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1. Which pair has equally likely...

1. Which pair has equally likely outcomes? List the letters of the two choices below which have equal probabilities of success, separated by a comma. A standard deck of cards has 12 face cards and four Aces (Aces are not face cards).

A. rolling a sum of 9 on two fair six sided dice

B. drawing the Queen of Diamonds out of a standard 52 card deck given it’s a face card.

C. rolling a sum of 11 on two fair six sided dice

D. drawing a king out of a standard 52 card deck given it’s a face card.

E. rolling a sum of 10 on two fair six sided dice

(Points : 3)

a. rolling a sum of 9 on two fair six sided dice "

There are 36 (6^2) possible outcomes.

(3,6)

(4,5)

(5,4)

(6,3)

Four of them sum to 9.

P(A given B) = P(A|B) = P(A and B)/P(B)

P(9)= 4/36 = 1/9

b. Drawing the Queen of diamonds out of a standard 52 card deck given it’s a face card"

There are 12 face cards.

There is 1 Queen of Diamonds.

P(QD,F)= 1/12

c. Rolling a sum of 11 on two fair six sided dice.

(5,6)

(6,5)

P(11)=2/36 = 1/18

d. Drawing a king out of a standard 52 card deck given it’s a face card.

There are 12 face cards.

There are 4 Kings.

P(k,F)=4/12=1/3

e. Rolling a sum of 10 on two fair six sided dice.

(4,6)

(5,5)

(6,4)

P(10)=3/36=1/12

f.No two pairs match in their probabilities.

2. A mini license plate for a toy car must consist of a two numbers followed by a vowel. Each number must be a 3 or a 6. Repetition of digits is permitted.

• Use the counting principle to determine the number of points in the sample space.

• Construct a tree diagram to represent this situation and submit it to the W5: Assignment 2 Dropbox ( need help)

• List the sample space.

• Determine the exact probability of creating a mini license plate with a 3. Give solution exactly in reduced fraction form.

Ans:

•Use the counting principle to determine the number of points in the sample space.

pick a vowel: 5 ways

pick a number: 2 ways

pick a number: 2 ways

===

# XXXXX points in the sample space = 5*2*2 = 20

------

•Construct a tree diagram to represent this situation and submit it to the

•List the sample space.

a33,a37,a73,a77

Same pattern with e,i,o,u

-----------------------------------

•Determine the exact probability of creating a mini license plate with an A.

Give solution exactly in reduced fraction form.

Ans: 4/20 = 1/5

(Points : 12)

3. Given that P(A) = 0.7 and P(B) = 0.2, and P(A and B) = 0.14, determine P(B|A). (Points : 3)

0.90

0.70

0.50

0.20

P(B|A) = [P(A and B)/P(A)] = 0.14/0.70 = .20

4. An identification code is to consist of two letters followed by six digits. How many different codes are possible if repetition is permitted? (Points : 2)

Code: LLDDDDDD

There are 26 letters

There are 10 digits

# XXXXX codes = 26^2*10^6 = 676,000,000

5. Show all work. A disc jockey has 7 songs to play. Three are slow songs, and four are fast songs. Each song is to be played only once. In how many ways can the disc jockey play the 7 songs if

• The songs can be played in any order.

• The first song must be a slow song and the last song must be a slow song.

• The first two songs must be fast songs.

(Points : 3)

In how many ways can the disc jockey play the 7 songs if

the songs can be played in any order:::7! = 5040 ways

-------------------------------------------------------------------

• The first song must be a slow song and the last song must be a slow song.

1st song: 3 ways

last song: 2 ways

Middle 5 songs: 5! = 120 ways

Total ways: 3*2*120 =720 ways

------------------------------------

• The first two songs must be fast songs.

1st song: 4 ways

2nd song: 3 ways

last 5 songs: 5! = 120

Total ways:4*3*120 = 1440 ways

==================================

6. Evaluate 10C8 (Points : 2)

45

90

40320

2

10C8 is the same as 10C2

10C2 = [10*9]/[1*2] = 45

======================

7. Show all work. Mary purchased a package of 20 different plants, but she only needed 16 plants for planting. In how many ways can she select the 20 plants from the package to be planted? (Points : 3)

Note: # XXXXX ways of selecting 16 is the

same as # XXXXX ways of not selecting 4.

20C16 = 20C4 = (20*19*18*17)/(1*2*3*4) = 4845

=============================================================================

8. Each of the numbers 0 through 24 is written on a piece of paper and all of the pieces of paper are placed in a hat. One number is XXXXX at random. Determine the probability that the number selected is even. Note: 0 is considered an even number.

(Points : 2)

13/25

13/24

11/24

1/2

Even digits: 0,2,4,6,8,10,12,14,16,18,20,22,24

P(even digit) = 13/25

9. Consider the Venn diagram below. The numbers in the regions of the circle

indicate the number of items that belong to that region.

Determine

• n(A)

• n(B)

• P(A)

• P(B)

• P(A|B)

• P(B|A)

Answer:

n(A) = 30+20 = 50

n(B) = 90+20 = 110

P(A) = 50/160

P(B) = 110/160

P(A|B)= 20/110

P(B|A)= 20/50

====================================================================================

1. A bag contains 8 pink marbles, 5 green marbles and 10 brown marbles. What is the chance of drawing a brown marble? If a brown marble is drawn then placed back into the bag, and a second marble is drawn, what is the probability of drawing a pink marble? Give solutions exactly in reduced fraction form, separated by a comma.

(Points : 3)

Ans:

(10/23),(8/23)

==================== =====================================

2. A mini license plate for a toy car must consist of two letters followed by a one digit odd number. Each letter must be a K or a Q. Repetition of letters is not permitted.

• Use the counting principle to determine the number of points in the sample space.

• Construct a tree diagram and submit it to the W5: Assignment 3 Dropbox.

• List the sample space.

• Determine the exact probability of creating a mini license plate with a 3. Give solution exactly in reduced fraction form.

(Points : 12)

1. Use the counting principle to determine the number of points in the sample space.

# XXXXX points = 2*1*3 = 6

-----------------------------------

2. Construct a tree diagram

-------

3. List the sample space.

KQ3,kQ5,KQ7,QK3,QK5,QK7

----------------

4. Determine the exact probability of creating a

mini license plate with a 3.

2/6=1/3

===================

2. A parent can choose from 5 types of protein, 6 vegetables and 4 desserts. If the parent serves a meal of 1 protein, 1 vegetable and 1 dessert to the family, how many different meals can be served?

(Points : 2)

Ans: 5*6*4 = 120 different means

=======================================================

3. A card is selected from a standard deck of 52 playing cards. A standard deck of cards has 12 face cards and four Aces (Aces are not face cards). Find the probability of selecting

•an odd prime number under 10 given the card is a club. (1 is not prime.)

•a Jack, given that the card is not a heart.

•a King given the card is not a face card.

Show step by step work. Give all solutions exactly in reduced fraction form.

(Points : 3)

5. In how many ways can 8 instructors be assigned to six sections of a course in mathematics?

(Points : 2)

Let the sections select the instructors:

8C6 = 8C2 = (8*7)/(1*2) = 28 ways

6. In how many different ways can the top eight new indie bands be ranked on a top eight list? The top hit song for each of the eight bands will compete to receive monetary awards of $1000, $500, $250 and $100, respectively. In how many ways can the awards be given out?

(Points : 3)

i. Ans: 8! = 40320

ii. 1st prize 8 ways

2nd prize 7 ways

3rd prize 6 ways

4th prize 5 ways

Ans: 8P4 = 8!/(8-4)! = 8*7*6*5 = 1680

7. A bag contains a total of 9 batteries, of which five are defective. Selecting two at random, without replacement, determine the probability that none of the batteries you select are good.

(Points : 2)

5/18

5/9

2/9

20/81

Combination=nCr=n!/((n-r)!*r!)

P=5C2/9C2

=10/36

= 5/18 the probability that none of the batteries you select are good.

4 are not defective.

P(2 good) = 4C2/9C2 = 8/36

=2/9

8. Evaluate: 7C5

(Points : 3)

21

35

12

42

7C5 = 18C8 = [7*6]=42

9.

Given that P(A) = 0.4, P(B) = 0.8, and P(A and B) = 0.40, determine P(A|B)

(Points : 3)

0.67

0.80

0.50

0.75

P(A|B) = P(A and B)/P(B) = 0.4/0.80 = 1/2 = 0.5

10. If P(A or B) = 0.8, P(A) = 0.1 and P(B) = 0.9, determine

P(A and B).

(Points : 3)

0.90

0.20

0.80

0.08

P(A or B)= P(A) + P(B) - P(A and B)

0.8=0.1+0.9-x

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

0.1+0.9-x=0.8

Add 0.9 to 0.1 to get 1.

1-x=0.8

Since 1 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 1 from both sides.

-x=-1+0.8

Add 0.8 to -1 to get -0.2.

-x=-0.2

Multiply each term in the equation by -1.

-x*-1=-0.2*-1

Multiply -x by -1 to get x.

x=-0.2*-1

Multiply -0.2 by -1 to get 0.2.

x=0.2

A. rolling a sum of 9 on two fair six sided dice

B. drawing the Queen of Diamonds out of a standard 52 card deck given it’s a face card.

C. rolling a sum of 11 on two fair six sided dice

D. drawing a king out of a standard 52 card deck given it’s a face card.

E. rolling a sum of 10 on two fair six sided dice

(Points : 3)

a. rolling a sum of 9 on two fair six sided dice "

There are 36 (6^2) possible outcomes.

(3,6)

(4,5)

(5,4)

(6,3)

Four of them sum to 9.

P(A given B) = P(A|B) = P(A and B)/P(B)

P(9)= 4/36 = 1/9

b. Drawing the Queen of diamonds out of a standard 52 card deck given it’s a face card"

There are 12 face cards.

There is 1 Queen of Diamonds.

P(QD,F)= 1/12

c. Rolling a sum of 11 on two fair six sided dice.

(5,6)

(6,5)

P(11)=2/36 = 1/18

d. Drawing a king out of a standard 52 card deck given it’s a face card.

There are 12 face cards.

There are 4 Kings.

P(k,F)=4/12=1/3

e. Rolling a sum of 10 on two fair six sided dice.

(4,6)

(5,5)

(6,4)

P(10)=3/36=1/12

f.No two pairs match in their probabilities.

2. A mini license plate for a toy car must consist of a two numbers followed by a vowel. Each number must be a 3 or a 6. Repetition of digits is permitted.

• Use the counting principle to determine the number of points in the sample space.

• Construct a tree diagram to represent this situation and submit it to the W5: Assignment 2 Dropbox ( need help)

• List the sample space.

• Determine the exact probability of creating a mini license plate with a 3. Give solution exactly in reduced fraction form.

Ans:

•Use the counting principle to determine the number of points in the sample space.

pick a vowel: 5 ways

pick a number: 2 ways

pick a number: 2 ways

===

# XXXXX points in the sample space = 5*2*2 = 20

------

•Construct a tree diagram to represent this situation and submit it to the

•List the sample space.

a33,a37,a73,a77

Same pattern with e,i,o,u

-----------------------------------

•Determine the exact probability of creating a mini license plate with an A.

Give solution exactly in reduced fraction form.

Ans: 4/20 = 1/5

(Points : 12)

3. Given that P(A) = 0.7 and P(B) = 0.2, and P(A and B) = 0.14, determine P(B|A). (Points : 3)

0.90

0.70

0.50

0.20

P(B|A) = [P(A and B)/P(A)] = 0.14/0.70 = .20

4. An identification code is to consist of two letters followed by six digits. How many different codes are possible if repetition is permitted? (Points : 2)

Code: LLDDDDDD

There are 26 letters

There are 10 digits

# XXXXX codes = 26^2*10^6 = 676,000,000

5. Show all work. A disc jockey has 7 songs to play. Three are slow songs, and four are fast songs. Each song is to be played only once. In how many ways can the disc jockey play the 7 songs if

• The songs can be played in any order.

• The first song must be a slow song and the last song must be a slow song.

• The first two songs must be fast songs.

(Points : 3)

In how many ways can the disc jockey play the 7 songs if

the songs can be played in any order:::7! = 5040 ways

-------------------------------------------------------------------

• The first song must be a slow song and the last song must be a slow song.

1st song: 3 ways

last song: 2 ways

Middle 5 songs: 5! = 120 ways

Total ways: 3*2*120 =720 ways

------------------------------------

• The first two songs must be fast songs.

1st song: 4 ways

2nd song: 3 ways

last 5 songs: 5! = 120

Total ways:4*3*120 = 1440 ways

==================================

6. Evaluate 10C8 (Points : 2)

45

90

40320

2

10C8 is the same as 10C2

10C2 = [10*9]/[1*2] = 45

======================

7. Show all work. Mary purchased a package of 20 different plants, but she only needed 16 plants for planting. In how many ways can she select the 20 plants from the package to be planted? (Points : 3)

Note: # XXXXX ways of selecting 16 is the

same as # XXXXX ways of not selecting 4.

20C16 = 20C4 = (20*19*18*17)/(1*2*3*4) = 4845

=============================================================================

8. Each of the numbers 0 through 24 is written on a piece of paper and all of the pieces of paper are placed in a hat. One number is XXXXX at random. Determine the probability that the number selected is even. Note: 0 is considered an even number.

(Points : 2)

13/25

13/24

11/24

1/2

Even digits: 0,2,4,6,8,10,12,14,16,18,20,22,24

P(even digit) = 13/25

9. Consider the Venn diagram below. The numbers in the regions of the circle

indicate the number of items that belong to that region.

Determine

• n(A)

• n(B)

• P(A)

• P(B)

• P(A|B)

• P(B|A)

Answer:

n(A) = 30+20 = 50

n(B) = 90+20 = 110

P(A) = 50/160

P(B) = 110/160

P(A|B)= 20/110

P(B|A)= 20/50

====================================================================================

1. A bag contains 8 pink marbles, 5 green marbles and 10 brown marbles. What is the chance of drawing a brown marble? If a brown marble is drawn then placed back into the bag, and a second marble is drawn, what is the probability of drawing a pink marble? Give solutions exactly in reduced fraction form, separated by a comma.

(Points : 3)

Ans:

(10/23),(8/23)

==================== =====================================

2. A mini license plate for a toy car must consist of two letters followed by a one digit odd number. Each letter must be a K or a Q. Repetition of letters is not permitted.

• Use the counting principle to determine the number of points in the sample space.

• Construct a tree diagram and submit it to the W5: Assignment 3 Dropbox.

• List the sample space.

• Determine the exact probability of creating a mini license plate with a 3. Give solution exactly in reduced fraction form.

(Points : 12)

1. Use the counting principle to determine the number of points in the sample space.

# XXXXX points = 2*1*3 = 6

-----------------------------------

2. Construct a tree diagram

-------

3. List the sample space.

KQ3,kQ5,KQ7,QK3,QK5,QK7

----------------

4. Determine the exact probability of creating a

mini license plate with a 3.

2/6=1/3

===================

2. A parent can choose from 5 types of protein, 6 vegetables and 4 desserts. If the parent serves a meal of 1 protein, 1 vegetable and 1 dessert to the family, how many different meals can be served?

(Points : 2)

Ans: 5*6*4 = 120 different means

=======================================================

3. A card is selected from a standard deck of 52 playing cards. A standard deck of cards has 12 face cards and four Aces (Aces are not face cards). Find the probability of selecting

•an odd prime number under 10 given the card is a club. (1 is not prime.)

•a Jack, given that the card is not a heart.

•a King given the card is not a face card.

Show step by step work. Give all solutions exactly in reduced fraction form.

(Points : 3)

5. In how many ways can 8 instructors be assigned to six sections of a course in mathematics?

(Points : 2)

Let the sections select the instructors:

8C6 = 8C2 = (8*7)/(1*2) = 28 ways

6. In how many different ways can the top eight new indie bands be ranked on a top eight list? The top hit song for each of the eight bands will compete to receive monetary awards of $1000, $500, $250 and $100, respectively. In how many ways can the awards be given out?

(Points : 3)

i. Ans: 8! = 40320

ii. 1st prize 8 ways

2nd prize 7 ways

3rd prize 6 ways

4th prize 5 ways

Ans: 8P4 = 8!/(8-4)! = 8*7*6*5 = 1680

7. A bag contains a total of 9 batteries, of which five are defective. Selecting two at random, without replacement, determine the probability that none of the batteries you select are good.

(Points : 2)

5/18

5/9

2/9

20/81

Combination=nCr=n!/((n-r)!*r!)

P=5C2/9C2

=10/36

= 5/18 the probability that none of the batteries you select are good.

4 are not defective.

P(2 good) = 4C2/9C2 = 8/36

=2/9

8. Evaluate: 7C5

(Points : 3)

21

35

12

42

7C5 = 18C8 = [7*6]=42

9.

Given that P(A) = 0.4, P(B) = 0.8, and P(A and B) = 0.40, determine P(A|B)

(Points : 3)

0.67

0.80

0.50

0.75

P(A|B) = P(A and B)/P(B) = 0.4/0.80 = 1/2 = 0.5

10. If P(A or B) = 0.8, P(A) = 0.1 and P(B) = 0.9, determine

P(A and B).

(Points : 3)

0.90

0.20

0.80

0.08

P(A or B)= P(A) + P(B) - P(A and B)

0.8=0.1+0.9-x

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

0.1+0.9-x=0.8

Add 0.9 to 0.1 to get 1.

1-x=0.8

Since 1 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 1 from both sides.

-x=-1+0.8

Add 0.8 to -1 to get -0.2.

-x=-0.2

Multiply each term in the equation by -1.

-x*-1=-0.2*-1

Multiply -x by -1 to get x.

x=-0.2*-1

Multiply -0.2 by -1 to get 0.2.

x=0.2

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