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Assessment [feedback page]
This is a reviewer version of this assessment. The page is broken into pools of questions.
Each individual pool has been carefully chosen to include questions covering similar topics, with
identical maximum marks and similar levels of difficulty.
This version of the assessment includes both questions that your students will never see outside
of this assessment plus questions that may also be available to your students in their e-workbook
for practice. This is designed to reward students who have been using their e-workbook to study.
In the student version of this assessment, only one question from each pool is randomly selected to be
presented to each student. In this way, your students will not each receive identical assessments yet their
assessments will nevertheless be of equal academic difficulty.
All numeric (and many non-numeric) questions are randomised such that numbers, concepts and answers change
for each student while retaining the same level of difficulty.
You will notice that all questions contain thorough worked feedback to ensure students understand the question
and solution. This page can be printed by students and kept for personal study.
Please speak to your Perdisco account manager to find out how you can use this assessment in your course.
POOL 1
This is the beginning of pool 1.
The questions in this pool may also be available to your students in their e-workbook for practice. This is designed
to reward students who have been using their e-workbook to study.
ID: MST.DPD.HD.02.0030
[3 points]
The following game is run at a carnival:
A jar contains 20 balls of identical size and weight. Three are green and 17 are red. One round of the game consists
of picking three balls from the jar without replacement. The player wins if one or more of the three balls are green.
a) Calculate the probability of a player winning in a single round of the game. Give your answer as a decimal to 2
decimal places.
P(win) = 1
b) Calculate the probability of a player enjoying three wins in three rounds of the game. Give your answer as a
decimal to 2 decimal places.
P(three wins) =
c) Calculate the probability that a player has no wins in three rounds of the game. Give your answer as a decimal
to 2 decimal places.
P(no wins) =
Feedback
[0 out of 3]
a) This is not correct.
P(win) = 0.40
b) This is not correct.
P(three wins) = 0.07
c) This is not correct.
P(no wins) = 0.21
Calculation
a) A player wins a single round if one or more of the three randomly chosen balls are green. So they lose a
round if all three balls are not green. The probability of a win can be calculated as one minus the probability
of a loss: show variables
P(win) = 1 - p(0)
A
=
Cx × N-ACn-x
1-
N
Cn
3
C0 × 17C3
20
C3
=
1-
=
3!
17!
×
0!(3 - 0)!
3!(17 - 3)!
120!
3!(20 - 3)!
=
1-
1 × 680
1,140
= 1 - 0.59649123...
= 0.40350877...
= 0.40
Rounded as last step
b) Since separate rounds will be independent, the probability of three wins can be found as the product of the
probability of three individual wins: show variables
P(three wins)
= (P(win))3
= (0.40350877...)3
= 0.06569903...
= 0.07
Rounded as last step
c) Since separate rounds will be independent, the probability of no wins can be found as the product of the
probability of three individual losses: show variables
P(no wins) = (P(loss))3
= (0.59649123...)3
= 0.21223264...
= 0.21
Rounded as last step
POOL 2
This is the beginning of pool 2.
The questions in this pool may also be available to your students in their e-workbook for practice. This is designed
to reward students who have been using their e-workbook to study.
ID: MST.DPD.HD.02.0040
[1 point]
A school has 110 students, 59 of which are girls and 51 are boys. A group of 10 students are to be selected at
random (without replacement) to participate in an inter-school competition.
Calculate the probability that at least 4 of the students in the group are girls. Give your answer as a decimal to 2
decimal places.
P(at least 4 girls) = 1
Feedback
[0 out of 1]
This is not correct.
P(at least 4 girls) = 0.89
Calculation
Let X be the random variable describing the number of girls in the group selected. Then X has the
hypergeometric distribution.
The probability that at least 4 of the students in the group are girls is then equivalent to the probability that X is
at least 4. This can be found as 1 minus the probability that X is not greater than 4 as follows: show variables
P(X ≥ 4) = P(X = 4 or X = 5 or …)
= 1 - P(X < 4)
= 1 - P(X = 0 or X = 1 or X = 2 or X = 3)
= 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3))
59
=
1-
(
59
59
59
C0 × 51C10
C1 × 51C9
C2 × 51C8
C3 × 51C7
+
+
+
)
110
110
110
110
C10
C10
C10
C10
= 1 - (0.00027246... + 0.00382741... + 0.02323148... + 0.08025421...)
= 0.89241444...
= 0.89
ID: MST.DPD.PD.04.0050
Rounded as last step
[1 point]
Visitor arrivals at a web site are Poisson distributed with a mean of 6 arrivals per second.
Calculate the probability that in a random second there are at most 4 visitor arrivals to the web site. Give your
answer as a decimal to 2 decimal places.
P(at most 4 arrivals) = 1
[0 out of 1]
Feedback
This is not correct.
P(at most 4 arrivals) = 0.29
Calculation
Let X be the random variable that is the number of visitor arrivals to the web site in a second. The X is Poisson
distributed with mean 6. Then the probability that in a random second there are at most 4 is the same as the
probability that X is at most 4. That is, that X is less than or equal to 4.
The probability that X takes a value less than or equal to 4 can be calculated using the following formula: show
variables
P(X ≤ 4) = P(X = 0 or X = 1 or X = 2 or X = 3 or X = 4)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
=
(6)0e-6
(6)1e-6
(6)2e-6
(6)3e-6
(6)4e-6
+
+
+
+
0!
1!
2!
3!
4!
=
1 × 0.00247875...
6 × 0.00247875...
36 × 0.00247875...
+
+
1
1
2
+
216 × 0.00247875...
1,296 × 0.00247875...
+
6
24
= 0.00247875... + 0.01487251... + 0.04461754... + 0.08923508... + 0.13385262...
= 0.2850565...
= 0.29
Rounded as last step
POOL 3
This is the beginning of pool 3.
The questions in this pool may also be available to your students in their e-workbook for practice. This is designed
to reward students who have been using their e-workbook to study.
ID: MST.DPD.DRV.02.0030
[12 points]
Geoff is running a carnival game. He has 13 marbles in a bag: there are 3 green marbles, 7 red marbles and 3
yellow marbles. To play a round of the game, a player randomly takes out 2 marbles (without replacement) from
the bag. Green marbles win 5 points, red marbles win 1 point and yellow marbles lose 2 points.
Let X be the random variable that describes the number of points won by a player playing a single round of Geoff's
marble game. Find the probability distribution for X. Give values for X as whole numbers and probabilities as decimal
values to 3 decimal places. Enter the values for X in ascending order (lowest to highest) from left to right in the
table.
1
X
P(X = x)
Feedback
[0 out of 12]
This is not correct.
The probability distribution for X is:
X
-4
-1
2
3
6
10
P(X = x)
0.038
0.269
0.269
0.115
0.269
0.038
Calculation
In order to find the probability distribution for your you need to find the possible values that X can take and then
find the probability that X takes each value.
There are three colors of marble that can be chosen and a player chooses two marbles, and there are 3 C2 = 3
possible ways that this can be done. Therefore, there are 3 possible values that X can take. The value for each
choice is determined by the given points values for each marble type: green marbles win 5 points, red marbles
win 1 point and yellow marbles lose 2 points. So the values for X then are:
Player
picks
2 yellow
1 red and 1
yellow
2 red
1 green and 1
yellow
1 green and
1 red
2 green
X=
points
value
2 × -2 = -4
1 + -2 = -1
2×1=2
5 + -2 = 3
5+1=6
2 × 5 = 10
The probability for each value of X is then found as the number of possible combinations that realize that
particular selection of marbles divided by the total number of possible combinations which is equal to
13
C2 = 78. There are 3 green marbles, 7 red marbles and 3 yellow marbles. The combinations and probabilities
are:
Player picks
Number of
combinations
P(X = x)
1 red and 1
yellow
2 yellow
3
C2 = 3
3 ≈
78 0.038
C1×3C1 = 21
1 green and
1 yellow
2 red
C1×3C1 = 9
1 green and
1 red
7
7
3
3
21
≈ 0.269
78
21 ≈
78 0.269
9
≈ 0.115
78
21
≈ 0.269
78
C2 = 21
C1×7C1 = 21
2 green
3
C2 = 3
3 ≈
78 0.038
POOL 4
This is the beginning of pool 4.
The questions in this pool may also be available to your students in their e-workbook for practice. This is designed
to reward students who have been using their e-workbook to study.
ID: MST.DPD.HD.01.0040
[5 points]
Select the distribution that would apply to each situation:
Binomial
a)
The 200 senior students of Eschire High School have
taken part in an anonymous survey, and it is found
that 74 of the students have already chosen a future
career path. Twenty senior students are selected at
random to speak at an upcoming presentation night.
The principal calculates the probability that at least 10
of these students have already chosen a future career
path.
b)
The quality control officer of the Paper Piper paper
company knows that 1% of the sheets of paper
produced are faulty in some way and must be rejected.
A stack of 200 sheets of paper have been bound
without having been checked. The officer calculates the
probability that this stack contains no faults.
c)
If a piece of paper from the Paper Piper paper company
is in fact faulty, there is an average of 0.4 visible faults
per 10 square inches of paper (and faults occur
independently of one another). The quality control
officer calculates the probability of finding at least 3
visible faults on an area of 10 square inches on a
faulty piece of paper
d)
A multiple choice test with 50 questions has just been
returned to Wendy. Wendy received 39 out of 50. The
teacher randomly chooses 5 questions to cover in front
of the whole class. Wendy calculates the probability
Poisson
Hypergeometric
that at most 2 of the questions covered are ones that
she had correct.
e)
Jack did the same test, but did not study for it and
knew nothing about the subject at all. Each multiple
choice question had 4 options (A, B, C, D) and Jack's
answer was selected purely at random for each. Jack
gets Wendy to calculate the probability that he will
receive at least 25 out of 50. Wendy informs Jack that
the prospects do not look good.
Feedback
[0 out of 5]
a) This is not correct.
In this case, the distribution that would apply is the hypergeometric distribution.
b) This is not correct.
In this case, the distribution that would apply is the binomial distribution.
c) This is not correct.
In this case, the distribution that would apply is the Poisson distribution.
d) This is not correct.
In this case, the distribution that would apply is the hypergeometric distribution.
e) This is not correct.
In this case, the distribution that would apply is the binomial distribution.
Discussion
The three distributions in this question apply to situations that appear similar and yet are fundamentally
different.
The binomial distribution
The binomial distribution applies to situations where a trial (which may result in a 'success' or 'failure') is
repeated a fixed number of times, and the number of successes is recorded. The binomial distribution is in fact
used to calculate the probability that a certain number of successes will occur when the probability of success
per trial is known. A key assumption for the binomial distribution is that these trials are independent: the
success or failure of one trial is independent of the success or failure of another trial. Because of this, the
binomial distribution may be used in situations where items are drawn and replaced from a finite population of
'successes' and 'failures', and the number of successes is recorded.
The Poisson distribution
The Poisson distribution applies to situations where the number of times that a particular event occurs in a
defined area of interest is observed. The Poisson distribution is used to calculate the probability that the event is
observed a certain number of times given that the average number of observations in an area is known.
The hypergeometric distribution
The hypergeometric distribution applies to situations where a fixed number of items are drawn (and not
replaced) from a finite population of 'successes' and 'failures', and the number of successes is recorded. The
hypergeometric distribution is in fact used to calculate the probability that a certain number of successes will be
drawn when the total number of successes is known. For this reason, the hypergeometric and binomial
distributions can be confused. They key difference is that the hypergeometric distribution accounts for the fact
that previous successes and failures will affect future successes and failures. Drawing a successful item from the
population will decrease the number of successful items in the population and therefore decrease the probability
that later items will be successes. Therefore if items are drawn from a finite population without replacement,
the hypergeometric distribution applies.
a) There is a finite population of students from which 74 have chosen their career. Drawing 20 of these
students, which involves drawing 20 items from the population without replacement, and calculating the
probability that at least 10 have already chosen their careers would require the hypergeometric
distribution.
b) This experiment involves repeating a trial, where the probability of 'success' is known. In particular, there
are 200 sheets of paper, and the probability of a piece of paper being faulty is 1%. Therefore this situation
calls for using the binomial distribution. Note that even though there are a finite number (200) of sheets
of paper, the hypergeometric distribution is not used. This is a subtlety that is worth noting: the 200 sheets
of paper do not form the population that is drawn from. This is because '1% of the sheets of paper produced
are faulty' is interpreted as 1% of all sheets of paper are faulty. Hence the probability of each sheet being
faulty is independent of whether or not any other sheets are faulty.
c) In this question the average number of times an event (visible fault) occurs in a certain area (10 inches
squared) is given, and the probability of the event occurring at least 3 times in the area is calculated.
Therefore the Poisson distribution would be used.
d) There is a finite population of 50 questions from which Wendy got 39 correct. A random sample of 5 have
been chosen from the 50, and Wendy calculates the probability that this sample contains at most 2 of the
39 that she got correct. Because the sample of 5 are chosen from the population of 50 without replacement,
the hypergeometric distribution is used.
e) In comparison to part d), the binomial distribution is used for this situation. This is because the 50 questions
are now 50 repeated trials where the probability of success is 0.25. Jack wants to calculate the probability
that at least 25 successes occur in these 50 trials. The success on any one trial is independent of success on
any other trial. Therefore the binomial distribution is used.
POOL 5
This is the beginning of pool 5.
The questions in this pool may also be available to your students in their e-workbook for practice. This is designed
to reward students who have been using their e-workbook to study.
ID: MST.DPD.HD.03.0030
[1 point]
A new brand of marbles has been produced by the We R Toys toy company. In this brand there is a 'Super Silver'
marble, which only comes in some marble bags and is being marketed as the best marble. An inspector wishes to
know how prevalent these silver marbles are. At the factory there is a vat of 12,746 marbles. After doing some
tests, the inspector concludes that if you draw 100 marbles at random from the vat, the expected number of silver
marbles drawn would be 14. Assuming this information is correct, calculate the number of silver marbles in the vat.
Give your answer to the nearest whole number.
Number of silver marbles = 1
Feedback
This is not correct.
Number of silver marbles = 1,784
[0 out of 1]
Calculation
The number of silver marbles can be calculated using the following formula: show variables
E(X) =
A
nA
N
=
N × E(X)
n
=
12,746 × 14
100
= 1,784.44
= 1,784
Rounded as last step
POOL 6
This is the beginning of pool 6.
The questions in this pool may also be available to your students in their e-workbook for practice. This is designed
to reward students who have been using their e-workbook to study.
[2 points]
ID: MST.DPD.PD.05.0010
The confectionery company Chocoholly make chocolate chips cookies as part of their
production line. Chocolate chips in the cookies are randomly and independently
distributed according to a Poisson distribution with an average of 12 chocolate chips per
cookie.
a) Calculate the probability that a cookie selected at random contains exactly 10
chocolate chips. Give you answer as a decimal to 3 decimal places.
P(a random cookie has exactly 10 chocolate chips) = 1
b) Calculate the probability that in 17 randomly selected cookies at least 3 have exactly
10 chocolate chips in them. Give you answer as a decimal to 3 decimal places.
P(in 17 randomly selected cookies at least 3 have exactly 10 chocolate chips) =
Feedback
[0 out of 2]
a) This is not correct.
P(a random cookie has exactly 10 chocolate chips) = 0.105
b) This is not correct.
P(in 17 randomly selected cookies at least 3 have exactly 10 chocolate chips) = 0.261
Calculation
a) The number of chocolate chips in the cookies are distributed according to a Poisson distribution with an
average of 12 chocolate chips per cookie. The probability that a cookie selected at random contains exactly
10 chocolate chips can be calculated using the following formula: show variables
P(X = x)
=
λxe-λ
x!
P(X = 10)
=
(12)10e-12
10!
=
61,917,364,224 × 0.00000614...
3,628,800
= 0.10483726...
= 0.105
Rounded as last step
b) In a randomly selected batch of 17 cookies, the number that contain exactly 10 chocolate chips is binomially
distributed with probability of success in a single trial given by the value found in part a).
Therefore, the probability that in 17 randomly selected cookies at least 3 have exactly 10 chocolate chips in
them can be calculated using the following formula: show variables
P(Y ≥ 3) = 1 - P(Y < 3)
= 1 - P(Y = 0 or Y = 1 or Y = 2)
= 1 - (P(Y = 0) + P(Y = 1) + P(Y = 2))
Addition rule for mutually exclusive events
= 1 - (nC0p0(1 - p)(n-0) + nC1p1(1 - p)(n-1) + nC2p2(1 - p)(n-2))
=
1-
(17C0(0.10483726...)0(1 - 0.10483726...)(17-0) + 17C1(0.10483726...)1(1 - 0.10483726.
+ 17C2(0.10483726...)2(1 - 0.10483726...)(17-2))
= 1 - (0.15217173... + 0.30296785... + 0.28385737...)
= 0.26100304...
= 0.261
Rounded as last step
Alternatively, you could use software to arrive at the same answer. You would use software to compute
P(Y < 3). Be aware though that most software programs would give the cumulative distribution function
value P(Y ≤ 3). However, you can use the fact that P(Y < 3) = P(Y ≤ 3-1) = P(Y ≤ 2) to find the required
probability.
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