Problem

1
Problem 6

• If a students takes a test consisting of 20
• true-false questions and randomly guesses at
• all of the answers, what is the probability that
• all 20 guesses will be correct?
• 0
• (1/2)20
• 1/(220)
• 1/2

• PRAXIS - WI
• Jarod Hart
• Maren Lau
• Kristin Radermacher
• Cheslea Simon

2
If a students takes a test consisting of
20 true-false questions and randomly guesses at
all of the answers, what is the probability that
all 20 guesses will be correct? a. 0 b.
(1/2)20 c. 1/(220) d. 1/2
What is the probability of getting 2 right out
of 3 true-false questions? The possible
combinations are (R right, W
wrong) (R,R,W),(R,W,R),(W,R,R) The probability
of getting 2 out of 3 right is pp(R,R,W)p(R,W,R
)p(W,R,R) 1/81/81/83/8 The user should be
able to put a number 0, 1, 2, or 3 in place of
the red underlined number. All of the blue
output should change when that number is entered.
If a 1 is entered, the only thing in the output
that will change is the combinations. All of the
values will stay the same, just replace the
combinations from two right to only 1.

• Incorrect, there is positive probability of
  answering all 20 correctly
• Correct
• Incorrect, remember that each question is an
  independent event
• Incorrect, that is the probability of getting one
  question right

The probability of multiple independent events is
the product of the probabilities of the events.
Problem Name
3
If a students takes a test consisting of
20 true-false questions and randomly guesses at
all of the answers, what is the probability that
all 20 guesses will be correct? a. 0 b.
(1/2)20 c. 1/(220) d. 1/2
What is the probability of getting 0 right out
of 3 true-false questions? The possible
combinations are (W,W,W) The probability of
getting 2 out of 3 right is pp(W,W,W)1/8 This
is the output that should be produced when a 0 is
entered. Similarly, if a 3 is entered, all that
will change is (W,W,W) will become (R,R,R).

• Incorrect, there is positive probability of
  answering all 20 correctly
• Correct
• Incorrect, remember that each question is an
  independent event
• Incorrect, that is the probability of getting one
  question right

The probability of multiple independent events is
the product of the probabilities of the events.
Problem Name
4
Tutorial
To find this probability, we can find the total
number of ways to get all problems correct, and
then divide that by the total number of possible
outcomes. This technique applies when you are
working with equally likely outcomes. Our
equally likely outcomes are getting an individual
problem right or wrong, each with probability ½.
Problem Name
5
Tutorial
To find this probability, we can find the total
number of ways to get all problems correct, and
then divide that by the total number of possible
outcomes. This technique applies when you are
working with equally likely outcomes. Our
equally likely outcomes are getting an individual
problem right or wrong, each with probability
½. There is only one possible combination of
events that results with all 20 questions right.
That is 20 right and 0 wrong.
Problem Name
6
Tutorial
To find this probability, we can find the total
number of ways to get all problems correct, and
then divide that by the total number of possible
outcomes. This technique applies when you are
working with equally likely outcomes. Our
equally likely outcomes are getting an individual
problem right or wrong, each with probability
½. There is only one possible combination of
events that results with all 20 questions right.
That is 20 right and 0 wrong. Now we have to find
the total number of possible outcomes. For the
first problem, there are two outcomes, right or
wrong.
Problem Name
7
Tutorial
Then for two problems there are four possible
outcomes.
To find this probability, we can find the total
number of ways to get all problems correct, and
then divide that by the total number of possible
outcomes. This technique applies when you are
working with equally likely outcomes. Our
equally likely outcomes are getting an individual
problem right or wrong, each with probability
½. There is only one possible combination of
events that results with all 20 questions right.
That is 20 right and 0 wrong. Now we have to find
the total number of possible outcomes. For the
first problem, there are two outcomes, right or
wrong.
Problem Name
8
Tutorial
Then for two problems there are four possible
outcomes.
To find this probability, we can find the total
number of ways to get all problems correct, and
then divide that by the total number of possible
outcomes. This technique applies when you are
working with equally likely outcomes. Our
equally likely outcomes are getting an individual
problem right or wrong, each with probability
½. There is only one possible combination of
events that results with all 20 questions right.
That is 20 right and 0 wrong. Now we have to find
the total number of possible outcomes. For the
first problem, there are two outcomes, right or
wrong.
For each question, the number of possible
outcomes doubles.
Problem Name
9
Tutorial
Then for two problems there are four possible
outcomes.
To find this probability, we can find the total
number of ways to get all problems correct, and
then divide that by the total number of possible
outcomes. This technique applies when you are
working with equally likely outcomes. Our
equally likely outcomes are getting an individual
problem right or wrong, each with probability
½. There is only one possible combination of
events that results with all 20 questions right.
That is 20 right and 0 wrong. Now we have to find
the total number of possible outcomes. For the
first problem, there are two outcomes, right or
wrong.
For each question, the number of possible
outcomes doubles.
Then for 20 questions the total number of
outcomes will be 222220. Then the
probability of getting all 20 problems correct is
1/(220)(1/2)20. So the correct answer is B.
Problem Name
10
Tutorial
Then for two problems there are four possible
outcomes.
To find this probability, we can find the total
number of ways to get all problems correct, and
then divide that by the total number of possible
outcomes. This technique applies when you are
working with equally likely outcomes. Our
equally likely outcomes are getting an individual
problem right or wrong, each with probability
½. There is only one possible combination of
events that results with all 20 questions right.
That is 20 right and 0 wrong. Now we have to find
the total number of possible outcomes. For the
first problem, there are two outcomes, right or
wrong.
For each question, the number of possible
outcomes doubles.
Then for 20 questions the total number of
outcomes will be 222220. Then the
probability of getting all 20 problems correct is
1/(220)(1/2)20. So the correct answer is B.
Problem Name
11

• If a students takes a test consisting of 20
  true-false questions and randomly guesses at all
  of the answers, what is the probability that all
  20 guesses will be correct?
• 0
• (1/2)20
• 1/(220)
• 1/2

Since the result of any question has no impact
on the result of any of the other questions, the
two event are said to independent R and W are
independent Then we know that the probability
of two independent events is the product of the
probabilities of the individual events. Then the
probability of getting two problems right 2R
two out of two questions right P(2R)P(R)P(R)(P(
R))2(1/2)2 Then we can find the probability of
getting all 20 right in the same way 20R 20 out
of 20 questions right P(20R)(P(R))20(1/2)20 Th
en the answer is B, ½ to the twentieth B
(1/2)20
We can define two events for this problem. One
event is getting a problem right. The other
event will be getting a problem wrong R
Answer a question right W Answer a question
wrong The probability of each of these events
is one half P(R)1/2 P(W)1/2
Problem Name
12
A
Jill has two different pairs of pants, 3
different pairs of sock and 5 different pairs of
shoes. Assuming Jill wears matching socks and
matching shoes, how many different combinations
of pants, socks and shoes can she wear?
A. 30
C. 21
B. 10
D. 38
Thats right.
Remember the multiplier rule.
Remember the multiplier rule.
Remember the multiplier rule.
Problem Name
13
B
What is the probability of rolling a 4 or a 5
with one role of a fair 6 sided die?
A. 1/6
C. 1/36
B. 1/3
D. 1/18
The probability of rolling a 4 is 1/6 and the
probability of rolling a 5 is 1/6
A 4 or a 5, not and.
Thats right.
The probability of each number on the die is 1/6
Problem Name
14
C
John rolls two dice, and examines the sum of the
rolled values. Which value is the least likely
to be the sum of the two dice?
A. 6
C. 8
B. 5
D. 12
Think of the possible ways to roll each value
Think of the possible ways to roll each value
Think of the possible ways to roll each value
Thats right
Problem Name
15
D
A student is taking a 20 question true false
test. The students knows the material well
enough to have a 2/3 chance of getting each
problem right. What is the probability that the
students gets all 20 correct?
A. 2/3
C. (1/2)20
B. (2/3)20
D. 202/3
That is the probability of getting an individual
question right, not all
The probability of getting each problem right is
2/3
Thats right
The probability of a series of independent events
is the product of their probabilities
Problem Name
16
E
David and Todd are playing the card game Hearts.
In the game of Hearts, it is typically a
disadvantage to be dealt the Queen of Spades.
What is the probability that in all 6 hands they
play, that Todd will be dealt the Queen of Spades?
A. 1/52
C. 52(1/6)
B. 0
D. (1/52)6
The probability of a series of independent events
is the product of their probabilities
That is the probability for one hand
There is a positive probability of being dealt
the Queen of Spades
Thats right
Problem Name
17
F
A bag contains 50 different colored balls 10
green, 20 pink, 15 blue, and 5 yellow. If you
draw 5 balls from the bag without replacement,
what is the probability that all of the balls you
draw out are yellow?
A. 5/50
C. (5/50)(4/49)(3/48)(2/47)(1/46)
B. (5/50)5
D. (5/50)(4/50)(3/50)(2/50)(1/50)
That is the probability of the first yellow ball
Thats right
The balls are not being put back into the bag
When a ball is taken out, the total number of
balls changes
Problem Name
18
G
If you roll a dice 10 times in a row, what is the
probability of rolling ten 1s?
A. 1/10
C. (1)10
B. 1/60
D. (1/6)10
The probability of rolling a 1 on a given roll is
1/6
The probability of rolling a 1 on a given turn is
1/6
The probability of a series of independent events
is the product of their probabilities
Thats right
Problem Name
19
H
If you roll a dice 10 times in a row, what is the
probability of not rolling any 6s?
A. 0
C. (1/6)10
B. (5/6)10
D. 5/6
It is possible to not roll any 6s
The probability of rolling a 1 on a given turn is
1/6, what is the probability of not rolling a 1.
Thats right
That is the probability of not rolling a 1 on one
roll
Problem Name
20
I
An MM bag contains 10 red candies, 14 brown
candies, and 8 green candies. You reach in and
grab 1 candy, record the color then put it back
in the bag. Then you repeat the process 3 more
times. What is the probability that you recorded
4 brown candies?
A. (14/32)4
C. 4(1/3)
B. (1/3)4
D. 4/32
Type feedback for answers in text boxes below.
Thats right
The probability of a series of independent event
is the product of the probability of the events
The three colors are not equally likely to be
chosen
That is the probability of recording a brown with
one draw, not 4
Problem Name
21
J
When playing poker, a flush is a hand of 5 cards
of the same suit. With a standard deck of cards,
you are dealt 5 cards at random. What is the
probability that you have a flush of Hearts?
A. (1/52)5
C. (13/52)(12/51)(11/50)(10/49)(9/48)
B. (13/52)5
D. (13/52)(12/52)(11/52)(10/52)(9/52)
There are 13 Hearts in a standard deck of cards
Thats right
The cards are being dealt without replacement
The total number of cards is changing as the
cards are being dealt
Problem Name