CH2 - Counting Methods

CH2.2 - Counting Methods

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So far, S has been small enough that we could list all the outcomes. However, as S grows, it becomes impractical to list all outcomes. We need to develop methods to count the number of outcomes in a sample space.

There are some counting methods that we will use in this chapter:

We will also use the following:

$p(A)=\frac{favorableoutcomes}{totaloutcomes}$

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Basic Counting Rule

1. There are n ways to make choice 1.
2. For each of these n ways, there are n ways to make choice 2.

⇒ Total number of ways to make both choices is n * n = n^2

graph TD
    A[Choice 1] --> C[Total number of ways n^2]
    B[Choice 2] --> C

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Permutations

Permutations are arrangements of objects in a specific order, where ordering of “n” objects is important.

$P(n,r)=\frac{n!}{(n-r)!}$

Permutations Example

how many ways can you arrange the letters in the word math?

$P(4,4)=\frac{4!}{(4-4)!}=4!=4\times 3\times 2\times 1=24$

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Combinations

ORDER DOES NOT MATTER

Combinations are selections of objects in which the order of selection is not important.

Combinations Example

50 students apply for a scholarship. How many ways can 5 students be selected to receive the scholarship?

We can use the combination formula:

$C(n,r)=\frac{n!}{r!(n-r)!}$

And plug in the numbers:

$C(50,5)=\frac{50!}{5!(50-5)!}=\frac{50!}{5!45!}=2118760$

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Bigger Examples

There are 17 broken lightbulbs in a box of 100 lightbulbs. A random sample of 3 lightbulbs is chosen without replacement.

1. How many ways can the sample be chosen?

C(100,3) = \frac{100!}{3!(100-3)!} = 161700

2. How many samples contain no broken lightbulbs?

C(83,3) = \frac{83!}{3!(83-3)!} = 91881

3. What is the probability that a sample contains no broken lightbulbs?

P(x=0) = \frac{91881}{161700} = .5682