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information.
Solution
Since \(\text{G}\) and \(\text{H}\) are independent, knowing that a person is taking a science class does not change the chance that he or she is taking a math class. If the two events had not been independent (that is, they are dependent) then knowing that a person is taking a science class would change the chance he or she is taking math. For practice, show that \(P(\text{H|G}) = P(\text{H})\) to show that \(\text{G}\) and \(\text{H}\) are independent events. Exercise \(\PageIndex{6}\) In a bag, there are six red marbles and four green marbles. The red marbles are marked with the numbers 1, 2, 3, 4, 5, and 6. The green marbles are marked with the numbers 1, 2, 3, and 4.
\(\text{S}\) has ten outcomes. What is \(P(\text{G AND O})\)? Answer Event \(\text{G}\) and \(\text{O} = \{G1, G3\}\) \(P(\text{G and O}) = \dfrac{2}{10} = 0.2\) Example \(\PageIndex{7}\) Let event \(\text{C} =\) taking an English class. Let event \(\text{D} =\) taking a speech class. Suppose \(P(\text{C}) = 0.75\), \(P(\text{D}) = 0.3\), \(P(\text{C|D}) = 0.75\) and \(P(\text{C AND D}) = 0.225\). Justify your answers to the following questions numerically.
Solution
Exercise \(\PageIndex{7}\) A student goes to the library. Let events \(\text{B} =\) the student checks out a book and \(\text{D} =\) the student checks out a DVD. Suppose that \(P(\text{B}) = 0.40\), \(P(\text{D}) = 0.30\) and \(P(\text{B AND D}) = 0.20\).
Answer
Example \(\PageIndex{8}\) In a box there are three red cards and five blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card. Let
The sample space \(S = R1, R2, R3, B1, B2, B3, B4, B5\). \(S\) has eight outcomes.
Exercise \(\PageIndex{8}\) In a basketball arena,
Let \(\text{A}\) be the event that a fan is rooting for the away team. Let \(\text{B}\) be the event that a fan is wearing blue. Are the events of rooting for the away team and wearing blue independent? Are they mutually exclusive? Answer
So \(P(\text{B})\) does not equal \(P(\text{B|A})\) which means that \(\text{B} and \text{A}\) are not independent (wearing blue and rooting for the away team are not independent). They are also not mutually exclusive, because \(P(\text{B AND A}) = 0.20\), not \(0\). Example \(\PageIndex{9}\) In a particular college class, 60% of the students are female. Fifty percent of all students in the class have long hair. Forty-five percent of the students are female and have long hair. Of the female students, 75% have long hair. Let \(\text{F}\) be the event that a student is female. Let \(\text{L}\) be the event that a student has long hair. One student is picked randomly. Are the events of being female and having long hair independent?
The choice you make depends on the information you have. You could use the first or last condition on the list for this example. You do not know \(P(\text{F|L})\) yet, so you cannot use the second condition. Solution 1 Check whether \(P(\text{F AND L}) = P(\text{F})P(\text{L})\). We are given that \(P(\text{F AND L}) = 0.45\), but \(P(\text{F})P(\text{L}) = (0.60)(0.50) = 0.30\). The events of being female and having long hair are not independent because \(P(\text{F AND L})\) does not equal \(P(\text{F})P(\text{L})\). Solution 2 Check whether \(P(\text{L|F})\) equals \(P(\text{L})\). We are given that \(P(\text{L|F}) = 0.75\), but \(P(\text{L}) = 0.50\); they are not equal. The events of being female and having long hair are not independent. Interpretation of Results The events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair. Exercise \(\PageIndex{9}\) Mark is deciding which route to take to work. His choices are \(\text{I} = \text{the Interstate}\) and \(\text{F} = \text{Fifth Street}\)
What is the probability of \(P(\text{I OR F})\)? Answer Because \(P(\text{I AND F}) = 0\), \(P(\text{I OR F}) = P(\text{I}) + P(\text{F}) - P(\text{I AND F}) = 0.44 + 0.56 - 0 = 1\) Example \(\PageIndex{10}\)
Solution
Exercise \(\PageIndex{10}\) A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let \(\text{T}\) be the event of getting the white ball twice, \(\text{F}\) the event of picking the white ball first, \(\text{S}\) the event of picking the white ball in the second drawing.
Answer
References
ReviewTwo events \(\text{A}\) and \(\text{B}\) are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent. In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other. Formula Review
Exercise \(\PageIndex{11}\) \(\text{E}\) and \(\text{F}\) are mutually exclusive events. \(P(\text{E}) = 0.4\); \(P(\text{F}) = 0.5\). Find \(P(\text{E∣F})\). Exercise \(\PageIndex{12}\) \(\text{J}\) and \(\text{K}\) are independent events. \(P(\text{J|K}) = 0.3\). Find \(P(\text{J})\). Answer \(P(\text{J}) = 0.3\) Exercise \(\PageIndex{13}\) \(\text{U}\) and \(\text{V}\) are mutually exclusive events. \(P(\text{U}) = 0.26\); \(P(\text{V}) = 0.37\). Find:
Exercise \(\PageIndex{14}\) \(\text{Q}\) and \(\text{R}\) are independent events. \(P(\text{Q}) = 0.4\) and \(P(\text{Q AND R}) = 0.1\). Find \(P(\text{R})\). Answer \(P(\text{Q AND R}) = P(\text{Q})P(\text{R})\) \(0.1 = (0.4)P(\text{R})\) \(P(\text{R}) = 0.25\) Bringing It TogetherExercise \(\PageIndex{16}\) A previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data are compiled into Table. Shirt#≤ 210211–250251–290290≤1–332150034–666187466–99612225For the following, suppose that you randomly select one player from the 49ers or Cowboys. If having a shirt number from one to 33 and weighing at most 210 pounds were independent events, then what should be true about \(P(\text{Shirt} \#1–33|\leq 210 \text{ pounds})\)? Exercise \(\PageIndex{17}\) The probability that a male develops some form of cancer in his lifetime is 0.4567. The probability that a male has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51. Some of the following questions do not have enough information for you to answer them. Write “not enough information” for those answers. Let \(\text{C} =\) a man develops cancer in his lifetime and \(\text{P} =\) man has at least one false positive.
Answer
Exercise \(\PageIndex{18}\) Given events \(\text{G}\) and \(\text{H}: P(\text{G}) = 0.43\); \(P(\text{H}) = 0.26\); \(P(\text{H AND G}) = 0.14\)
Exercise \(\PageIndex{19}\) Given events \(\text{J}\) and \(\text{K}: P(\text{J}) = 0.18\); \(P(\text{K}) = 0.37\); \(P(\text{J OR K}) = 0.45\)
Answer
GlossaryDependent EventsIf two events are NOT independent, then we say that they are dependent.Sampling with ReplacementIf each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once.Sampling without ReplacementWhen sampling is done without replacement, each member of a population may be chosen only once.The Conditional Probability of One Event Given Another EventP(A|B) is the probability that event A will occur given that the event B has already occurred.The OR of Two EventsAn outcome is in the event A OR B if the outcome is in A, is in B, or is in both A and B.This page titled 3.3: Independent and Mutually Exclusive Events is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. What are the events when the outcome affects the outcome of other events?When the outcome affects the second outcome, which is what we called dependent events. Dependent events: Two events are dependent when the outcome of the first event influences the outcome of the second event.
What are the 3 types of events in probability?The different types of events in probability are: Sure event. Impossible event. Independent events.
Which of the following are the events in which the probability of occurrence of any one event is not affected by the occurrence of any other events?Mutually Exclusive Events are when the occurrence of one event has no effect on the probability of occurrence of the other.
What is the probability of occurrence of an event?Number of trials in which the event happened divided by the total number of trials.
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