Page 2Independent and mutually exclusive do not mean the same thing.
Two events are independent if the following are true:
Two events A and 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 may be done with replacement or without replacement.
If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise.
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, J (jack), Q (queen), K (king) of that suit. a. Sampling with replacement: b. Sampling without replacement:
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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, J (jack), Q (queen), K (king) of that suit. Three cards are picked at random.
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, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs.
Which of a. or b. did you sample with replacement and which did you sample without replacement?
a. Without replacement; b. With replacement
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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, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, 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 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) = 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.
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.
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Draw two cards from a standard 52-card deck with replacement. Find the probability of getting at least one black card.
Flip two fair coins. Find the probabilities of the events.
Look at the sample space in (Figure).
J and H have nothing in common so P(J AND H) = 0. J and H are mutually exclusive.
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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:
Roll one fair, six-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event A = a face is odd. Then A = {1, 3, 5}. Let event B = a face is even. Then B = {2, 4, 6}.
Are C and E mutually exclusive events? (Answer yes or no.) Why or why not?
No. C = {3, 5} and E = {1, 2, 3, 4}. P(C AND E) = . To be mutually exclusive, P(C AND E) must be zero.
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Let event A = learning Spanish. Let event B = learning German. Then A AND B = learning Spanish and German. Suppose P(A) = 0.4 and P(B) = 0.2. P(A AND B) = 0.08. Are events A and B independent? Hint: You must show ONE of the following:
P(A|B) = The events are independent because P(A|B) = P(A).
Let event G = taking a math class. Let event H = taking a science class. Then, G AND H = taking a math class and a science class. Suppose P(G) = 0.6, P(H) = 0.5, and P(G AND H) = 0.3. Are G and H independent? If G and H are independent, then you must show ONE of the following:
NOTE The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information.
a. Show that P(G|H) = P(G).
P(G|H) = = = 0.6 = P(G)
b. Show P(G AND H) = P(G)P(H).
P(G)P(H) = (0.6)(0.5) = 0.3 = P(G AND H) Since G and 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(H|G) = P(H) to show that G and H are independent events.
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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.
S has ten outcomes. What is P(G AND O)?
Let event C = taking an English class. Let event D = taking a speech class. Suppose P(C) = 0.75, P(D) = 0.3, P(C|D) = 0.75 and P(C AND D) = 0.225. Justify your answers to the following questions numerically.
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A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(B AND D) = 0.20.
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 R = red card is drawn, B = blue card is drawn, E = even-numbered card is drawn. The sample space S = R1, R2, R3, B1, B2, B3, B4, B5. S has eight outcomes.
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In a basketball arena,
Let A be the event that a fan is rooting for the away team.
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 F be the event that a student is female. Let 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?
NOTE 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(F|L) yet, so you cannot use the second condition. Solution 1Check whether P(F AND L) = P(F)P(L). We are given that P(F AND L) = 0.45, but P(F)P(L) = (0.60)(0.50) = 0.30. The events of being female and having long hair are not independent because P(F AND L) does not equal P(F)P(L). Solution 2Check whether P(L|F) equals P(L). We are given that P(L|F) = 0.75, but P(L) = 0.50; they are not equal. The events of being female and having long hair are not independent. Interpretation of ResultsThe 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.
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Mark is deciding which route to take to work. His choices are I = the Interstate and F = Fifth Street.
What is the probability of P(I OR F)?
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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 T be the event of getting the white ball twice, F the event of picking the white ball first, S the event of picking the white ball in the second drawing.
Use the following information to answer the next 12 exercises. The graph shown is based on more than 170,000 interviews done by Gallup that took place from January through December 2012. The sample consists of employed Americans 18 years of age or older. The Emotional Health Index Scores are the sample space. We randomly sample one Emotional Health Index Score.
Find the probability that an Emotional Health Index Score is 82.7.
Find the probability that an Emotional Health Index Score is 81.0.
Find the probability that an Emotional Health Index Score is more than 81?
Find the probability that an Emotional Health Index Score is between 80.5 and 82?
If we know an Emotional Health Index Score is 81.5 or more, what is the probability that it is 82.7?
What is the probability that an Emotional Health Index Score is 80.7 or 82.7?
What is the probability that an Emotional Health Index Score is less than 80.2 given that it is already less than 81.
What occupation has the highest emotional index score?
What occupation has the lowest emotional index score?
What is the range of the data?
Compute the average EHIS.
If all occupations are equally likely for a certain individual, what is the probability that he or she will have an occupation with lower than average EHIS?
P(Occupation < 81.3) = 0.5 |