Chapter 2 Midterm Review PDF

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Chapter 2 Midterm Review for Mathematical Thinking MATH 1215...

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Chapter 2 Midterm Review

1.) Suppose you have set A = {2, 3, 4, {5, 6, 7}, 8} True or false? (a) 2 ∈ A

(b) 4 ∈ A

(e) {2} ⊆ A

(f) {5} ⊆ A

(i) ∅ ⊆ A

(j) {∅, 2} ⊆ A

(c) 5 ∈ A

(d) {5, 6, 7} ∈ A

(g){5, 6, 7} ⊆ A

(h) {{5, 6, 7}} ⊆ A

How many subsets does A have? 2.) U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, X = {1, 2, 3, 4, 5, 6}, Y = {1, 3, 5, 7, 9} Find the following sets. (a) X ∩ Y (b) X ∪ Y (c) X ′ (d) X ′ ∩ Y (e) X ′ ∩ Y ′ (f) X ′ ∪ Y ′

3.) Shade in the part of the Venn diagram that represents the set A ∩ (B ∪ C ′ ) A

B

C

4.) Fill in the rest of the Venn diagram: (a) n(U ) = 29, n(A ∪ B) = 17, n(A) = 9, n(B ) = 13 A

B

(b) n(A′ ∩ B ′ ) = 20, n(B) = 15, n(B ′ ) = 27, n(A ∩ B) = 3 A

B

5.) A group goes apple picking with the following results: 21 picked Red apples

19 picked Yellow apples

18 picked Green apples

9 picked Red and Yellow

12 picked Green and Yellow

6 picked all three kinds

7 picked no apples

21 did not get Green apples

Fill in the Venn diagram R

G

Y

6.) A card is drawn. Find the probabilities: (a) A black 9

(b) A face card

7.) Two dice are rolled. Find the probabilities: (a) sum is ≥ 8 (b) sum is ≥ 8 or at least one die is a 4 (c) sum is ≥ 4 and ≤ 6 (d) not getting a sum ≥ 9

8.) Which of the following are mutually exclusive? You pick a single card from a 52-card deck: (a) picking a queen and picking a red card (b) picking a queen and picking a king You roll two dice: (c) sum is 5 and at least one die is a 4 (d) sum of 5 and at least one is a 6

(c) A red card or a 10

9.) Two dice are tossed. What are the odds in favor of the sum being greater than 10?

10.) If the odds in favor of event A are 7 to 12, what is P (A)?

11.) Draw 3 cards from a standard deck. Find the probability that the first is red, the second is black, and the third is red.

12.) Roll 2 dice. Find P (sum = 8 | at least one die is a 5)

13.) Pick a single card. Find P (face card | diamond). Are the events independent?

14.) You have three jars. Jar 1 has 2 red balls and 1 green ball. Jar 2 has 3 red balls and 1 green ball. Jar 3 has 1 red ball and 2 green balls. You roll a die. If it is 1, 2, or 3, you pick a ball from Jar 1. If it is 4, pick a ball from Jar 2. If it is 5 or 6, pick a ball from Jar 3. Suppose that you ended up picking a red ball. Find the probability that it came from Jar 2.

Solutions

1.) (a) T

(b) T

(c) F

(d) T

(e) T

(f) F

(g) F

(h) T

(i) T

(j) F

5 elements in the set, so 25 = 32 subsets.

2.) (a) X ∩ Y = {1, 3, 5} (b) X ∪ Y = {1, 2, 3, 4, 5, 6, 7, 9} (c) X ′ = {7, 8, 9} (d) X ′ ∩ Y = {7, 9} (e) X ′ ∩ Y ′ = {8} (f) X ′ ∪ Y ′ = {2, 4, 6, 7, 8, 9}

3.) A

B

C

4.) (a) A

B

4

5

8

12

(b) A

B

7

3

12

20

5.) R

G 5 7 3

7

1

6

Y

6

4

6.) (a)

2 52

=

7.) (a)

15 36

=

5 12

(b)

15 36

+

11 36

(c)

12 36

=

1 3

1 26

− 365 =

21 36

=

(b)

12 52

= 133

10 36

=

26 36

(c)

26 52

+

1 2 (a) 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 4 5 6 7 5 6 7 8 6 7 8 9 7 8 9 10 8 9 10 11 9 10 11 12

1 2 (c) 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 4 5 6 7 5 6 7 8 6 7 8 9 7 8 9 10 8 9 10 11 9 10 11 12

=

=

7 13

1 2 (b) 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 4 5 6 7 5 6 7 8 6 7 8 9 7 8 9 10 8 9 10 11 9 10 11 12

1 2 (d) 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 4 5 6 7 5 6 7 8 6 7 8 9 7 8 9 10 8 9 10 11 9 10 11 12

(b) P (Q ∩ K) = 0 → mutually exclusive (c) P (sum is 5 ∩ at least one 4) 6= 0 → not M.E. (d) P (sum is 5 ∩ at least one 6) = 0 → M.E.

10.)

28 52

13 18

8.) (a)P (Q ∩ R) 6= 0 → not mutually exclusive

3 , 36

− 522 =

7 12

(d) 1 − P (sum ≥ 9) = 1 −

9.) P (sum >10) =

4 52

so 3 to 33 → 1 to 11

7 19

11.) P (Red1 ∩ Black2 ∩ Red3 )=

26 52

26 · 51 ·

25 50

=

1 2

·

26 1 51 2

=

13 102

12.) P (sum = 8|at least one die = 5) = 1 2 3 4 5 6 7

1 2 3 4 5 6

2 3 4 5 6 7 8

3 4 5 6 4 5 6 7 5 6 7 8 6 7 8 9 7 8 9 10 8 9 10 11 9 10 11 12

13.) P (face|♦) = P (face) =

n(sum = 8∩at least one die=5) n(at least one die=5

12 52

=

3 13

n(face∩♦) n(♦)

=

3 13

= P (face|♦) → Independent

14.) 2 3

R

1 3

G

3 4

R

1 4

G

1 3

R

Jar 1 1 2

start

1 6

Jar 2

1 3

Jar 3 2 3

P (J2|R) =

P (J 2∩R) P (R)

=

1 3 · 6 4 1 2 · + 1· 3 + 1· 1 2 3 6 4 3 3

G =

1 8 1 1 + 8 + 91 3

=

9 41

=

2 11...