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11Answers
1. a) These are not Bernoulli trials. The possible outcomes are 1, 2, 3, 4, 5, and 6. There are more than two possible outcomes.
b) These may be considered Bernoulli trials. There are only two possible outcomes, Type A and not Type A. Assuming the 120
donors are representative of the population, the probability of having Type A blood is 43%. The trials are not independent,
because the population is finite, but the 120 donors represent less than 10% of all possible donors.
c) These are not Bernoulli trials. The probability of getting a heart changes as cards are dealt without replacement.
d) These are not Bernoulli trials. We are sampling without replacement, so the trials are not independent. Samples without
replacement may be considered Bernoulli trials if the sample size is less than 10% of the population, but 500 is more than 10% of
3000.
e) These may be considered Bernoulli trials. There are only two possible outcomes, sealed properly and not sealed properly. The
probability that a package is unsealed is constant, at about 10%, as long as the packages checked are a representative sample of all
packages. Finally, the trials are not independent, since the total number of packages is finite, but the 24 packages checked
probably represent less than 10% of the packages.
2. a) These may be considered Bernoulli trials. There are only two possible outcomes, getting a 6 and not getting a 6. The
probability of getting a 6 is constant at 1/6. The rolls are independent of one another, since the outcome of one die roll doesn’t
affect the other rolls.
b) These are not Bernoulli trials. There are more than two possible outcomes for eye color.
c) These can be considered Bernoulli trials. There are only two possible outcomes, properly attached buttons and improperly
attached buttons. As long as the button problem occurs randomly, the probability of a doll having improperly attached buttons is
constant at about 3%. The trails are not independent, since the total number of dolls is finite, but 37 dolls is probably less than
10% of all dolls.
d) These are not Bernoulli trials. The trials are not independent, since the probability of picking a council member with a
particular political affiliation changes depending on who has already been picked. The 10% condition is not met, since the sample
of size 4 is more than 10% of the population of 19 people.
e) These may be considered Bernoulli trials. There are only two possible outcomes, cheating and not cheating. Assuming that
cheating patterns in this school are similar to the patterns in the nation, the probability that a student has cheated is constant, at
74%. The trials are not independent, since the population of all students is finite, but 481 is less than 10% of all students.
3. a) Answers will vary. A component is the simulation of the picture in one box of cereal. One possible way to model this
component is to generate random digits 0-9. Let 0 and 1 represent Tiger Woods and 2-9 a picture of another sports star. Each run
will consist of generating random numbers until a 0 or 1 is generated. The response variable will be the number of digits
generated until the first 0 or 1.
b) Answers will vary.
c) Answers will vary. To construct your simulated probability model, start by calculating the simulated probability that you get a
picture of Tiger Woods in the first box. This is the number of trials in which a 0 or 1 was generated first, divided by the total
number of trials. Perform similar calculations for the simulated probability that you have to wait until the second box, the third
box, etc.
d) Let X = the number of boxes opened until the first Tiger Woods picture is found.
9
X
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8
2
3
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7
0 8 0 2
0 8
0 2
0 8
0 2
0 8
0 2
0 8
0 2
0 8
0 2
0 8
0 2
.
.
.
.
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.
.
.
.
.
P(X)
0.20
0.168
0 16
0 128
0 1024
0 082
0 066
0 052
0 042
.
.
.
.
.
.
.
e) Answers will vary.
4. a) Answers will vary. A component is the simulation of one die roll. One possible way to model this component is to generate
random digits 1-6. Let 1 represent getting 1 (the roll you need and let 2-6 represent not getting the roll you need. Each run will
consist of generating random numbers until 1 is generated. The response variable will be the number of digits generated until the
first 1.
b) Answers will vary.
c) Answers will vary. To construct your simulated probability model, start by calculating the simulated probability that you roll a
1 on the first roll. This is the number of trials in which a 1 was generated first divided by the total number of trials. Perform
similar calculations for the simulated probability that you have to wait until the second roll, the third roll, etc.
d) Let X = the number of rolls until the first 1 is rolled.
9
X
1
2
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8
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1
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1
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1
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1
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1
5
1
5
1
1
P(X)
0.233
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6
6
6
6
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6
6
6
0 139
0 116
0 096
0 80
0 067
0 056
0 047
.
.
.
.
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.
.
e) Answers will vary.
5. a) Answers will vary. A component is the simulation of the picture in one box of cereal. One possible way to model this
component is to generate random digits 0-9. Let 0 and 1 represent Tiger Woods and 2-9 a picture of another sports star. Each run
will consist of generating five random numbers. The response variable will be the number of 0s and 1s in the five random
numbers.
b) Answers will vary.
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