CK-12 Probability and Statistics - Advanced

http://www.ck12.org Chapter 3. An Introduction to Probability

RY,BR,BB,BY,Y R,Y B.

Next, we want to calculate the probabilities of each outcome.

P(R 1 standR 2 st) =P(RR) = 2 / 6 · 1 / 5 = 2 / 30

P(R 1 standB 2 st) =P(RB) = 2 / 6 · 3 / 5 = 6 / 30

P(RY) = 2 / 6 · 1 / 5 = 2 / 30

P(BR) = 3 / 6 · 2 / 5 = 6 / 30

P(Y B) = 3 / 6 · 2 / 5 = 6 / 30

P(Y B) = 3 / 6 · 1 / 5 = 3 / 30

P(Y B) = 1 / 6 · 2 / 5 = 2 / 30

P(Y B) = 1 / 6 · 3 / 5 = 3 / 30

Notice that all the probabilities must add up to 1, as they should.

The method used to solve the example above can be generalized to any number of stages. This method is called the
Multiplicative Rule of Counting.

The Multiplicative Rule of Counting

(I)If there arenpossible outcomes for eventAandmpossible outcomes for eventB, then there are a total ofnm
possible outcomes for the series of eventsAfollowed byB.

Another way of stating it:

(II)You haveksets of elements,n 1 in the first set,n 2 in the second set,..., andnkin thekth set. Suppose you want to
take one sample from each of theksets. The number of different samples that can be formed is the product

n 1 n 2 n 3 ...nk

Example:

A restaurant offers a special dinner menu every day. There are three entrées to choose from, five appetizers, and four
desserts. A costumer can only select one item from each category. How many different meals can be ordered from
the special dinner menu?

Solution:

Let’s summarize what we have.

Entrees:3

Appetizer: 5

Dessert: 4

We use the multiplicative rule above to calculate the number of different dinner meals that can be selected. We
simply multiply all the number of choices per item together: