4.5 Probability and Counting Rules

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Sec 4.5
Probability and Counting
Rules
Bluman, Chapter 4
Review questions: Calculate the following on your
calculators
A.
B.
C.
D.
E.
F.
G.
H.
c5
12p5
nc1 for any positive values of n
np1 for any positive values of n
ncn for any positive values of n
npn for any positive values of n
nc0 for any positive values of n
np0 for any positive values of n
12
Bluman, Chapter 4
Roger’s parents plan to visit 4 of
his 8 classes.
A.
In How many ways can his parents visit
his classes at Back to School Night?
B.
In How many ways can his parents visit
his classes at Parent conferences.
Bluman, Chapter 4
Bluman, Chapter 4
Multiple choice examples
http://www.indiabix.com/aptitude/permutati
on-and-combination/
๏ฎ Comboniation and Permuation practice
problems
๏ฎ
Bluman, Chapter 4
4.5 Probability and Counting Rules
The counting rules can be combined with the
probability rules in this chapter to solve
many types of probability problems.
By using the fundamental counting rule, the
permutation rules, and the combination rule, you
can compute the probability of outcomes of many
experiments, such as getting a full house when 5
cards are dealt or selecting a committee of 3
women and 2 men from a club consisting of 10
women and 10 men.
Bluman, Chapter 4
Chapter 4
Probability and Counting Rules
Section 4-5
Example 4-50
Page #237
Bluman, Chapter 4
7
Four Aces
๏ฎ
Find the probability of drawing four aces
when five cards are drawn from a deck of
cards.
Make problem simpler!
Bluman, Chapter 4
Bluman, Chapter 4
4 ๐‘Ž๐‘๐‘’๐‘  ๐‘Ž๐‘›๐‘‘ ๐‘Ž๐‘›๐‘œ๐‘กโ„Ž๐‘’๐‘Ÿ ๐‘๐‘Ž๐‘Ÿ๐‘‘
๐‘ƒ ๐ด๐‘๐‘’๐‘  =
๐‘Ž๐‘›๐‘ฆ 5 ๐‘๐‘Ž๐‘Ÿ๐‘‘๐‘ 
๐‘ƒ ๐‘Ž๐‘๐‘’๐‘  =
๐‘
4 4
โˆ™
๐‘
52 5
1 โˆ™ 48
=
2,598,960
Bluman, Chapter 4
๐‘
48 1
Final answer and rounding
1.84689 โˆ™ 10−5 ๐‘‚๐‘… 0.0000184689
OR 0.000018
Bluman, Chapter 4
Four of a Kind
๏ฎ
Find the probability of getting four of a kind
when five cards are drawn from a deck of
cards.
The answer to the last problem
multiplied by 13; since there are 13 four
of a kind hands possible.
๐Ÿ. ๐Ÿ’๐ŸŽ๐ŸŽ๐Ÿ—๐Ÿ”๐ŸŽ๐Ÿ‘๐Ÿ–๐Ÿ’ โˆ™ ๐Ÿ๐ŸŽ−๐Ÿ’ = ๐ŸŽ. ๐ŸŽ๐ŸŽ๐ŸŽ๐Ÿ๐Ÿ’
Bluman, Chapter 4
Chapter 4
Probability and Counting Rules
Section 4-5
Example 4-51
Page #238
Bluman, Chapter 4
13
Example 4-51
๏ฎA
box contains 24 transistors, 4 of
which are defective. If 4 are sold at
random, find the following
probabilities.
๏ฎ A.) Exactly 2 are defective.
๏ฎ B.) None is defective.
๏ฎ C.) All are defective.
๏ฎ D.) At least 1 is defective.
Example 4-52: Committee Selection
A store has 6 TV Graphic magazines and 8 Newstime
magazines on the counter. If two customers purchased a
magazine, find the probability that one of each magazine
was purchased.
TV Graphic: One magazine of the 6 magazines
Newstime: One magazine of the 8 magazines
Total: Two magazines of the 14 magazines
6
C1 ๏ƒ—8 C1 6 ๏ƒ— 8 48
๏€ฝ
๏€ฝ
91
91
14 C 2
Bluman, Chapter 4
15
Chapter 4
Probability and Counting Rules
Section 4-5
Example 4-52
Page #238
Bluman, Chapter 4
16
Chapter 4
Probability and Counting Rules
Section 4-5
Example 4-53
Page #239
Bluman, Chapter 4
17
Example 4-53: Combination Locks
A combination lock consists of the 26 letters of the
alphabet. If a 3-letter combination is needed, find the
probability that the combination will consist of the letters
ABC in that order. The same letter can be used more
than once. (Note: A combination lock is really a
permutation lock.)
There are 26·26·26 = 17,576 possible combinations.
The letters ABC in order create one combination.
1
P ๏€จ ABC ๏€ฉ ๏€ฝ
17,576
Bluman, Chapter 4
18
On your own:
๏ฎ Read
“Speaking
of Statistics” on
page 240
๏ฎ Sec
4-5 page
240
๏ฎ Exercises #1-17
odds
Bluman, Chapter 4
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