This is a graded in-class assignment with peer review. One submission per group on paper. Do a clean work, your style will be evaluated too. Take a snapshot of your work after peer review. Check the details of peer review guidelines on Bilgi Learn or course webpage.
Suppose we draw three cards from a deck and roll two dice. Answer the following questions.
What is the experiment?
The experiment is “drawing three cards from a deck and rolling two dice”.
What is “getting two-sixes and three-kings or five-one (in any order) one queen one king one ace”? Pick one (Event / Outcome / Sample Space)
Event.
Give an example of two mutually exclusive events.
Event A: Queen of Hearts / Queen of Spades / Queen of Diamonds / 6 / 5 Event B: Ace of Clubs / King of Clubs / Queen of Clubs / 4 / 4
What is the probability of getting four-three (in any order) in dice roll and three queens in card draw?
#First roll can either be 3 or 4 and second roll should be the other
# 2/6 * 1/6
#Getting the first Queen has probability of 4/52
#Getting the second Queen has probability of 3/51
#Getting the third Queen has probability of 2/50
2/6*1/6*4/52*3/51*2/50
## [1] 1.00553e-05
How many different outcomes can there be? This time assume ordering is important (e.g. 6-1 and 1-6 are different outcomes).
#Six outcomes per die
#52 outcomes for the first card draw
#51 outcomes for the second card draw
#50 outcomes for the second card draw
#Multiplication rule
#You can also use permutation rule for cards
6*6*52*51*50
## [1] 4773600
In how many ways can you arrange the letters of “HOUSEPARTY”?
Any order.
the_phrase<-"HOUSEPARTY"
#No repetitive letters
#Permutation rule
factorial(nchar(the_phrase))
## [1] 3628800
Vowels together?
#4 vowels, 6 consonants
#Assume all vowels are a single "letter". So 8 characters.
#But vowels permutate within the single "letter".
#Multiplication rule
factorial(6+1)*factorial(4)
## [1] 120960
Vowels in alphabetical order?
#We start with all the permutations 10!
#For any permutation there can be only one ordering of vowels.
#For instance HOUSEPARTY is not valid but HAESOPURTY is valid
#So remove invalid permutations with division
factorial(10)/factorial(4)
## [1] 151200
There should be no consecutive vowels?
#There are 6 consonants, 4 vowels.
# Assume Xs are consonants and .s are potent vowel places.
# .X.X.X.X.X.X.
#Consonants can permutate in any order so 6! there
#7 places for vowels but only 4 vowels.
# So it is a permutation of 4 out of 8 places.
factorial(6)*(factorial(7)/factorial(7-4))
## [1] 604800
In how many ways can you arrange the letters of “CAMARADERIE”?
# 11 characters.
# 6 vowels, 5 consonants
# 3 As, 2 Es, 2 Rs
Any order.
#By the formula of permutation with repetitive letters
#Assign the value to all_perms object
all_perms<-factorial(11)/(factorial(3)*factorial(2)*factorial(2))
all_perms
## [1] 1663200
Vowels together?
#Assume all vowels are single "character" again. So 6 characters
(factorial(5+1)/(factorial(2)))*(factorial(6)/(factorial(3)*factorial(2)))
## [1] 21600
Vowels in alphabetical order?
#Same as the last question. But be careful about identical vowels.
all_perms/(factorial(6)/(factorial(3)*factorial(2)))
## [1] 27720
There should be no consecutive vowels?
#Same as the last question. But be careful about identical vowels.
(factorial(5)/factorial(2))*(factorial(6)/factorial(6-6))/(factorial(3)*factorial(2))
## [1] 3600
Suppose you are putting the top 12 basketball teams in 4 groups evenly (each group should consist of 3 teams). In how many different ways can you arrange the teams?
#It is either a chain of combinations or just grouping combination
choose(12,3)*choose(9,3)*choose(6,3)
## [1] 369600
There are 18 people; 10 from Izmir, 8 from Mugla.
Suppose you want to form a group of 5 people with at least 1 person from Izmir and Mugla. In how many ways can you form such a group?
#Calculate as if no rules. It is the combination of 18 to 5.
#Then remove the combinations of all Izmir or all Mugla people
choose(18,5) - choose(10,5) - choose(8,5)
## [1] 8260
In how many ways can you form a group of 3 people from Izmir and 4 people from Mugla?
#Simply separate combinations with multiplication rule.
choose(10,3)*choose(8,4)
## [1] 8400