Tutorial Quiz 1_Version A_Solutions
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School
University of Waterloo *
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Course
135
Subject
Mathematics
Date
Jan 9, 2024
Type
Pages
9
Uploaded by MasterPorpoiseMaster881
Instructions Marking Scheme: 1.
All electronic devices are to be turned off and put away. 2.
This is a closed-book quiz. 3.
Answer the questions in the spaces provided. 4.
Final answers to written questions should be rounded to THREE DECIMAL PLACES, or left in fraction form, where appropriate 5.
Only question pages will be marked. 6.
You may tear off the last page and use it for rough work. All pages will be collected at the end of the quiz. 7.
Only a non-programmable, non-graphical calculator with a pink-tie or blue-goggle sticker will be allowed. No other aids are allowed. 8.
DO NOT WRITE ON THIS COVER PAGE. Questions Out of 1 –
9 9 10 11 11 6 Total 26
*** This page was intentionally left blank. ***
Part 1: Multiple Choice –
Each question is worth 1 mark Please CLEARLY CIRCLE your answer selection on this page. Do NOT simply write your choice next to the question. If you do, the question will not be graded. Each question has one correct answer. Choose the best answer. 1.
Even though he is not a fan of the Toronto Raptors (professional basketball team), Alex does follow the team’s progress. Alex has many friends who believe that the Raptors will definitely win a championship within the next five years. He does not believe this to be true. However, Alex does believe that there is a 40% chance that the Raptors will win a championship within the next five years. The type of probability being used here is: A) Classical probability B) Subjective probability C) Relative frequency probability •
There is really no rational basis underlying this probability statement. It is essentially based on a feeling or opinion. As such, it would be an example of a subjective probability statement. 2.
To determine whether the waiting time to be served is less than 7 minutes at a local restaurant, the waiting times of 300 customers were observed over a 4-hour period. It was found that 186 customers out of the 300 experienced a wait of less than 7 minutes. The manager of the restaurant declares that the probability of waiting less than 7 minutes is 186/300 = 0.62. Which definition of probability best describes the above scenario? A) Classical probability B) Subjective probability C) Relative frequency probability •
The statement is based on a long series of repetitions of an experiment or process -- in this case, the observation of waiting times for a large number of customers. For this reason, the relative frequency definition is the best choice to describe the situation.
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3.
A fair six-sided die is to be rolled two times. Each time the die is rolled, the number of pips (dots) on the upturned face is recorded. What is the probability that number of dots will be different in each of the two rolls? Note:
“Fair” implies that each face is equally likely to be observed on any
roll. A) 5/6
B) 1/4 C) 1/3 D) 1/12 E) None of these •
In this case: •
S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), ……,
(6,5), (6,6)}. •
As such, we see that there are a total of 36 outcomes in the sample space. •
To solve this probability, it may be easier to look at the complement. •
The outcomes “the number of dots will be THE SAME in the two rolls are given by: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). There are 6 outcomes in this case. •
Thus, the probability that the number of dots will be different in each of the two rolls (36 –
6)/36 = 30/36 = 5/6. 4.
Consider a well shuffled standard deck of 52 playing cards. A card is drawn from the deck and its face value is observed. What is the probability that its face value is a prime number? Notes:
Let Jack = 11, Queen = 12, King = 13, and Ace = 1. The face value of all other cards is equal to the number on the card. 1 is not a prime number. A) 4/13 B) 5/13 C) 6/13
D) 7/13 •
There are 13 possible face values in the deck, of which 6 are primes (2, 3, 5, 7, 11, and 13). So, the probability of selecting a card with a face value that is a prime number is 6/13. Use the following information to answer the next TWO questions: Two numbers from {1, 2, 3, 4, 5, 6} are to be randomly chosen with replacement. Each number is equally likely to be chosen, and each number is recorded after each selection. 5.
How many outcomes are in the sample space for this experiment? A) 6 B) 12 C) 24 D) 36
•
In this case, S = {(1, 1), (1, 2), …, (1,6), (2, 1), (2, 2), …, (2,6),
(3, 1), (3,2), (3,3), …(3, 6), (6,1), (6,2), …., (6, 6)}. •
We see that there are a total of 36 outcomes in the sample space.
6.
What is the probability that their product is even? A) 5/6 B) 1/2 C) 1/4 D) 3/4
E) None of these •
If either of the numbers selected is even, then the product will be even. •
So, this is equivalent to finding P(at least one of the selected numbers is even). •
It may be easier to determine the complement of this probability to answer the question. •
Let’s look
at the event space for A = no selected numbers are even. •
A = {(1, 1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)}. •
There are 9 outcomes corresponding to event A. •
So, P(at least one of the selected numbers is even) •
= 1 –
P(no selected numbers are even) = ? − (
𝟗
??
) =
??
??
=
?
?
. •
So, P(the product is even) = ?
?
7.
Suppose that you were invited to play the following matching game on a game show: You are shown three prizes and then given three price tags. The object of this game is to put the correct price tag on each prize. You will win every prize for which you correctly place the tag. Unfortunately, you have no knowledge of the price of any of the prizes, so you simply place the tags on the prizes randomly. What is the chance that you will win more than one prize? A) 1/6
B) 1/3 C) 1/2 D) 2/3 •
We can set up a table of possibilities here. Let’s assume that a perfect match will occur when Prize A has correct price tag A, Prize B has correct price tag B, and Prize C has correct price tag C (or sequence A, B, C). •
We can list the 6 possible arrangements in the sample space and organize the results in a table. The possible sequences are given in the table below: Sequence # of Matches A, B, C 3 A, C, B 1 B, A, C 1 B, C, A 0 C, A, B 0 C, B, A 1 •
From the table, we can see that P(more than one prize) = P(two matches) + P(three matches) = 0 + 1/6.
•
You could also obtain this probability using a complementary event approach. •
In this case, P(more than one prize) = 1 –
P(no more than 1 prize) •
= 1 –
[P(no prizes) + P(one prize)] = 1 –
(2/6 + 3/6) = 1 –
5/6 = 1/6. •
You will get the same answer. 8.
A student is asked ten true/false questions. They can answer each question using T (true) or F (false). Which of the following sets below is a possible sample space for the experiment? A) S={the event that there are at least four F’s, the event that there are at most four T’s}
B) S={the event that there are no F’s, the event that there is at least one F} C) S={the event that there are more F’s than T’s, the event that there are fewer F’s than T’s}
D) S={the event that there are at least five F’s, the event that there are less than four F’s}
•
A) has overlap with 6, 7, 8, 9, and 10 F’s, and has not included 0, 1,
2, or 3 F’s.
•
B) is the only valid sample space for this experiment. •
C) is incomplete, missing the same number of F’s as T’s
•
D) is incomplete, missing the case where there are 4 F’s. 9.
Let S = {A, B, C, D, E} be a sample space. Let F be the event {A, C}. Determine P(F) if the following relationships hold: P(A) = 2*P(B) P(C) = 3*P(A) P(D) = 4*P(B) P(E) = 2*P(C) A) 1/5 B) 4/5 C) 8/25 D) 12/25 Hint:
Express the elements in your sample space in terms of B. •
As noted in the Hint, we can find out what each is worth in terms of B. •
Recall that S = {A, B, C, D, E} •
So, based on the given information, we have •
1 = P(S) = P(A) + P(B) + P(C) + P(D) + P(E) •
1 = 2P(B) + P(B) + 6P(B) + 4P(B) + 12P(B) = 25P(B)
•
So P(B) = 1/25 •
F = {A, C} = {2B, 6B} => P(F) = 8/25
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Part 2: Written Answer / Short Answer Questions IMPORTANT NOTES: Final answers can be left in fraction form OR rounded to THREE DECIMAL PLACES. For full credit, you need to show all necessary work, writing out answers in full sentences. (i.e. Don’t just write 4. This is not a complete answer). Instead, you can write: There are a total of 4 outcomes in the sample space. Also note that unjustified answers will not receive full credit. 10. A box contains 3 marbles numbered 1, 2, and 3. Two marbles are to be randomly selected without replacement.
a.
Write out the possible outcomes in the sample space and state the number of outcomes in the sample space. (3 marks)
•
From the given information, we can write out the sample space as follows: •
S = {(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)}. •
We see that there are a total of 6 outcomes in the sample space. b. If we let event A = the sum of the numbers is four, write out the event space for A, in terms of all possible outcomes. (2 marks)
•
From part a., S = {(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)}. •
The outcomes corresponding to A = the sum of the numbers is four are: •
A = {(1, 3), (3, 1)}. c. If we assume that all outcomes in the sample space are equally likely, what is P(A)? (2 marks)
•
From part a., we know that there are 6 outcomes in the sample space. •
From part b., the outcomes corresponding to A = the sum of the numbers is four are: A = {(1, 3), (3, 1)}. •
If all outcomes in S are equally likely, then P(A) = 2/6 = 1/3 (= 0.333 to 3 d.p.) d. If we let event B = the first number drawn is less than the second number drawn, write out the event space for B, in terms of all possible outcomes. (2 marks)
•
S = {(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)}. •
The outcomes corresponding to B = the first number drawn is less than the second number drawn: •
B = {(1, 2), (1, 3), (2, 3)}.
e. If we assume that all outcomes in the sample space are equally likely, what is P(B)? (2 marks)
•
From part d., the outcomes corresponding to B = the first number drawn is less than the second number drawn are given by: B = {(1, 2), (1, 3), (2, 3)}. •
If all outcomes in S are equally likely, then P(B) = 3/6 (= 0.5) 11.
You are given information below regarding average grades for a sample of 200 Undergraduate (U) and Graduate (G) students: U(ndergraduate) G(raduate) Total A 24 38 62 B 31 43 74 C 25 8 33 D 16 4 20 F 10 1 11 Total 106 94 200 A student is randomly selected from the sample. (James Note: The Totals will not be given). a.
What is the probability that a student with a B average is selected? (2 marks)
•
From the given information, we see that there are a total of 200 students in the sample •
There are a total of 74 students with a B average. •
So, P(student with a B average is selected) = 74/200 = 0.37 b.
What is the probability that the student is not an Undergraduate student with an A average? (2 marks)
•
From the given information, we see that there are a total of 200 students in the sample. •
Of these, 24 are Undergraduate students with an A average. •
This means that P(student is not an Undergraduate with an A average) = 1 –
24/200 = 176/200 = 0.88
c.
If a randomly selected student is an Undergraduate student, what is the probability that they have a C average? (2 marks)
•
From the given information, we see that there are a total of 106 Undergraduate students. •
There are a total of 25 Undergraduates with a C average. •
So, the required probability is 25/106 = 0.236 (to 3 d.p.)
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