Review Question

1
Review Question 1

• How many ways can I fill a box of 50 chocolates
  from 10 types if I must have at least 2 of each
  type in the box?
• Answer
• Since there must be 2 of each type, 20 of the 50
  chocolates are already chosen. Therefore, the
  problem reduces to calculating the number of ways
  to choose 30 chocolates from 10 types.
• This is a combination with repetition represented
  with bars and starsn 10 types of candy, r
  30 to choose
• ways C(n r 1 , r) C(10 30 - 1, 30)
• C(39, 30) 39!/30!9!

2
Review Question 2

• What is the probability that a 4-bit integer is
  prime, given the first (most-significant) bit is
  1?
• Note that 4-bit primes 2, 3, 5, 7, 11, 13
• AnswerThis is a conditional probability.
• Let E 4-bit primes and F 4-bits w/
  MSB1, then E n F 1011, 1101. (11012 11,
  10112 13)
• Thus, P(E F) P(E n F) / P(F)
• (2/16) / (1/2)
• 4/16 1/4.

3
Review Question 3

• Determine whether or not a 4-bit integer is prime
  is independent of its first bit being 1?
• Note that the 4-bit primes 2, 3, 5, 7, 11, 13
  0010, 0011, 0101, 0111, 1011, 1101 in binary
• Answer
• Let E 4-bit primes and F 4-bit w/ MSB1,
  then E and F are independent if and only if
  P(E)P(F) P(E n F).
• There are 24 16 4-bit integers.
• So, P(E) 6/16 3/8 and P(F) 1/2 and P(E n
  F) 2/16 1/8.
• Since 3/8 1/2 ? 1/8, E and F are NOT
  independent.

4
Review Question 4

• How many ways can a teacher create a team of 8
  students from a class of 13 Boys and 16 Girls, if
  she must choose at least 6 boys?
• Answer
• each team is created by choosing some number of
  boys and some number of girls. The order in
  which they are chosen is not important, so we use
  COMBINATIONS.
• The number of ways to choose at least 6 boys
  is the number of ways to choose 6 boys and 2
  girls number of ways to choose 7 boys and 1
  girl number of ways to choose 8 boys and no
  girls
• ways C(13, 6)C(16, 2) C(13, 7)C(16, 1)
  C(13, 8)C(16, 0)

5
Review Question 5

• How many unique orderings are there of the
  letters of the word SURROUNDSOUND ?
• Answer
• This is a permutation with indistinguishable
  objects problem.
• The indistinguishable objects are the letter
  which appear more than once.
• SURROUNDSOUND contains 13 letters of which
• 2 are S, 2 are R, 3 are U, 2 are O, 2 are N, 2
  are D
• Therefore, by the given formula
• ways 13! / 2! 2! 2! 2! 2! 3!

6
Review Question 6

• What is the probability that a bit string of
  length 10 will have no more than two 1s?
• Answer
• The number of bit strings with no more than two
  1s is the number of bit strings with no 1s
  plus number of bit strings with one 1, the number
  of bit strings with two 1s.
• Since the 1s may appear anywhere in the bit
  string, to calculate the number of bit strings
  with N ones, we must choose which of the 10 bit
  positions will be a 1 C(10, N).
• bit string with no 1s C(10, 0) 1 (all
  0s) bit strings with one 1 C(10, 1) 10
  bit strings with two 1s C(10, 2) 45Note
  that there are 210 1024 bit strings of length
  10.
• So, prob( no more than two 1s) C(10, 0)
  C(10, 1) C(10, 2)
  ----------------------------------------
• 210

7
6 Continued

• We could also count number of bit strings with
  zero, one or two 1s as follows using
  permutations if we consider each of these as
  string with indistinguishable objects.
• bit strings with no 1s is the number
  permutations of 0000000000 1
• bit strings with one 1 is the number
  permutations of a 10-bit string which contains
  one 1 and nine 0s. The 0s are
  indistinguishable, so by formula,ways 10! / 1!
  9! 10, which is the same as C(10, 1)
• Similarly, bit strings with two 1s is the
  number permutations of a 10-bit string which
  contains two 1s and eight 0s. The 1s and 0s
  are indistinguishable, so by formula, ways
  10! / 2! 8! which is the same as C( 10, 2 ).

8
Review Question 7

• How many ways can I select either a RED card or a
  FACE card (King, Queen, or Jack) from a standard
  deck of 52 cards?
• Answer
• Let R set of red cards (13 hearts and 13
  diamonds).
• Let F set of face cards (K, Q, J from each of
  the 4 suits)
• By the inclusion-exclusion rule R U F R
  F - R n F
• Whats R n F? Thats the number of RED FACE
  cards.
• There are 6 of these Kh, Qh, Jh, Kd, Qd, Jd
• So R U F 26 12 6 32

9
Review Question 8

• In Maryland, some license plates consist of 3
  letters followed by 3 digits. How many license
  plates can be created if no letter or digit can
  be duplicated?
• Answer
• The number of ways to arrange 3 letters from 26
  without duplication is P(26, 3) 26! / 23!
  26 25 24.(Note that a direct application of
  the product rule gives the same result)
• Similarly, the number of ways to arrange 3
  digits from 10 is P(10, 3) 10! / 7! 10 9
  8
• Therefore the total number of license plates is
• 26 25 24 10 9 8

10
Review Question 9

• A book shelf contains 24 math books, 25 computer
  science books, 21 biology books and 14 economic
  books. How many books must you place in your
  backpack to assure there are at least 7 books on
  the same subject?
• AnswerThere are 4 different kinds of books
  these are pigeon holes.
• We need to find the required number of books
  (pigeons) such that at least one pigeon hole has
  7 pigeons. By the generalized pigeon hole
  principle, we are looking for the solution to
  ceiling(x/4) 7. The solution is x 25 since
  25 / 4 6.25 and ceiling(6.25) 7.

11
Review Question 10

• How many different pizzas can be ordered if a
  pizza can contain any combination of the
  following 5 toppings onion, pepperoni, ham,
  mushrooms, and sausage?
• AnswerOne approach to this problem is to
  consider each possible combination as a bit
  string of length 5. If a topping is selected,
  its represented by a 1. If not selected, its
  represented by a 0. There are 25 32 bit
  strings of length 5, so 32 pizzas are possible.
• Another approach is to count the number of ways
  to select 0, 1, 2, 3, 4 or 5 toppings from the 5
  possible toppings. In this way,
• pizzas C(5, 0) C(5,1) C(5,2) C(5,3)
  C(5, 4) C(5, 5)
• 1 5 10
  10 5 2 32

12
Review Question 11

• Given a pair of fair dice, what is the
  probability of rolling a total of 7 three times
  in ten rolls?
• AnswerIf we define roll total of 7 as
  success, this is a Bernoulli experiment with p
  1/6 and q 1 p 5/6
• By formula the probability of three 7s in 10
  rolls is
• C(10, 3) (1/6)3 (5/6)7

13
Review Question 12

• Suppose we randomly choose one of the
  permutations of the integers 1, 2, 3, n with
  n 4.
• a. What is the probability that 1 comes before
  2?
• b. What is the probability that 1 comes
  immediately before 2?
• c. What is the probability that n comes before
  1 and before 2?
• Answera. either 1 comes before 2 or it doesnt
  and each of these is equally likely, so prob ½
• b. if 1 comes immediately before 2, we can treat
  12 as a separate entity. We then count how to
  permute n-1 integers (n-1)!. Since there are
  n! permutations of 1 n, the prob (n-1)! / n!
  1/n
• c. we need only look at the relative positions
  of n, 1, 2. Clearly n will be the first of these
  three 1/3 of the time

14
Review Question 13

• Suppose two cards are drawn from a regular deck
  of 52 cards. If two cards are drawn at random,
  what is the probability that the 2nd card is a
  spade if the first card is also a spade?
• Answer 1After the 1st spade is drawn, there are
  51 cards of which 12 are spades, so prob(2nd card
  is spade) 12 / 51
• Answer 2As a conditional probability problem,
  let S1 1st card is spade and S2 2nd card is
  spade. We are asked to compute P( S2 S1).By
  formula P(S2 S1 ) P(S1 n S2) / P(S1).
• Clearly P(S1) 13/52 since 13 of the 52 cards
  are spades. P(S1 n S2) 13 12 / 52 51
• and therefore P(S2 S1 ) (1312)/(5251) /
  (13/52) 12 / 51

15
Review Question 14

• In some lottery games, the state will choose 6
  balls from a set of balls numbered 1 40. You
  can win a prize if you correctly choose 4 of the
  6 numbers selected by the state. What is the
  probability of winning the 4 out of 6 prize?
• Answer
• We can correctly choose 4 of the 6 winning
  numbers in C(6,4) ways. We can choose 2 losing
  numbers in C(34, 2) ways. There are therefore
  C(6,4)C(34,2) ways to correctly choose 4 out of
  6. There are C(40, 6) possible sets of winning
  numbers, so the probability of winning is C(6,4)
  C(34, 2) / C(40, 6)

16
Review Question 15

• An octal die has eight faces numbered 1 8. If
  a pair of fair octal dice is rolled, what is the
  expected value of the sum of the two dice?
• Answer1
• Since the die are fair, the probability of any
  value is 1/8. So for a single die, the expected
  value is
• 1/8 (1 2 3 4 5 6 7 8) 36 / 8
  9 / 2Since two dice are rolled, the expected
  value is 9 / 2 9 /2 9
• Answer 2 There are 64 outcomes with 5 possible
  sum ( 2 16). We calculate the probability of
  each sum, then apply the formula
• P(sum 2) 1 / 64 P(9) 8 / 64
• P(sum 3) 2 / 64 P(10) 7 / 64
• P(sum 4) 3 / 64 P(11) 6 / 64
• P(5) 4 / 64 P(12) 5 / 64
• P(6) 5 / 64 P(13) 4/ 64
• P(7) 6 / 64 P(14) 3/ 64
• P(8) 7 / 64 P (15) 2/64
• P(16) 1/64
• E(X) 2 1/64 3 2/64 4 3/64 15
  2/64 16 1/64 576 / 64 9
• Also note that 9 has the most combinations and
  therefore the highest probability of any sum