Math 574 — Review Exam #1

1. How many 8-card hands, dealt from a standard deck of 52, consist of: (a). 5 of one suit and 3 of another suit? (b). 4 each of two suits. (c). 3 of one suit, 3 of a second suit, and two of a third suit. (d). 2 of each suit.

Solution: (a).

1213

5

!

" #

$

% &

13

3

!

" #

$

% & (b).

613

4

!

" #

$

% &

2

(c).

1213

3

!

" #

$

% &

213

2

!

" #

$

% & (d).

13

2

!

" #

$

% &

4

2. Suppose that A and B are events in a sample space S such that

P(A) = 0.4 ,

P(A | B) = 0.3

P(A!B) = 0.24 . Then evaluate the following. (a).

P(A!B) (b).

P(B | A) (c).

P(A ' |B) (d). P(A '!!!B ') (e).

P(A | B ') Solution: (a) 0.96 (b) 0.6 (c) 0.7 (d) 0.04 (e) 0.8

3. Suppose that there are five boxes labeled V, W, X, Y, and Z.

Box V contains 4 red marbles. Box W contains 3 red and 1 blue marble. Box X contains 2 of each color. Box Y contains 1 red and 3 blue. Box Z contains 4 blue marbles. A box is chosen at random and a marble randomly drawn from it. (Note that by symmetry, a red marble and blue marble are both equally likely to be chosen.) Given that the marble is red, what is the probability that it came from box W?

Solution: P(W | R) =P(W ! R)

P(R)=P(R |W )P(W )

P(R)=0.75 " 0.2

0.5= 0.3

4. How many ways are there to distribute 18 identical marbles into 6 distinct boxes

if all the boxes are to be non-empty and the first three boxes must contain half the marbles and the final three boxes must contain the remaining half. Solution: First put a marble into each box. Then distribute 6 of the remaining marbles into the first three boxes and the other six into the other three boxes.

So we get

8

2

!

" # $

% &

2

= 784

5. If a subset of

1,2,3,4,5,6{ } is chosen at random, what is the probability that it contains exactly 3 elements?

Solution:

6

3

!

" # $

% &

26

=20

64=5

16

6. How many ways are there to choose a set consisting of 5 letters and 4 digits? (There are 26 letters and 10 digits) For example: {a, e, g, s, t, 3, 7, 0, 9}

Solution:

26

5

!

" #

$

% &

10

4

!

" #

$

% &

7. How many ways are there to arrange 4 A’s, 5 B’s, 6 C’s and 7 D’s in a row with

no two adjacent A’s? Solution: First place the B’s, C’s and D’s. Then put the A’s into the 19 spaces

produced by those 18 letters. So we get 18

5

!"#

$%&13

6

!"#

$%&19

4

!"#

$%&

.

8. How many ways are there to arrange the ten letters a, b, c, d, e, f, g, h, i, j

in a row so that (i). a appears somewhere before b, and c appears somewhere before d. (ii). abc appear together in some order and ghij appear together in some order. Example: d c a b f e i j h g (iii). Each of a, b and c appears somewhere before f and after g? (iv). Have a, b and c appearing together in some order and before both of d and e?

Solution: (i).

10

2

!

" #

$

% &

8

2

!

" # $

% & 6!=

10!

4 (ii).

5! ! 4! ! 3!

(iii).

10

5

!

" #

$

% & ' 3!'5!=

10!

20 (iv).

8

3

!

" # $

% & ' 3!'2 '5!= 2 ' 8!

9. (a). How many strings of length 16 are there that can be formed from the letters in { a, b, c, d, e } Example: b c d d d d e d a a b a c c d c (b). How many of these strings contain exactly 4 a's and 5 b's?

Solution: (a). 516 (b).

16

4

!

" #

$

% &

12

5

!

" #

$

% & 3

7

10. (a). How many subsets of { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } contain 3, 4, and 5

but not 6 and 7? (b). How many strings of length n using the letters a, b, and c use at least one a? Solution: (a). 128 (b). 3n ! 2

n

11. How many ways are there to place 10 red, three blue and one green marble into 8 boxes if no box can contain marbles of different colors? Solution: First place the green marble in 8 ways. Then place the blue marbles. They can all be in one box, all in separate boxes. or two in one and one in another. In each case, after placing the blue marbles, place the red marbles into the

remaining boxes. We get: 8 7 !15

5

"#$

%&'+7

3

"#$

%&'!13

3

"#$

%&'+ 7 !6 !

14

4

"#$

%&'

(

)*

+

,-

12. Suppose that a closet contains 10 pairs of shoes. If 6 shoes are chosen at random,

what is the probability that some two (or more) of them form a pair?

Solution: 1!

10

6

"#$

%&'26

20

6

"#$

%&'

13. Suppose that there are two cartons of eggs. The first contains 8 good and 4 bad

eggs. The second contains 10 good and 2 bad eggs. A single die is rolled and if a 1 or a 2 appears, then an egg is picked from the first carton and otherwise an egg is picked from the second carton. (a). What is the probability that the egg is good? (b). Given that the egg is good, what is the probability that it was chosen from the first carton? Solution: (a). Let G denote ‘Good Egg Drawn”, C1 denote “from Carton 1” etc. P G( ) = P G!C1( ) + P G!C2( ) = P G |C1( )P C1( ) + P G |C2( )P C2( )

=2

3"1

3+5

6"2

3=7

9

(b). P C1 |G( ) =P G!C1( )

P(G)=P G |C1( )P C1( )

P(G)=

2

3"1

37

9

=2

7.

14. If four distinct dice are rolled, what is the probability that they form two pairs?

Example: 3, 3, 5, 5 OR 6, 6, 2, 2?

Solution:

6

2

!"#

$%&4

2

!"#

$%&

64

=5

72

15. How many dice must you roll to have at least a 95% chance of rolling at least one 6? Hint: You’ll need a calculator for this one. Answer: You must roll 17 or more dice.

16. How many solutions are there to 8 ! x + y + z + w ! 15 with x, y, z, and w non-negative integers? Solution: This is (# solutions to x + y + z + w ! 15) " (#solutions to x + y + z + w ! 7) , which is equal to (# solutions to x + y + z + w + t = 15) ! (#solutions to x + y + z + w + t = 7)

and this evaluates to 19

4

!"#

$%&'11

4

!"#

$%&

.

17. Suppose that

(1! 2x + 3x

2)8= a

0+ a

1x + a

2x + a

3x3+!+ a

16x16 .

What is the value of a0+ a

2+ a

4+ a

6+!+ a

16?

Solution: Letting x = 1on both sides gives a0+ a

1+ a

2+!+ a

16= 2

8 . Also, using x = !1 ,

a0! a

1+ a

2! a

3+ a

4! a

5+ a

6!!+ a

14! a

15+ a

16= 6

8 . Adding these two expressions together and then dividing by 2

gives

a0+ a

2+ a

4+ a

6+!+ a

16=28+ 6

8

2.

18. How many strings of length 6 using the letters a, b, c, d, e, and f

(i). Contain at most two a’s? (ii). Contain an even number of a’s? (iii). Contain at least two a’s? (iv). Are missing at least one of a or b? Solution: (i). 56 + 6 ! 55 +15 ! 54 (ii). 56 +15 ! 54 +15 ! 52 +1 (iii).

6

6! 5

6+ 6 "5

5( ) (iv). 2 !56 " 46

19. In how many ways can three non-overlapping blocks of 4 consecutive seats be chosen from a block of 20 consecutive seats? Examples:

Solution: 11

3

!

" #

$

% & =165

20. (a). How many numbers from 0 to 999,999 are there that contain exactly three 1’s? (b). How many of the numbers from 0 to 999,999 contain at least one 2?

Solution: (a). 6

3

!

" # $

% & ' 93 =14,580 (b). 106 ! 96

21. How many ways are there to put at most 12 identical eggs into 4 distinct cartons if

the first carton must contain at most 3 eggs? Solution: This is the same as the number of solutions to a + b + c + d ! 12 with a ! 3 . Which is turn is the same as the number of solutions to a + b + c + d + t = 12 minus the number of solutions to

a + b + c + d + t = 8 . So we get: 16

4

!"#

$%&'12

4

!"#

$%&

.

22. In the grid below, how many paths (usual rules apply!) are there from A to B that:

(i). Do not pass through C. (ii). Pass through at least one of C or D. (iii). Make no two consecutive moves upward?

Solution: (i).

18

8

!

" #

$

% & '

6

2

!

" # $

% &

12

6

!

" #

$

% & (ii).

6

2

!

" # $

% &

12

6

!

" #

$

% & +

10

4

!

" #

$

% &

8

4

!

" # $

% & '

6

2

!

" # $

% &

4

2

!

" # $

% &

8

4

!

" # $

% & (iii).

11

3

!

" #

$

% &

23. How many permutations of a, b, c, d, e, f, g, h, i

(a). Contain at least one of the strings beg and acdf? [Example: cdbegafih idfbeghac acdfbgehi beghacdfi (b). Contain at most one of the strings beg and acdf? (c). Contain at least one of the strings beg and abe? Solution: (a). 7! + 6!- 4! (b). 9! – 4! (c). 2 !7!"6!

24. How many solutions are there to the equation: a + b + c + d = 23 a,b,c,d all non-negative integers with a ! 5, b ! 7, c " 3 ? Solution: This is the total number of non-negative integer solutions to

a + b + c + d = 20 with no restrictions on the variables minus the number of solutions in which either a ≥ 6 or b ≥ 8. The total number of non-negative integer

solutions with no restrictions is

23

3

!

" #

$

% & . The number of solutions in which either a

≥ 6 or b ≥ 8 is: # solutions with a ≥ 6 + # solutions with b ≥ 8 - # solutions with a ≥ 6 and b ≥ 8.

This is

17

3

!

" #

$

% & +

15

3

!

" #

$

% & '

9

3

!

" # $

% & . So the final answer is

23

3

!

" #

$

% & '

17

3

!

" #

$

% & '

15

3

!

" #

$

% & +

9

3

!

" # $

% & .

25. By counting the number of k-element subsets of {1, 2, 3,…, n, a, b, c,!d} in two

different ways, complete the following identity: n

k

!"#

$%&+ 4

n

k '1!"#

$%&+ ? + ?+ ? = ? .

Solution: Both sides count the number k-element subsets of this n + 4 element set. Each subset either has 0 letters, 1 letter, 2 letters, 3 letters, or all 4 letters.

n

k

!"#

$%&+ 4

n

k '1!"#

$%&+ 6

n

k ' 2!"#

$%&+ 4

n

k ' 3!"#

$%&+

n

k ' 4!"#

$%&=

n + 4

k

!"#

$%&

.

26. How many ways are there to place 20 identical marbles into 4 distinct boxes if

each box must have an even number of marbles?

Solution: 13

3

!"#

$%&

.

27. Suppose that 4 cards are dealt at random from a standard deck of 52. What is the

probability that they are each one of two suits (they cannot be all of one suit).

Solution:

4

2

!"#

$%&213

1

!"#

$%&13

3

!"#

$%&+13

2

!"#

$%&

2'

())

*

+,,

52

4

!"#

$%&

28. (a). What is the coefficient of x6 in the expansion of 1+ 3x + 2x2( )12

?

(b). What is the coefficient of x4 in the expansion of 1+ 2x + x2( )6

? [Can you express your answer as a single binomial coefficient?]

Solution: (a). 3612

6

!"#

$%&+ 24 ' 34

11

4

!"#

$%&+ 36

12

2

!"#

$%&10

2

!"#

$%&+ 8

12

3

!"#

$%&

(b). Note that 1+ 2x + x2( )6

= 1+ x( )12 and so the answer is

12

4

!"#

$%&

.