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Determine whether the events are independent or dependent ..., Study Guides, Projects, Research of French Literature

University of the Philippines Baguio (UPB)French Literature

A number cube is rolled twice, and an odd number is rolled each time. SOLUTION: • Events A and B are independent events if the probability that A occurs does ...

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Determinewhethertheeventsareindependent
ordependent.Explain.
1.Anumbercubeisrolledtwice,andanoddnumberis
rolledeachtime.
SOLUTION:
•EventsAandBareindependenteventsifthe
probabilitythatAoccursdoesnotaffectthe
probabilitythatBoccurs.
•EventsAandBaredependenteventsifthe
probabilitythatAoccursinsomewaychangesthe
probabilitythatBoccurs.
Theoutcomeofthefirstrolldoesnotaffectthe
probabilitiesoftheoutcomesforthesecondroll.
Therefore,theseeventsareindependent.
ANSWER:
Theoutcomeofthefirstrolldoesnotaffectthe
probabilitiesoftheoutcomesforthesecondroll.
Therefore,theseeventsareindependent.
2.Abagcontainsseveralmarbles.Alitaselectsablack
marble,doesnotreplaceit,andthenselectsayellow
marble.
SOLUTION:
•EventsAandBareindependenteventsifthe
probabilitythatAoccursdoesnotaffectthe
probabilitythatBoccurs.
•EventsAandBaredependenteventsifthe
probabilitythatAoccursinsomewaychangesthe
probabilitythatBoccurs.
Becausethefirstmarbleisnotreplaced,theoutcome
ofthefirstselectionaffectstheprobabilitiesofthe
outcomesforthesecondselection.Therefore,these
eventsaredependent.
ANSWER:
Becausethefirstmarbleisnotreplaced,theoutcome
ofthefirstselectionaffectstheprobabilitiesofthe
outcomesforthesecondselection.Therefore,these
eventsaredependent.
Tellwhethertheeventsareindependent.
Explainusingprobability.
3.José,Pippa,andRaymondarescheduledtogive
speeches.Theorderofthespeechesisassignedat
random.Joséisassignedtogofirst,andRaymondis
assignedtogosecond.
SOLUTION:
Theeventsarenotindependent.Thesamplespace
has6equallylikelyoutcomes:{JPR,JRP,PJR,PRJ,
RJP,RPJ}.SoP(J1standR2nd)= ,P(J2st)= ,
andP(R2nd)= ,but .
ANSWER:
Theeventsarenotindependent.Thesamplespace
has6equallylikelyoutcomes:{JPR,JRP,PJR,PRJ,
RJP,RPJ}.SoP(J1standR2nd)= ,P(J2st)= ,
andP(R2nd)= ,but .
4.Saraspinsaspinnertwice.Thespinnerhas5equally
sizedsectionsnumbered1to5.Thespinnerlandson
anoddnumberfirstandanevennumbersecond.
SOLUTION:
Theeventsareindependent.Thesamplespacehas
25equallylikelyoutcomes:{(1,1),(1,2),(1,3),(1,4),
(1,5),(2,1),(2,2),(2,3),(2,4),(2,5),(3,1),(3,2),
(3,3),(3,4),(3,5),(4,1),(4,2),(4,3),(4,4),(4,5),
(5,1),(5,2),(5,3),(5,4),(5,5)}.SoP(oddandeven)
= ,P(odd)= ,P(even)= ,and .
ANSWER:
Theeventsareindependent.Thesamplespacehas
25equallylikelyoutcomes:{(1,1),(1,2),(1,3),(1,4),
(1,5),(2,1),(2,2),(2,3),(2,4),(2,5),(3,1),(3,2),
(3,3),(3,4),(3,5),(4,1),(4,2),(4,3),(4,4),(4,5),
(5,1),(5,2),(5,3),(5,4),(5,5)}.SoP(oddandeven)
= ,P(odd)= ,P(even)= ,and .
eSolutionsManual-PoweredbyCogneroPage1
12-5ProbabilityandtheMultiplicationRule
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Determine whether the events are independent or dependent. Explain.

  1. A number cube is rolled twice, and an odd number is rolled each time. SOLUTION:
    • Events A and B are independent events if the probability that A occurs does not affect the probability that B occurs.
    • Events A and B are dependent events if the probability that A occurs in some way changes the probability that B occurs. The outcome of the first roll does not affect the probabilities of the outcomes for the second roll. Therefore, these events are independent. ANSWER: The outcome of the first roll does not affect the probabilities of the outcomes for the second roll. Therefore, these events are independent.
  2. A bag contains several marbles. Alita selects a black marble, does not replace it, and then selects a yellow marble. SOLUTION:
    • Events A and B are independent events if the probability that A occurs does not affect the probability that B occurs.
    • Events A and B are dependent events if the probability that A occurs in some way changes the probability that B occurs. Because the first marble is not replaced, the outcome of the first selection affects the probabilities of the outcomes for the second selection. Therefore, these events are dependent. ANSWER: Because the first marble is not replaced, the outcome of the first selection affects the probabilities of the outcomes for the second selection. Therefore, these events are dependent. Tell whether the events are independent. Explain using probability.
  3. José, Pippa, and Raymond are scheduled to give speeches. The order of the speeches is assigned at random. José is assigned to go first, and Raymond is assigned to go second. SOLUTION: The events are not independent. The sample space has 6 equally likely outcomes: {JPR, JRP, PJR, PRJ, RJP, RPJ}. So P (J 1st and R 2nd) = , P (J 2st) = , and P (R 2nd) = , but. ANSWER: The events are not independent. The sample space has 6 equally likely outcomes: {JPR, JRP, PJR, PRJ, RJP, RPJ}. So P (J 1st and R 2nd) = , P (J 2st) = , and P (R 2nd) = , but.
  4. Sara spins a spinner twice. The spinner has 5 equally sized sections numbered 1 to 5. The spinner lands on an odd number first and an even number second. SOLUTION: The events are independent. The sample space has 25 equally likely outcomes: {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5)}. So P (odd and even) = , P (odd) = , P (even) = , and. ANSWER: The events are independent. The sample space has 25 equally likely outcomes: {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5)}. So P (odd and even) = , P (odd) = , P (even) = , and.
  1. CARDS A card is randomly chosen from a deck of 52 cards, replaced, and a second card is chosen. What is the probability of choosing both of the cards shown in the order shown? SOLUTION: These events are independent since a first card is replaced before choosing a second card. First we consider the outcome of 4 and then 5. Let F represent a first card and S a second card. The probability of choosing both cards is or 3. × 10-4. ANSWER: or 3.7 × 10- 6. TRANSPORTATION Isaiah is getting on the bus after work. It costs $0.50 to ride the bus to his house. If he has 3 quarters, 5 dimes, and 2 nickels in his pocket, find the probability that he will randomly pull out two quarters in a row. Assume that the events are equally likely to occur. SOLUTION: The probability of drawing first quarter of 3 quarters out of 10 coins is. The chosen quarter is not replaced before the second draw. Therefore, the total number of coins becomes 9 and the number of quarters is 2. The probability of drawing an another quarter of 2 quarters out of 9 coins is. ANSWER: or 0.
  1. An ace is drawn, without replacement, from a deck of 52 cards. Then, a second ace is drawn. SOLUTION: These events are dependent since an ace does not replace before the second draw. If two events A and B are dependent, then P ( A and B ) = P ( A ) · P ( B | A ). ANSWER: dependent; or 0.5%
  2. In a bag of 3 green and 4 blue marbles, a blue marble is drawn and not replaced. Then, a second blue marble is drawn. SOLUTION: These events are dependent since a blue marble is not replaced before the second draw. If two events A and B are dependent, then P ( A and B ) = P ( A ) · P ( B | A ). ANSWER: dependent; or about 29%
  3. You roll two dot cubes and get a 5 each time. SOLUTION: Since the probability of the fist event does not affect the probability of the second event, these are independent events. If two events A and B are independent, then P ( A and B ) = P ( A ) · P ( B ). ANSWER: independent; or about 3% Describe whether the events are independent. Explain using probability.
  4. A computer program randomly assigns a letter from A to E and a number from 1 to 4 as a user's temporary password. The program assigns C2 as Rebecca's temporary password. SOLUTION: The events are independent. The sample space has 20 equally likely outcomes: {A1, A2, A3, A4, B1, B2, B3, B4, C1, C2, C3, C4, D1, D2, D3, D4, D5, E1, E2, E3, E4}. So P (C2) = , P (C) = , P (2) = , and . ANSWER: The events are independent. The sample space has 20 equally likely outcomes: {A1, A2, A3, A4, B1, B2, B3, B4, C1, C2, C3, C4, D1, D2, D3, D4, D5, E1, E2, E3, E4}. So P (C2) = , P (C) = , P (2) = , and .
  1. A box contains a blueberry muffin, a pumpkin muffin, and a cinnamon muffin. Without looking, Andy takes 2 muffins from the box at the same time. One of Andy's muffins is pumpkin, and the other is cinnamon. SOLUTION: The events are not independent. The sample space has 3 equally likely outcomes: {BP, BC, PC}. So P (PC) = , P (P) = , and P (C) = , but. ANSWER: The events are not independent. The sample space has 3 equally likely outcomes: {BP, BC, PC}. So P (PC) = , P (P) = , and P (C) = , but.

14. In a large city, 70% of the residents are adults, 84%

of the residents have traveled out of state, and 9%

are adults who have not traveled out of state. What

is the probability that a randomly selected resident

has not traveled out of state, given that the resident

is an adult?

SOLUTION:

It is given that 70% or the residents are adults, so this represents the possible outcomes in the sample space. P (not traveled out of state|adult) = ANSWER:

15. WEATHER Tomorrow’s weather forecast calls

for a 25% chance of rain, an 80% chance that the

temperature will exceed 80°F, and a 15% chance of

both. What is the probability of rain, given that the

temperature exceeds 80°F?

SOLUTION:

P (rain|temperature exceeds 80°F) = ANSWER:

16. PROM In Armando’s senior class of 100 students,

91 went to the senior prom. If two people are

chosen at random from the entire class, what is the

probability that at least one of them did not go to

prom?

SOLUTION:

These are two independent events, because the outcome of the first event does not affect the probability of the outcome of the second event. P ( A and B ) = P (A) · P ( B ) = 0.91 · 0.91 = 0. P [not( A and B )] = 1 – P ( A and B ) = 1 – 0.8281 =

So, the probability is about 17.2%. ANSWER: 17.2%

  1. CLASSES The probability that a student takes geometry and French at Satomi’s school is 0.064. The probability that a student takes French is 0.45. What is the probability that a student takes geometry if the student takes French? SOLUTION:

Add the probability that a student takes both

geometry and French (0.064) and the probability

that a student takes French (0.45). That sum is

0.514. This is the denominator.

P (G | F) = or 0.

ANSWER:

CANDY A box of chocolates contains 10 milk chocolates, 8 dark chocolates, and 12 white chocolates. Sophie randomly chooses a chocolate and eats it and then randomly chooses another chocolate. Find each probability.

  1. P (milk and dark) SOLUTION: P (milk and dark) = or about 9% ANSWER: or about 9%
  2. P (dark and white) SOLUTION: P (dark and white) = or about 11% ANSWER: or about 11% 23. P (white and dark) SOLUTION: P (white and dark) = or about 11% ANSWER: or about 11% 24. P (milk and milk) SOLUTION: P (milk and milk) = or about 10% ANSWER: or about 10% 25. SOCKS Damon has 14 white socks, 6 black socks, and 4 blue socks in his drawer. He chooses two socks at random. What is the probability that both socks are white? SOLUTION: P (W and W) = or about 33% ANSWER: or about 33%
  1. PROJECTS Angela, Emery, Rico, and Taylor are working on a class project. Each student is assigned one of the following project roles at random, with no roles repeated: leader, presenter, recorder, and timekeeper. a. Is the assignment of project roles a set of independent events or dependent events? Explain. b. Find the probability that Angela is the presenter, Emery is the timekeeper, Rico is the leader, and Taylor is the recorder by using the multiplication rule for independent events or for dependent events. c. Find the probability in part b by using combinations or permutations. Compare your result to your answer for part b. SOLUTION: a. Dependent; The assignment of one student as leader affects the probabilities that the other students will be assigned to the other roles. b. or about 4% c. There are 4 P 4 = 24 equally likely ways the students could be assigned roles, so each has probability or about 4%; the result is the same using either the multiplication rule for dependent events or permutations. ANSWER: a. Dependent; sample answer: The assignment of one student as leader affects the probabilities that the other students will be assigned to the other roles. b. or about 4% c. There are 4 P 4 = 24 equally likely ways the students could be assigned roles, so each has probability or about 4%; the result is the same using either the multiplication rule for dependent events or permutations.
    1. TENNIS A double fault in tennis is when the serving player fails to land his or her serve "in" without stepping on or over the service line in two chances. Kelly’s first serve percentage is 40%, and her second serve percentage is 70%. a. Draw a probability tree that shows each outcome. b. What is the probability that Kelly will double fault? c. Design a simulation using a random number generator that can be used to estimate the probability that Kelly double-faults on her next serve. SOLUTION: a. b. A double fault is back-to-back faults, or P ( F , F ) = or 18%.

c. See students' work.

ANSWER:

a. b. 0.18 or 18% c. See students' work.

  1. VACATION A random survey was conducted to determine where families vacation. The results indicated that P ( B ) = 0.6, P ( B M ) = 0.2, and the probability that a family did not vacation at either destination is 0.1.
  1. REASONING Consider whether the multiplication rule for dependent events can be used to find the probability of independent events. a. Choose three different pairs of independent events. Find the probability of each pair of events using both multiplication rules. What do you notice? b. Use mathematical reasoning to explain your observation, and make a conjecture about whether the multiplication rule for dependent events can be used to find the probability of independent events. SOLUTION: a. Independent events will vary. Students should notice that both rules give the same result. b. The multiplication rule for dependent events can be used for independent events. When two events A and B are independent, P ( B following A ) = P ( B ) because A has no effect on the probability of B. Substituting P ( B ) for P ( B following A ) in the rule for dependent events results in the rule for independent events, P ( A and B )= P ( A ) · P ( B ). ANSWER: a. Independent events will vary. Students should notice that both rules give the same result. b. The multiplication rule for dependent events can be used for independent events. Sample explanation: When two events A and B are independent, P ( B following A ) = P ( B ) because A has no effect on the probability of B. Substituting P ( B ) for P ( B following A ) in the rule for dependent events results in the rule for independent events, P ( A and B )= P ( A ) · P ( B ).
  2. REASONING Consider whether the rule for finding the probability of two dependent events A and B , P ( A and B ) = P ( A ) · P ( B following A ), can also be written as P ( A and B ) = P ( B ) · P ( B following A ). a. Suppose that a card is randomly selected from a standard deck and that a second card is selected without replacing the first card. Use both rules given above to find the probability that the first card is a 3 and the second card is an 8. How do the results from the two rules compare?

b. Choose two other pairs of dependent events that

do not involve cards. Find the probability of each

pair of events using both rules. What do you notice?

c. Make a conjecture about whether the rules are

equivalent.

SOLUTION:

a. Both rules show that P(3 and 8) = or about

b. Sample answer: A jar contains 5 pennies and 6

dimes. One coin is chosen at random, and then a

second coin is chosen without replacing the first.

Find the probability that the first coin is a penny and

the second coin is a dime.

P ( P and D ) = P ( P ) · P ( D following P ) =

P ( P and D ) = P ( D ) · P ( P following D ) =

c. P ( A and B ) = P ( A ) · P ( B following A ) and P ( A

and B ) = P ( B ) · P ( A following B ) are equivalent.

ANSWER:

a. Both rules show that P(3 and 8) = or about

b. Sample answer: A jar contains 5 pennies and 6

dimes. One coin is chosen at random, and then a

second coin is chosen without replacing the first.

Find the probability that the first coin is a penny and

the second coin is a dime.

P ( P and D ) = P ( P ) · P ( D following P ) =

P ( P and D ) = P ( D ) · P ( P following D ) =

c. P ( A and B ) = P ( A ) · P ( B following A ) and P ( A

and B ) = P ( B ) · P ( A following B ) are equivalent.

  1. SENSE-MAKING In some cases, if one bulb in a string of holiday lights fails to work, the whole string will not light. If each bulb in a set has a 99.5% chance of working, what is the maximum number of lights that can be strung together with at least a 90% chance that the whole string will light? SOLUTION: 21 lights can be strung together. ANSWER: 21
    1. CONSTRUCT ARGUMENTS There are n different objects in a bag. The probability of drawing object A and then object B without replacement is about 1.4%. What is the value of n? Explain. SOLUTION:

9; The probability of drawing object A is , and the

probability of drawing object B when object A is

not replaced is.

Since we know that the probability is about 1.4%,

Solve this equation using the quadratic formula to

determine n is 9.

ANSWER:

9; Sample answer: The probability of drawing object A is , and the probability of drawing object B when object A is not replaced is. Since we know that the probability is 1.4%,. Solve this equation to determine that n is 9.

  1. Francisco reaches into the bag shown here and chooses one of the marbles without looking. He puts the marble aside, and then Nevaeh reaches into the bag and chooses a marble without looking. What is the probability that they both choose a black marble?

A

B

C

D

SOLUTION:

A black marble is selected at random. The chosen marble is not replaced before the second draw. Therefore, the number of black marbles becomes 2 and the total number of marbles in the bag is 8.

P ( B and B ) =

A is the correct answer choice.

ANSWER:

A

  1. A deck contains 10 cards that are numbered 1 through 10. Alex chooses a card at random. What is the probability that the number on Alex's card is 5, given that it is a multiple of 5? A B C D SOLUTION: There are 2 outcomes that are multiples of 5, 5 and
    P(5|multiple of 5) = The correct choice is C. ANSWER: C
  1. A box contains 4 black tiles and x white tiles. Michelle chooses a tile without looking, puts it aside, and then chooses a second tile. Which expression represents the probability that both of the tiles that Michelle chooses are white? A B C D SOLUTION: The events are dependent since Michelle does not replace the first tile she chose from the box. The probability that Michelle chooses a white tile is P (white) =. Since Michelle does not return the tile to the box, both the number of tiles and the number of white tiles decrease by 1. Hence, the probability that she chooses a white tile is P (white|white) = . P (white and white) = P (white) · P (white|white) = The correct choice is D. ANSWER: D
  2. LeBron rolls a number greater than or equal to 4 on a dot cube. What is the probability that LeBron has rolled a 5? SOLUTION: There are 3 outcomes greater than or equal to 4: 4, 5, and 6. The probability that LeBron rolled a 5, given that he rolled a number greater than or equal to 4 =. ANSWER:
    1. Lila has a standard deck of cards. She selects 4 cards at random, one at a time without replacing them. Is the probability that she selects 4 aces greater than or less than 1 out of 1 million? SOLUTION: P (4 aces) = and P (4 aces) is greater than 1 out of 1 million. ANSWER: greater than
    2. At a carnival game, a wading pool contains 50 rubber ducks, and 20 ducks are marked on the bottom with a star. A player selects a duck, does not replace it, and then selects another duck. If both ducks are marked with a star, the player wins a prize. Bella decides to play only if her probability of winning is at least 20%. Should she play the game? Use probability to justify your answer. SOLUTION: No; let S represent a duck with a star. P (S and S) = or or about 16%, which is less than 20%. ANSWER: No; let S represent a duck with a star. P (S and S) = or or about 16%, which is less than 20%.