Problems Probability Models Worksheets Page 6
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11c) Answers will vary. To construct your simulated probability model, start by calculating the simulated probability that you get no
pictures of Tiger Woods in the five boxes. This is the number of trials in which neither 0 nor 1 were generated divided by the total
number of trials. Perform similar calculations for the simulated probability that you would get one picture, 2 pictures, etc.
d) Let X = the number of Tiger Woods pictures in 5 boxes.
X
0
1
2
3
4
5
5
5
5
5
5
5
0
5
1
4
2
3
3
2
4
1
5
0
0 2
0 8
0 2
0 8
0 2
0 8
0 2
0 8
0 2
0 8
0 2
0 8
.
.
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P(X)
0
1
2
3
4
5
0 33
0 41
0 20
0 05
0 01
0 0001
.
.
.
.
.
.
e) Answers will vary.
6. a) Answers will vary. A component is the simulation of one driver in a car. One possible way to model this component is to
generate pairs of random digits 00-99. Let 01-75 represent a driver wearing his or her seatbelt and let 76-99 and 00 represent a
driver not wearing his or her seatbelt. Each run will consist of generating five pairs of random digits. The response variable will
be the number of pairs of digits that are 00-75.
b) Answers will vary.
c) Answers will vary. To construct your simulated probability model, start by calculating the simulated probability that none of
the five drivers are wearing seatbelts. This is the number of trials in which no pairs of digits were 00-75, divided by the total
number of trials. Perform similar calculations for the simulated probability that one driver is wearing his or her seatbelt, two
drivers, etc.
d) Let X = the number of drivers wearing seatbelts in 5 cars.
X
0
1
2
3
4
5
5
5
5
5
5
5
0
5
1
4
2
3
3
2
4
1
5
0
0 75
0 25
0 75
0 25
0 75
0 25
0 75
0 25
0 75
0 25
0 75
0 25
.
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P(X)
0
1
2
3
4
5
0 0001
0 01
0 09
0 26
0 40
0 24
.
.
.
.
.
.
e) Answers will vary.
7. The player’s shots may be considered Bernoulli trials. There are only two possible outcomes (make or miss), the probability of
making a shot is constant (80%), and the shots are independent of one another (making, or missing, a shot does not affect the
probability of making the next).
Let X = the number of shots until the first missed shot.
Let Y = the number of shots until the first made shot.
Since these problems deal with shooting until the first miss (or until the first made shot), a geometric model, either Geom (0.8) or
Geom (0.2), is ap0propriate.
4
a) Use Geom (0.2). P ( X = 5) =
0 8
0 2
0 08192
(Four shots made, followed by a miss.)
.
.
.
3
b) Use Geom (0.8). P ( Y = 4) =
0 2
0 8
0 0064
(Three misses, then a made shot.)
.
.
.
2
c) Use Geom (0.8). P ( Y = 1) + P ( Y = 2) + P ( Y = 3) =
0 8
0 2 0 8
0 2
0 8
0 992
.
.
.
.
.
.
8. The selection of chips may be considered Bernoulli trials. There are only two possible outcomes (fail testing and pass testing).
Provided that the chips selected are a representative sample of all chips, the probability that a chip fails testing is constant at 2%.
The trials are not independent, since the population of chips is finite, but we won’t need to sample more than 10% of all chips.
Let X = the number of chips required until the first bad chip. The appropriate model is Geom(0.0 2).
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a) P(X = 5) =
0 98
0 02
0 0184
(Four good chips, then a bad one.)
.
.
.
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b) P(1 ≤ X ≤ 10) =
0 02
0 98 0 02
0 98
0 02
0 98
0 02
0 183
.
.
.
.
.
...
.
.
.
(Use the geometric model on a calculator or computer for this one!)
9. As determined in Exercise 3, the shots can be considered Bernoulli trials, and since the player is shooting until the first miss,
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1
Geom(0.2) is the appropriate model.
5
shots. The player is expected to take 5 shots until the first miss.
E x
0 2
p
.
10. As determined in Exercise 4, the selection of chips can be considered Bernoulli trials, and since the company is selecting until the
1
1
first bad chip, Geom(0.0 2) is the appropriate model.
50
chips. The first bad chip is expected to be the 50th
E x
0 02
p
.
chip selected.
11. These may be considered Bernoulli trials. There are only two possible outcomes, Type AB and not Type AB. Provided that the
donors are representative of the population, the probability of having Type AB blood is constant at 4%. The trials are not
independent, since the population is finite, but we are selecting fewer than 10% of all potential donors. Since we are selecting
people until the first success, the model Geom(0.0 4) may be used. Let X = the number of donors until the first Type AB donor is
found.
1
1
a)
people. We expect the 25th person to be the first Type AB donor.
25
E x
0 04
p
.
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