Praxis Probability

7/27/2019 Praxis Probability

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Problem #6

PRAXIS - WI

Jarod Hart

Maren Lau

Kristin Radermacher

Cheslea Simon

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/(2*20)

d. 1/2


2/21


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/(2*20)

d. 1/2

Problem Name

The probability of multiple

independent events is the product

of the probabilities of the events.

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:

p=p(R,R,W)+p(R,W,R)+p(W,R,R)

=1/8+1/8+1/8=3/8

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.

a. Incorrect, there is positive

probability of answering all 20

correctly

b. Correctc. Incorrect, remember that each

question is an independent

event

d. Incorrect, that is the probability

of getting one question right


3/21


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/(2*20)

d. 1/2

Problem Name

The probability of multiple

independent events is the product

of the probabilities of the events.

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:

p=p(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).

a. Incorrect, there is positive

probability of answering all 20

correctly

b. Correctc. Incorrect, remember that each

question is an independent

event

d. Incorrect, that is the probability

of getting one question right


4/21

Problem Name

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 workingwith equally likely outcomes. Our equally

likely outcomes are getting an individual

problem right or wrong, each with

probability .

Tutorial


5/21

Problem Name








probability .

There is only one possible combination of

events that results with all 20 questionsright. That is 20 right and 0 wrong.

Tutorial


6/21

Problem Name








probability .



Now we have to find the total number of

possible outcomes. For the first problem,

there are two outcomes, right or wrong.

Tutorial

right

wrong

1st question

2 outcomes


7/21Problem Name








probability .






Tutorial

right

wrong

1st question

2 outcomes

right

wrong

1st question

2 outcomes 22 outcomes

2nd question

right

right

wrong

wrong

Then for two problems there

are four possible outcomes.


8/21Problem Name








probability .






Tutorial

For each question, the number

of possible outcomes doubles.

right

wrong

1st question

2 outcomes

right

wrong

1st question


2nd question

right

right

wrong

wrong



right

wrong

1st question


2nd question

right

right

wrong

wrong

23 outcomes

right

right

right

right

wrong

wrong

wrong

wrong

3rd question

220 outcomes

20th question


9/21Problem Name








probability .






Tutorial

right

wrong

1st question


2nd question

right

right

wrong

wrong

23 outcomes

right

right

right

right

wrong

wrong

wrong

wrong

3rd question

220 outcomes

20th question



right

wrong

1st question

2 outcomes

right

wrong

1st question


2nd question

right

right

wrong

wrong



Then for 20 questions the total number of

outcomes will be 2*2**2=220.

Then the probability of getting all 20 problems

correct is 1/(220)=(1/2)20. So the correct answer

is B.


10/21Problem Name








probability .






Tutorial

right

wrong

1st question


2nd question

right

right

wrong

wrong

23 outcomes

right

right

right

right

wrong

wrong

wrong

wrong

3rd question

220 outcomes

20th question



right

wrong

1st question

2 outcomes

right

wrong

1st question


2nd question

right

right

wrong

wrong



Then for 20 questions the total number of

outcomes will be 2*2**2=220.

Then the probability of getting all 20 problems

correct is 1/(220)=(1/2)20. So the correct answer

is B.


11/21Problem Name

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 rightP(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 rightP(20R)=(P(R))20=(1/2)20

Then the answer is B, to the twentieth

B (1/2)20

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 20guesses will be correct?

a. 0

b. (1/2)20

c. 1/(2*20)

d. 1/2

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


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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?

Problem Name

A

Thats right.

Remember the multiplier rule.



A. 30 *

B. 10

C. 21

D. 38


13/21Problem Name

BWhat is the probability of rolling a 4 or a 5 with one role of a fair 6 sided

die?

The probability of rolling a 4 is 1/6 and

the probability of rolling a 5 is 1/6

Thats right.

A 4 or a 5, not and.

The probability of each number on the die is

1/6

A. 1/6

B. 1/3 *

C. 1/36

D. 1/18


14/21Problem Name

CJohn 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?

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

A. 6

B. 5

C. 8

D. 12 *


15/21Problem Name

DA 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?

That is the probability of getting an

individual question right, not all

Thats right The probability of a series of independent

events is the product of their probabilities

A. 2/3

B. (2/3)20 *

C. (1/2)20

D. 20*2/3

The probability of getting each problem right

is 2/3


16/21Problem Name

EDavid 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?

That is the probability for one hand

There is a positive probability of

being dealt the Queen of Spades

The probability of a series of

independent events is the product of

their probabilities

Thats right

A. 1/52

B. 0

C. 52(1/6)

D. (1/52)6 *


17/21Problem Name

FA 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?

That is the probability of the first yellow

ball

The balls are not being put back into

the bag

Thats right

When a ball is taken out, the total number of

balls changes

A. 5/50

B. (5/50)5

C. (5/50)*(4/49)*(3/48)*(2/47)*(1/46) *

D. (5/50)*(4/50)*(3/50)*(2/50)*(1/50)


18/21Problem Name

GIf you roll a dice 10 times in a row, what is the probability of rolling ten 1s?

The probability of rolling a 1 on a given

roll is 1/6


independent events is the product of

their probabilities

The probability of rolling a 1 on a

given turn is 1/6

Thats right

A. 1/10

B. 1/60

C. (1)10

D. (1/6)10 *


19/21

Problem Name

HIf you roll a dice 10 times in a row, what is the probability of not rolling any

6s?

It is possible to not roll any 6s

Thats right

The probability of rolling a 1 on a

given turn is 1/6, what is the

probability of not rolling a 1.

That is the probability of not rolling a 1 on one

roll

A. 0

B. (5/6)10 *

C. (1/6)10

D. 5/6


20/21

Problem Name

IAn M&M 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?

Thats right

The three colors are not equally

likely to be chosen


independent event is the product of

the probability of the events

That is the probability of recording a brown

with one draw, not 4

A. (14/32)4 *

B. (1/3)4

C. 4(1/3)

D. 4/32

Type feedback for answers in text boxes below.


21/21

Problem Name

JWhen 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?

There are 13 Hearts in a standard

deck of cards

The cards are being dealt without

replacement

Thats right

The total number of cards is changing as the

cards are being dealt

A. (1/52)5

B. (13/52)5

C. (13/52)*(12/51)*(11/50)*(10/49)*(9/48) *

D. (13/52)*(12/52)*(11/52)*(10/52)*(9/52)

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Praxis Probability

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