warm up a ferris wheel holds 12 riders. if there are 20 people waiting to ride it, how many ways can...
TRANSCRIPT
Warm up
A ferris wheel holds 12 riders. If there are 20 people waiting to ride it, how many ways can they ride it?
Solution
Since only 12 of the 20 people can ride the ferris wheel at a time, there are C(20,12) or 125 970 different groups of riders.
Each group can be placed on the ferris wheel 11! or 39 916 800 ways since it is a circular permutation.
So the total number of ways is:
125 970 x 39 916 800 = 5 028 319 296 000
or 5.03 x 1012
I hope they purchased the season pass!
Chapter 4 ReviewMDM 4U
Mr. Lieff
4.1 Intro to Simulations and Experimental Probability be able to design a simulation to investigate
the experimental probability of some event ex: design a simulation to determine the
experimental probability that more than one of 5 keyboards chosen in a class will be defective if we know that 25% are defective
1. get a shuffled deck of cards, choosing clubs to represent a defective keyboard
2. choose 5 cards with replacement and see how many are clubs
3. repeat a large number of times (e.g. 4 outcomes x 10 = 40) and calculate probability
4.2 Theoretical Probability
work effectively with Venn diagrams ex: create a Venn diagram illustrating the sets
of face cards and red cards S = 52red & face = 6
red = 20face = 6
4.2 Theoretical Probability
calculate the probability of an event or its complement
ex: what is the probability of randomly choosing a male from a class of 30 students if 10 are female?
P(A) = n(A)÷n(S) = 20÷30 = 0.67
4.2 Theoretical Probability
ex: calculate the probability of not throwing a total of four with 3 dice
there are 63 possible outcomes with three dice
only 3 outcomes produce a 4 probability of a 4 is:
3/63
probability of not throwing a sum of 4 is: 1- 3/63 = 0.986
4.3 Finding Probability Using Sets
recognize the different types of sets utilize the additive principle for unions of sets The Additive Principle for the Union of Two
Sets:n(A U B) = n(A) + n(B) – n(A ∩ B)P(A U B) = P(A) + P(B) – P(A ∩ B)
calculate probabilities using the additive principle
4.3 Finding Probability Using Sets
ex: what is the probability of drawing a red card or a face card
ans: P(A U B) = P(A) + P(B) – P(A ∩ B) P(red or face) = P(red) + P(face) – P(red and face) = 26/52 + 12/52 – 6/52
= 32/52 = 0.615
4.3 Finding Probability Using Sets
What is n(B υ C) 2+8+3+3+6+2+1+8+1 = 34 What is P(A∩B∩C)? n(A∩B∩C) = 3 = 0.07 n(S) 43
4.4 Conditional Probability
100 Students surveyedCourse Taken No. of students
English 80
Mathematics 33
French 68
English and Mathematics
30
French and Mathematics
6
English and French
50
All three courses 5
What is the probability that a student takes Mathematics given that he or she also takes English?
4.4 Conditional Probability
M
F
E
5
25
45
5
1
2
17
4.4 Conditional Probability
To answer the question in (b), we need to find P(Math|English).
We know... P(Math|English) = P(Math ∩ English)
P(English) Therefore…
P(Math|English) = 30 / 100 = 30 x 100 = 3
80 / 100 100 808
4.4 Conditional Probability
calculate a probability of events A and B occurring, given that A has occurred
use the multiplicative law for conditional probability
ex: what is the probability of drawing a jack and a queen in sequence, given no replacement?
P(J ∩ Q) = P(Q | J) x P(J) = 4/51 x 4/52 = 16/2652 = 0.006
4.5 Tree Diagrams and Outcome Tables
a sock drawer has a red, a green and a blue sock you pull out one sock, replace it and pull another out draw a tree diagram representing the possible outcomes what is the probability of drawing 2 red socks? these are independent events
R
R
R
R
B
B
B
BG
G
G
G
9
1
3
1
3
1
)()(
)(
redPredP
redandredP
4.5 Tree Diagrams and Outcome Tables Mr. Greer is going fishing he finds that he catches fish 70% of the time
when the wind is out of the east he also finds that he catches fish 50% of the
time when the wind is out of the west if there is a 60% chance of a west wind today,
what are his chances of having fish for dinner?
we will start by creating a tree diagram
4.5 Tree Diagrams and Outcome Tables
west
east
fish dinner
fish dinner
bean dinner
bean dinner
0.6
0.4 0.7
0.3
0.5
0.5
P=0.3
P=0.3
P=0.28
P=0.12
4.5 Tree Diagrams and Outcome Tables P(east, catch) = P(east) x P(catch | east) = 0.4 x 0.7 = 0.28 P(west, catch) = P(west) x P(catch | west) = 0.6 x 0.5 = 0.30 Probability of a fish dinner: 0.28 + 0.3 = 0.58 So Mr. Greer has a 58% chance of catching a
fish for dinner
4.6 Permutations
find the number of outcomes given a situation where order matters
calculate the probability of an outcome or outcomes in situations where order matters
recognizing how to restrict the calculations when some elements are the same
4.6 Permutations
ex: How many ways can 5 students be arranged in a line?
ans: 5! = 120 ex: How many ways are there if Jake must be first? ans: (5-1)! = 4! = 60 ex: in a class of 10 people, a teacher must pick 3 for
an experiment (students are tested in a particular order)
How many ways are there to do this? ans: P(10,3) = 10!/(10 – 3)! = 720
Permutations cont’d
How many ways are there to rearrange the letters in the word TOOLTIME?
8! / (2!2!) = 8! / (4) = 10 080
4.6 Permutations
ex: what is the probability of opening one of the school combination locks by chance? Second digit must be different from the first
ans: 1 in 60 x 59 x 59 = 1 in 208 860 Circular Permutations: (n-1)! ways to arrange
n objects in a circle.
4.7 Combinations
find the number of outcomes given a situation where order does not matter
calculate the probability of an outcome or outcomes in situations where order does not matter
ex: how many ways are there to choose a 3 person committee from a class of 20?
ans: C(20,3) = 20! ÷ [ (20-3)! 3! ] = 1140
4.7 Combinations
ex: from a group of 5 men and 4 women, how many committees of 5 can be formed with a. exactly 3 women b. at least 3 women
ans a:
ans b:
404103
4
2
5
454
4
1
5
3
4
2
5
Combinatorics (§4.6 & 4.7)
Permutations – order matters E.g. President
Combinations – order does not matter E.g. Committee
Test Review
pp. 268-269 #1, 4, (5-6) ace…, 8, 9, 11 p. 270 #1, 2