Download SOLUTIONS: Review for TEST #3 – Probability

SOLUTIONS: Review for TEST #3 – Probability
Part A. Multiple Choice
Identify the letter of the choice that best completes the statement or answers the question.
1.
a.
A student writes a multiple choice test that consists of 3 questions, each with 4 choices. What is
the probability of the student have a perfect mark on the test by guessing?
c.
b.
d. none of the above
Answer: “b”  ¼ x ¼ x ¼
2.
a.
A number between 1 and 50 is randomly chosen. The probability of the number being a perfect
square is
c.
b.
d. none of the above
Answer: “d” none of the above. There are 48 numbers between 1 and 50 (2 through 49). There are
six perfect squares in this range  4, 9, 16, 25, 36, 49. So the answer is 6/48, reduces to 1/8, which
is not one of the choices.
3. If and
a.
b.
Answer: “b”
4.
are mutually exclusive events, then
c.
d.
Mutually exclusive means there is no intersection.
a.
Find the probability that when a single card is drawn from a regular deck of cards a diamond or six
is chosen.
c.
b.
d. none of the above
Answer: “c” There are 13 diamonds (including the six), plus 3 other 6’s, for a total of 16 “winning”
cards. The probability is 16/52, reduces to
5.
a.
Out of a group of 100 people, 63 are adults and 37 are children. Of the adults, 31 have brown
eyes and of the children, 29 have brown eyes. If a person is selected at random, determine the
probability of that person being a child if the colour of eyes is known to be brown.
c.
b.
Answer: “c”
d.
P(child | brown) = n(child and brown eyes) = 29/60
n(brown eyes)
6.
In a school, 70% of the students listen to music. Of those, 60% listen to rock music, 20% listen to
dance music, and 10% listen to both types of music. Determine the probability that a student
chosen at random listens to dance music.
a. 0.2
c. 0.3
b. 0.14
d. 0.9
Answer: “b” 20% of the 70% who listen to music  0.2 x 0.7 = 0.14
7.
In a school, 70% of the students listen to music. Of those, 60% listen to rock music, 20% listen to
dance music, and 10% listen to both types of music. Determine the probability that a student
listens to music other than rock or dance music given that they listen to music.
a. 0.3
c. 0.21
b. 0.1
d. 0.07
Answer: “a” Of those who listen to music, 70% of them listen to rock or dance (60 + 20 – 10). This
means that 30% of them must listen to something else.
8.
a.
Determine the probability that when a coin is flipped 3 times, tails occurs, then heads, then tails.
c.
b.
d.
Answer: “d”  simply ½ x ½ x ½
Part B
9.
A six-sided die is rolled at the same time as a coin is tossed.
a) Draw a tree diagram that represents all possible results of the two actions
together.
b) Determine the theoretical P(odd number on die, dice toss is heads).
Tree
Diagram
H
1
T
H
2
T
H
3
T
H
4
T
H
5
T
H
6
T
b) Using the tree diagram, count the number of outcomes of “odd
die” and “H”. There are 3 such outcomes  1-H, 3-H, and 5-H
Altogether there are 12 possible outcomes.
So P(odd number on die, dice toss is heads). = 3/12 or
¼
Another method is to simply multiply the individual probabilities:
P(odd number on die) x P(dice toss is heads)
= ½x½
=¼
10.
Determine the theoretical probability of each of the following situations:
a) a red seven is drawn from a deck of 52 cards
Answer: 2/52 reduces to 1/26
b) two dice are rolled and both numbers are even
Solution: ½ x ½ =
¼
c) A spin wheel, numbered from 1 to 8, is spun twice  P(sum >= 12)
Solution: There are 64 possible outcomes. List those whose sum >= 12
8+8 8+7 8+6 8+5 8+4 7+8 7+7 7+6 7+5 6+8 6+7 6+6 5+8 5+7 4+8
There are 15 “successful” outcomes. Therefore P(sum >= 12) = 15/64.
11.
The universal set “S” is defined by S = {all natural numbers less than 100}.
Set A = {all perfect squares < 100}. Set B = {all odd natural numbers < 50}
a) List all members set A.
b) List all members of A ∩ B
c) Describe, in efficient wording, the set A U B.
d) Describe, in efficient wording, the set B’
e) List all members of B’∩A
Solutions: a) A = {1,4,9,16,25,36,49,64,81}
b) A ∩ B = {1,9,25,49}
c) A U B = {odds up to 49 and perfect squares 4, 16, 36, 64 and 81}
d) B’ = {evens from 2 through 50, and all numbers from 51 to 99}
e) B’∩A = {4,16,36,64,81}
12.
Two hundred students are enrolled in at least one 12U math course.
* 68 are in Advanced Functions
* 78 are in Data Management
* 60 are in both Adv. Func. and Calculus
* 35 are in both Calc. and D - M.
* 13 are in both Adv. Func. and D - M.
* 10 are in all 3 courses
a) Draw a Venn Diagram and fill in the missing values.
b) Determine P(a student takes Calculus | the student takes Data Management)
Solution: a)
b) 78 take Data Management. Of those, 35 take Calculus. Therefore,
P(a student takes Calculus | the student takes Data Management) =
35/78
13.
Illustrate, using an example, why n(AUB) = n(A) + n(B) – n(A∩B), for any sets A and B.
Solution: numerous examples. For example: Let A={1,2,3,4,5,6} B={2,4,6,8,10}
Then A∩B = {2,4,6} and AUB = {1,2,3,4,5,6,8,10} Using the formula:
n(AUB) = n(A) + n(B) – n(A∩B) and substituting the numbers in each set, we get
8 = 6 + 5 - 3 True statement.
14.
A survey indicated that 60% of the people watch television. Of those people, 70% watch
more than one hour per day and the rest watch less than one hour. Determine the
probability that a person, chosen at random, watches more than one hour of television
per day.
Solution: 70% of the 60%  0.7 x 0.6 = 0.42, or
15.
42%
A jar of marbles contains 40% black marbles, 30% red marbles, 10% yellow marbles,
and the remainder are green marbles. If a marble is drawn randomly, determine the
probability that it is yellow if is known that it is not green. ie -> P(yellow | not green)
Solution: P(not green) = 80%; P(yellow ∩ not green) aka P(yellow) = 10%;
P(yellow | not green) = 10%/80%
16.
The probability of event
= 1/8
occurring is
. The probability of event
. The probability of events
occurring given event
occurring is
and
occurring is
.
a) Determine P(B).
Solution:
P(A ∩ B)
P(A | B)
= 5/30 = 18/30 or 3/5
5/18
P(B)
=
b) Draw a Venn diagram to represent the numbers of items in each event.
Solution: Venn diagram will have 7/30 in A only, 13/30 in B only, and 5/30 in the
intersection.
17.
The probability of rain tomorrow is 0.7. The probability of rain the next day is 0.55. The
probability of rain the day after that is 0.4.
a) Determine P(it will rain on two of the next three days).
b) Describe how you would set up a simulation to determine this probability
experimentally, using i) the graphing calculator; and ii) Excel.
Solution: a) Consider all cases of raining on two of the next 3 days.
P(rain on two of 3 days) = P(rain, rain, not) + P(rain, not, rain) + P(not, rain, rain)
= 0.7 x 0.55 x 0.6 + 0.7 x 0.45 x 0.4 + 0.3 x 0.55 x 0.4
= .231 + .126 + .066
= .423
b) Graphing Calculator: randInt(1,100,3) get 3 numbers from 1 to 100.
1st number -> rain if 70 and under
2nd number -> rain if 55 and under
3rd number -> rain if 40 and under
Figure out how many times it rained out of the three days. Record it.
Repeat the experiment 100 times or so.
Keep track of how many times it rained twice.
Determine the probability of raining twice  How many times it rained twice
Number of total trials
Excel: Cell A1: +if(rand()<.7,”rain”,”not”)
Cell B1: +if(rand()<.55,”rain”,”not”)
Cell C1: +if(rand()<.4,”rain”,”not”)
Copy these down 1000 times or so.
Cell D1: +countif(A1:C1,”rain”)  how many times it rained in three days
Copy this down 1000 times or so
Cell E1: +countif(D1:D1000,2)  how many times it rained 2 times in the 1000 3-day trials
Cell E3: =E1/1000  the probability of it raining 2 days out of 3