Probability Distribution Sample Test
Multiple Choice
Identify the letter of the choice that best completes the statement or answers the question.
____
1. A student is preparing a probability distribution as shown below.
X
0
1
2
3
4
P(X)
0.3
0.3
0.3
0.3
A value is needed for P(3) to complete the table. Which statement below is true?
a. The required value for P(3) is 0.3.
b. The required value for P(3) is 0.2.
c. The required value for P(3) is –0.2.
d. There is no possible value for P(3) that can make this a valid probability distribution.
____
____
____
____
2. According to Pascal’s Identity, the single expression of the form
a.
c.
b.
d.
, which is equivalent to
3. According to Pascal’s Identity, what two expressions may be added to obtain
a.
c.
b.
d.
4. What is the coefficient of the third term in the expansion of
?
a. 10
c. 150
b. 15
d. 270
5. Which of the following is a correct expression of Pascal’s Identity?
a.
c.
?
?
b.
____
d.
6. Which expression describes the probability of k “3s” being rolled on 20 successive rolls of a six-sided die?
a.
c.
b.
d.
Short Answer
7. A soft drink company has a promotion. Each cap liner for 1 000 000 bottles is printed with either a prize or a
cheerful invitation to buy more product. The value of each liner along with its frequency is shown in the table.
Value of Liner
0
1
5
100
10 000
Number of Liners
949 779
50 000
200
20
1
What is the expected value for an individual liner?
8. Use Pascal’s Identity to write an expression equivalent to
9. Write the terms in the unsimplified expansion of
11. The expansion of
contains a term of the form
.
.
. What is the value of k?
12. A baseball player has an on base percentage of 0.425. What is the probability that he will reach base every
time in a game where he comes to the plate 5 times?
13. A fair die has four faces numbered one to four. What is the probability of rolling a two exactly three times in
ten rolls of the die?
14. What is the mean value for a binomial experiment with 45 repetitions and a probability of success of 0.25?
15. Calculate the standard deviation for the normal approximation to the binomial distribution for an experiment
with 300 trials and a probability of success of 0.27.
16. A student guesses at all 10 questions on a true/false test. What is the probability that he will get exactly 5
answers correct?
Problem
17. Two four-sided dice are rolled and the sum observed is recorded.
a) List the possible outcomes for the two dice and the resulting sums.
b) Create a table showing the probability distribution of the random variable representing the sum observed.
c) Calculate the expected value for the sum observed.
19. A game involves simultaneously tossing a 50 cent coin, a one dollar coin, and a two dollar coin. If two or
more heads show, you win all the coins that are showing heads.
a) What is the expected value of the random variable representing the amount of money won on a given
toss?
b) To the nearest cent, what price per play would make this a fair game?
20. An inventory of nine CD-RW drives is known to contain two defective drives. An order of three drives is
filled at random from the inventory.
a) X is a random variable representing the number of defective drives in the order. Calculate the probability
for all possible values of X and determine the expected value of X.
b) Describe a simulation that would use the numbered cards from one suit in a deck of playing cards to
model this situation and determine an experimental probability for the number of defective drives in the
order.
21. Apply the Binomial Theorem to expand (3a – 5b)6.
22. Find the 7th term in the expansion of
.
26. A student calculates the probability that exactly 8 parts out of a sample of 100 will be defective given that
11% of the total production is known to be defective. The next question he encounters asks what the
probability is for 4 parts out of a sample of 50 being defective. Having a sound grasp of the concept of
equivalent fractions, the student copies her answer to the first question and moves on. Calculate both
probabilities and comment on why the two situations are not equivalent.
27. A baseball player has a batting average of .315. What minimum number of hits can he be 78% certain of
getting in his next eight at bats? Solve using the binomial distribution rather than a normal approximation.
28. Calculate the probability that there will be between 7 and 12 defective parts in a production run of 50 parts if
the production process is known to have a defect rate of 16%.
29. A soft drink company labels its cans as containing 350 mL of beverage. Their quality control department has
determined that 90% of the cans contain at least this volume. A consumer watchdog group samples their
production and issues a public warning if more than 12% of a sample of 50 cans is below the stated volume.
What is the probability that a warning will be issued?
30. State the coefficient of the term containing x–11 in the expansion of (4x2 + x–3)7.
31. A soccer linesman will make the correct call for a possible offside pass 90% of the time. What is the
probability that he will make 2 or fewer incorrect calls in a game in which he sees 32 passes?
32. A coin is tossed 30 times. Calculate the probability that either 17 or 18 heads will be observed and then find
the value provided by the normal approximation for the same interval. What is the percentage error resulting
from the approximation?
33. A car salesman has a long-term average of selling a car to 7% of the potential customers he meets. What is the
probability he will sell at least 5 cars in a given week if he meets 80 potential customers?
34. Rudoph’s Electronics makes sets of Christmas lights. On average, 4% of the bulbs produced by their supplier
will be defective. Rudolph samples 300 bulbs from each shipment. If more than 3% of the bulbs are defective,
he refuses the shipment. What is the probability that a given shipment will be accepted?
Sample Test # 5
Answer Section
MULTIPLE CHOICE
1. ANS:
OBJ:
TOP:
2. ANS:
OBJ:
TOP:
3. ANS:
OBJ:
TOP:
4. ANS:
OBJ:
TOP:
5. ANS:
OBJ:
TOP:
6. ANS:
LOC:
D
REF: Knowledge and Understanding
5.1 Probability Distributions and Expected Value
LOC:
Probability Distributions and Predictions
C
REF: Knowledge and Understanding
5.2 Pascal's Triangle and the Binomial Theorem
LOC:
Probability Distributions and Predictions
B
REF: Knowledge and Understanding
5.2 Pascal's Triangle and the Binomial Theorem
LOC:
Probability Distributions and Predictions
D
REF: Knowledge and Understanding
5.2 Pascal's Triangle and the Binomial Theorem
LOC:
Probability Distributions and Predictions
C
REF: Knowledge and Understanding
5.2 Pascal's Triangle and the Binomial Theorem
LOC:
Probability Distributions and Predictions
A
REF: Knowledge and Understanding
OBJ:
CP2.05
TOP: Probability Distributions and Predictions
CP2.06
CP1.07
CP1.07
CP1.07
CP1.07
5.3 Binomial Distributions
SHORT ANSWER
7. ANS:
The expected value of a liner is 6.3 cents.
REF: Application
LOC: CP2.04
8. ANS:
OBJ: 5.1 Probability Distributions and Expected Value
TOP: Probability Distributions and Predictions
REF: Knowledge and Understanding
OBJ: 5.2 Pascal's Triangle and the Binomial Theorem
LOC: CP1.07
TOP: Probability Distributions and Predictions
9. ANS:
REF: Knowledge and Understanding
OBJ: 5.2 Pascal's Triangle and the Binomial Theorem
LOC: CP1.07
TOP: Probability Distributions and Predictions
11. ANS:
k = 12
REF: Knowledge and Understanding
OBJ: 5.2 Pascal's Triangle and the Binomial Theorem
LOC: CP1.07
TOP: Probability Distributions and Predictions
12. ANS:
0.0139
REF: Application OBJ: 5.3 Binomial Distributions
TOP: Probability Distributions and Predictions
13. ANS:
0.2503
LOC: CP2.05, CP2.06
REF: Knowledge and Understanding
OBJ: 5.3 Binomial Distributions
LOC: CP2.05
TOP: Probability Distributions and Predictions
14. ANS:
11.25
REF: Knowledge and Understanding
LOC: CP2.04, CP2.05
15. ANS:
7.69
OBJ: 5.4 Normal Approximation of the Binomial Distribution
TOP: Probability Distributions and Predictions
REF: Knowledge and Understanding
OBJ: 5.4 Normal Approximation of the Binomial Distribution
LOC: STV.03
TOP: Probability Distributions and Predictions
16. ANS:
The probability is 0.2461 or almost 25%.
REF: Knowledge and Understanding
OBJ: 5.4 Normal Approximation of the Binomial Distribution
LOC: CP2.05
TOP: Probability Distributions and Predictions
PROBLEM
17. ANS:
a) There are sixteen possible outcomes.
1+1=2
2+1=3
3+1=4
1+2=3
2+2=4
3+2=5
1+3=4
2+3=5
3+3=6
1+4=5
2+4=6
3+4=7
b)
X
2
3
4
5
P(X)
c) E(X) = 2
+3
+4
+5
REF: Knowledge and Understanding
LOC: CPV.01, CP2.03
19. ANS:
a)
4+1=5
4+2=6
4+3=7
4+4=8
6
+6
7
+7
8
+8
=5
OBJ: Chapter 5 Prob
TOP: Probability Distributions and Predictions
0.50
H
H
H
H
T
T
T
T
1.00
H
H
T
T
H
H
T
T
2.00
H
T
H
T
H
T
H
T
Prize
3.50
1.50
2.50
0.00
3.00
0.00
0.00
0.00
All eight possible outcomes are equally likely. The total winnings for all 8 outcomes is $10.50. The
expected value is $1.3125 per toss.
b) A fair game pays out as much as it takes in. The price per play should be the same as the expected value,
which is (to the nearest cent) $1.31 per play.
REF: Application OBJ: Chapter 5 Prob
TOP: Probability Distributions and Predictions
20. ANS:
LOC: CP2.04
a)
E(X) = 0
+1
+2
=
This is intuitively satisfying. Three drives is one-third of the inventory. We would expect to get one-third
of the defective drives as a long-term average.
b) If we take the nine numbered cards from one suit, we could let two cards (say the 5 and 10) represent the
defective drives. We could repeatedly draw three cards at random from the nine cards available and
record the number of times the 5 and/or 10 were observed in the cards drawn. For ease of comparison, the
simulation could be repeated 84 times and the frequencies for each possible outcome compared to the
numerators of the theoretical probabilities.
REF: Communication
LOC: CP2.03, CP2.04, CPV.03
21. ANS:
(3a – 5b)6 =
OBJ: Chapter 5 Prob
TOP: Probability Distributions and Predictions
REF: Knowledge and Understanding
OBJ: Chapter 5 Prob
LOC: CP1.07
TOP: Probability Distributions and Predictions
22. ANS:
The 7th term is
.
REF: Knowledge and Understanding
OBJ: Chapter 5 Prob
LOC: CP1.07
TOP: Probability Distributions and Predictions
26. ANS:
We can see that the probability of the second question is almost double that of the first. In the first example,
we have an expected value of 11 and are requesting the probability that we deviate from the expected value by
27% over 100 trials. In the second example, we have an expected value of 5.5. We are still requesting the
probability of deviating by 27% but it is for a sample that is twice as small. In general, we expect that the
more trials we undertake, the closer our outcomes will be to the expected value.
REF: Application OBJ: Chapter 5 Prob
LOC: CP2.05
TOP: Probability Distributions and Predictions
27. ANS:
We will consider the probabilities for small values of the random variable and stop when they total 22% or
somewhere close to it.
The combined probabilities for no hits or one hit is 0.2265. The player has a 77.35% likelihood of getting 2 or
more hits in his next 8 at bats.
REF: Application OBJ: Chapter 5 Prob
LOC: CP2.05
TOP: Probability Distributions and Predictions
28. ANS:
Converting from a discrete to a continuous variable, we are looking for the probability that 6.5 < X < 12.5 for
a binomial experiment with p = 0.16 and n = 50.
= np = 8 so we may use the normal approximation to the binomial distribution.
=
=
=
We need to find the z-scores for the left and right boundaries of the area required.
and z =
From tables, we obtain the area to the left of z = 1.736 is 0.9588 and the area to the left of z = –0.579 is
0.2814. The area between the two values is 0.6774.
The probability of between 7 and 12 defective parts is about 68%.
REF: Application OBJ: Chapter 5 Prob
LOC: STV.03
TOP: Probability Distributions and Predictions
29. ANS:
We are investigating the probability that 88% of a sample of 50 cans will be at or above the required volume,
given that 90% of production is at or above the required volume. Each can in the sample is part of a binomial
experiment for which n = 50 and p = 0.90. We have np = 45 and nq = 5, so the normal approximation may be
applied.
= np = 45 and  =
We require that at least 88% of 50 cans or 44 cans contain the required volume.
To convert to a continuous variable, we require P(X 43.5).
= –0.707
From tables, the area to the left of this value is about 0.24. There is about a 24% probability that a warning
would be issued.
REF: Application OBJ: Chapter 5 Prob
TOP: Probability Distributions and Predictions
30. ANS:
LOC: ST3.03
The general term for the expansion is
=
.
For the required term, 14 – 5r = –11, which provides r = 5 (the sixth term in the expansion).
The value of the coefficient is
.
REF: Knowledge and Understanding
OBJ: 5.2 Pascal's Triangle and the Binomial Theorem
LOC: CP1.07
TOP: Probability Distributions and Predictions
31. ANS:
We will simply evaluate the probabilities for 0, 1, and 2 incorrect calls and find their sum.
The total probability of 2 or fewer incorrect calls is 0.3667 or 36.7%.
REF: Application OBJ: 5.3 Binomial Distributions
TOP: Probability Distributions and Predictions
32. ANS:
LOC: CP2.05
Direct method: P(X = 17) =
= 0.111535 and P(X = 18) = 0.080553, for a total probability
of 0.192088.
Normal approximation P(16.5 X 18.5)
z=
= 0.5477 and z =
= 1.2780
From tables, P(0.5477  z 1.2780) = 0.8994 0.7080 = 0.1914 (linear interpolation applied).
The difference between solutions is 6 units out of 1920 for a 0.3% error.
REF: Thinking/Inquiry/PS
OBJ: 5.4 Normal Approximation of the Binomial Distribution
LOC: STV.03, CP2.05
TOP: Probability Distributions and Predictions
33. ANS:
We squeeze into the normal approximation with a mean value of 5.6. If we are interested in the total
probability of selling 5 or more cars, we will consider the probability of selling 4 cars or fewer. Making the
adjustment from discrete to continuous variables, we need P(X 4.5).
= –0.482
z=
From tables, this corresponds to an area of 0.3149 (by interpolation). The probability of selling less than 5 is
about 31.5%, so the probability of selling 5 or more is about 68.5%.
REF: Application OBJ: 5.4 Normal Approximation of the Binomial Distribution
LOC: STV.03
TOP: Probability Distributions and Predictions
34. ANS:
We have a binomial distribution with 300 trials and a probability of success of 4%. We may apply the normal
approximation since np = 12 and n(1 – p)  np. Rudoph will accept up to 3% of the sample as defective, so
we need to find the probability that X 9. In order to convert to a continuous variable, we will find the z-score
for X 9.5.
z=
=
= –0.737
From tables, this corresponds to about a 23% chance that Rudolph will accept the shipment.
REF: Application
LOC: STV.03
OBJ: 5.4 Normal Approximation of the Binomial Distribution
TOP: Probability Distributions and Predictions